If $A = \pmatrix{1 & 3 \\ 4 & 2}$ and $\boldsymbol{x} = \pmatrix{3 \\ 4}$, then $A\boldsymbol{x} = \pmatrix{15 \\ 20}$  $= 5\pmatrix{3 \\ 4}$.

Here we have that $A\boldsymbol{x} = 5 \boldsymbol{x}$ and so we say that $\boldsymbol{x} = \pmatrix{3 \\ 4}$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda = 5$ .

The geometric effect in this example is that the vector $\boldsymbol{x}$ has been stretched by a factor of 5 but its direction remains unchanged as $\lambda = 5$.

Note that any scalar multiple of the vector $\boldsymbol{x} = \pmatrix{3 \\ 4}$ is an eigenvector corresponding to the eigenvalue $\lambda = 5$.

Find the eigenvalues of the following matrices :

(i) $A = \pmatrix{5 & -2 \\ 7 & -4}$ (ii). $B = \pmatrix{13 & -4 \\ -4 & \;\;\;7}$(iii). $C = \pmatrix{0 & 2 \\ 2 & 0}$

Solution

(i). $A - \lambda I = \pmatrix{5 & -2 \\ 7 & -4} - \lambda \pmatrix{1 & 0 \\ 0 & 1} = \pmatrix{5 - \lambda & -2 \\ \;\;7 & -4-\lambda}$.

Hence, $\text{det}(A - \lambda I) = \bigg | \matrix{5 - \lambda & -2 \\ \;\;7 & -4-\lambda} \bigg | = (5 - \lambda)(-4-\lambda) - (-2)(7) = \lambda^2 - \lambda - 6$.

We call $\lambda^2 - \lambda - 6$ the characteristic polynomial of the matrix $A$.

The eigenvalues of $A$ are the roots of the characteristic equation $\text{det}(A - \lambda I) = 0$, i.e.

$\lambda^2 - \lambda - 6 = 0 \implies (\lambda - 3)(\lambda + 2) = 0 \implies \lambda = -2$ and $\lambda = 3$.

Hence, $\lambda_1 = -2$ and $\lambda_2 = 3$ are the eigenvalues of the matrix $A$.

(ii). $B - \lambda I = \pmatrix{\;13 & -4 \\ -4 & \;\;\;7 } - \lambda \pmatrix{1 & 0 \\ 0 & 1} = \pmatrix{13-\lambda & -4 \\ -4 & 7-\lambda}$

Hence, $\text{det}(B - \lambda I) = \bigg | \matrix{13-\lambda & -4 \\ -4 & \;7-\lambda} \bigg | = (13-\lambda)(7-\lambda) - (-4)(-4) = \lambda^2 - 20 \lambda + 75$.

Now solve $\det(B - \lambda I) = 0$ to find the eigenvalues of $B$, i.e.

$\lambda^2 - 20 \lambda + 75 - 0 \implies (\lambda - 5)(\lambda - 15) = 0 \implies \lambda = 5$ and $\lambda = 15$.

Hence, $\lambda_1 = 5$ and $\lambda_2 = 15$ are the eigenvalues of the matrix $B$.

(iii). $C - \lambda I = \pmatrix{0 & 2 \\2 & 0} - \lambda \pmatrix{1 & 0 \\ 0 & 1} = \pmatrix{-\lambda & \;\;2 \\ \;\;2 & -\lambda}$

Hence, $\text{det}(C - \lambda I) = \bigg | \matrix{-\lambda & \;\;2 \\ \;\;2 & -\lambda} \bigg | = (-\lambda)(-\lambda) - (2)(2) = \lambda^2 - 4$.

The eigenvalues of $C$ satisfy $\text{det}(C - \lambda I) = 0$, i.e. $\lambda^2 - 4 = 0 \implies \lambda = \pm 2$.

Hence, $\lambda_1 = -2$ and $\lambda_2 = 2$ are the eigenvalues of the matrix $C$.

Find the eigenvalues of the following matrices:

(i). $A = \pmatrix{5 & 0 \\ 0 & 8}$ (ii). $\pmatrix{3 & \;\;7 \\ 0 & -4}$ (iii). $C = \pmatrix{1 & 0 \\ 3 & 2}$.

In this example we note that:

matrix $A$ is a diagonal matrix (see Section 4.4) and has the property that all of its entries not on the main diagonal are $0$.

matrix $B$ is an upper-triangular matrix (see Section 4.5) and has the property that all of its entries below the main diagonal are $0$

matrix $C$ is a lower-triangular matrix (see Section 4.5) and has the property that all of its entries above the main diagonal are $0$.

Note that, in each case, some of the entries on the main diagonal can be zero.

Solution

In all three cases – diagonal, upper-triangular and lower triangular - the eigenvalues are simply the entries on the main diagonal and so we can just read them off without the need for any calculations. Hence,

The eigenvalues of matrix $A$ are: $\lambda_1 = 5,\; \lambda_2 = 8$.

The eigenvalues of matrix $B$ are: $\lambda_1 = 3,\; \lambda_2 = -4$.

The eigenvalues of matrix $C$ are: $\lambda_1 = 1, \; \lambda_2 = 2$.

We verify our answers using the method described earlier.

(i). Solving $\text{det}(A - \lambda I) = 0$ gives $\bigg | \matrix{5-\lambda & \;\,0 \\ \;\,0 & 8-\lambda} \bigg| = 0 \implies (5-\lambda)(8-\lambda) = 0 \implies \lambda_1 = 5, \lambda_2 = 8$.

(ii). Solving $\text{det}(B - \lambda I) = 0$ gives $\bigg | \matrix{3-\lambda & \;\,7 \\ \;\,0 & -4-\lambda} \bigg| = 0 \implies (3-\lambda)(-4-\lambda) = 0 \implies \lambda_1 = 3, \lambda_2 = -4$.

(iii). Solving $\text{det}(C - \lambda I) = 0$ gives $\bigg | \matrix{1-\lambda & \;\,0 \\ \;\,3 & 2-\lambda} \bigg| = 0 \implies (1-\lambda)(2-\lambda) = 0 \implies \lambda_1 = 1, \lambda_2 = 2$.

Find the eigenvalues of the matrix, $A = \pmatrix{3 & -9 \\ 1 & \;\;\;9}$ .

Solution

To find the eigenvalues we solve

$\text{det}(A - \lambda I) = \bigg | \matrix{3-\lambda & -9 \\ \;\;1 & 9-\lambda} \bigg | = 0$

$(3 - \lambda)(9 - \lambda) + 9 = 0$

$\lambda^2 - 12\lambda + 36 = 0$

$(\lambda - 6)(\lambda - 6) + 0$

$\lambda = 6$ (repeated).

The eigenvalue $\lambda = 6$ is said to have algebraic multiplicity $2$, i.e. the number of times it is a root of the characteristic equation.

Find the eigenvalues and eigenvectors of the matrix,

$A = \pmatrix{\;\;\;3 & -6 \\ -2 & \;\;\;4}$

Solution

To find the eigenvalues we need to solve

$\text{det}(A - \lambda I) = \bigg | \matrix{3-\lambda & -6 \\ -2 & 4-\lambda} \bigg | = 0$

$(3 - \lambda)(4 - \lambda) - 12 = 0$

$\lambda^2 - 7 \lambda = 0$

$\lambda(\lambda - 7) = 0$

$\lambda_1 = 0, \; \lambda_2 = 7$.

This example shows that it is possible for $0$ to be an eigenvalue of a matrix.

Note that if $0$ is an eigenvalue of a matrix then the matrix is not invertible. Hence, the matrix $A$ in this example cannot be inverted.

Find the eigenvalues of the matrices:

(i). $A = \pmatrix{0 & -1 \\ 1 & \;\;\;0}$ (ii). $B = \pmatrix{4 & -3 \\ 6 & -2}$.

Solution

(i). $\text{det}(A - \lambda I) = \bigg | \matrix{-\lambda & -1 \\ 1 & -\lambda} \bigg | = 0$

$\lambda^2 + 1 = 0$.

$\lambda_1 = j, \; \lambda_2 = -j$.

(ii). $\text{det}(B - \lambda I) = \bigg | \matrix{4-\lambda & -3 \\ 6 & -2-\lambda} \bigg | = \lambda^2 - 2\lambda + 10 = 0$

Solve using the quadratic formula, or by completing the square, to obtain,

$\lambda_1 = 1 + 3j, \; \lambda_2 = 1 - 3j$.

Note: For a matrix with real entries its complex eigenvalues always occur in complex conjugate pairs.

Find the eigenvalues and eigenvectors of the matrix, $A = \pmatrix{\;\;\;2 & \;\;\;7 \\ -1 & -6}$.

Solution

First we find the eigenvalues by solving:

$\text{det}(A - \lambda I) = \bigg | \matrix{2-\lambda & \;7 \\ -1 & -6-\lambda} \bigg | = 0$

$(2 - \lambda)(-6 - \lambda) + 7 = 0$

$\lambda^2 + 4 \lambda - 5 = 0$

$(\lambda + 5)(\lambda - 1) = 0$

$\lambda_1 = -5, \; \lambda_2 = 1$.

We now calculate the eigenvectors corresponding to the eigenvalues by solving Eq. (III).

Case 1: To find an eigenvector $\boldsymbol{x}_1$ corresponding to eigenvalue $\lambda_1 = -5$ we solve,

$A\;\boldsymbol{x}_1 = \lambda_1\;\boldsymbol{x}_1$.

$\pmatrix{\;\;\;2 & \;\;\;7 \\ -1 & -6}\pmatrix{x_1 \\ x_2} = -5 \pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{2x_1 + 7x_2 = -5x_1 \\ -x_1 - 6x_2 = -5x_2}$

$\bigg \{ \array{7x_1 + 7x_2 = 0 \; ........ \;(1) \\ -x_1 - x_2 = 0 \; ......... \;(2) }$

These are simultaneous equations and we note here that one equation will always be a multiple of the other - if not then you have made a mistake! Here Eq. (1) is $-7$ times Eq. (2).

Both equations give $x_1 = -x_2$. If we let $x_2 = \alpha$, say, for some non-zero real number $\alpha$, then $x_1 = - \alpha$ and we find the first eigenvector to be of the form

$\boldsymbol{x}_1 = \pmatrix{-\alpha \\ \;\;\; \alpha} = \pmatrix{-1 \\ \;\;\;1} \alpha$.

Note that there are infinitely many non-zero eigenvectors depending on the value chosen for $\alpha$. Setting $\alpha = 1$ gives an eigenvector corresponding to the eigenvalue $\lambda = -5$ as $\boldsymbol{x}_1 = \pmatrix{-1 \\ \;\;\;1}$.

We can check our answer by showing that $A\;\boldsymbol{x}_1 = -5\;\boldsymbol{x}_1$.

$A \;\boldsymbol{x}_1 = \pmatrix{\;\;\;2 & \;\;\; 7 \\ -1 & -6}\pmatrix{-1 \\ \;\;\; 1} = \pmatrix{\;\;\;5 \\ -5}$  and  $\lambda_1\;\boldsymbol{x}_1 = -5 \pmatrix{-1 \\ \;\;\;1} = \pmatrix{\;\;\;5 \\ -5}$.

Hence, $A\;\boldsymbol{x}_1 = \lambda_1 \; \boldsymbol{x}_1$ as required.

Case 2: To find an eigenvector $\boldsymbol{x}_2$ corresponding to eigenvalue $\lambda_2 = 1$ we solve,

$A\;\boldsymbol{x}_2 = \lambda_2\;\boldsymbol{x}_2$.

$\pmatrix{\;\;\;2 & \;\;\;7 \\ -1 & -6}\pmatrix{x_1 \\ x_2} = \pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{2x_1 + 7x_2 = x_1 \\ -x_1 - 6x_2 = x_2}$

$\bigg \{ \array{x_1 + 7x_2 = 0 \\ -x_1 - 7x_2 = 0}$

Both these equations give $x_1 = -7x_2$. Let $x_2 = \alpha$, say, for some non-zero real number $\alpha$, then $x_1 = -7\alpha$ and so

$\boldsymbol{x}_2 = \pmatrix{-7\alpha \\ \;\;\;\alpha} = \pmatrix{-7 \\ \;\;\;1}\alpha$.

Setting $\alpha = 1$ gives $\boldsymbol{x}_2 = \pmatrix{-7 \\ \;\;\;1}$. It is straightforward to check that $A\;\boldsymbol{x}_2 = 1\;\boldsymbol{x}_2$.

In summary, we therefore have the eigenvalue/eigenvector pairs,

$\lambda_1 = -5, \; \boldsymbol{x}_1 = \pmatrix{-1 \\ \;\;\;1}$; $\lambda_2 = 1, \; \boldsymbol{x}_2 = \pmatrix{-7 \\ \;\;\;1}$.

Find the eigenvalues and eigenvectors of the matrix, $B = \pmatrix{\;13 & -4 \\ -4 & \;\;\; 7}$.

Solution

In Example 2 part (ii) we found the eigenvalues of $B$ to be $\lambda_1 = 5$ and $\lambda_2 = 15$.

We now calculate the eigenvectors corresponding to these eigenvalues by solving the eigenvector equation, $A\; \boldsymbol{x} = \lambda \; \boldsymbol{x}$.

Case 1: To find an eigenvector $\boldsymbol{x}_1$ corresponding to eigenvalue $\lambda_1 = 5$ we solve,

$A \;\boldsymbol{x}_1 = \lambda_1 \; \boldsymbol{x}_1$

$\pmatrix{\;13 & -4 \\ -4 & \;\;\;7}\pmatrix{x_1 \\ x_2} = 5\pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{13x_1 - 4x_2 = 5x_1 \\ -4x_1 + 7x_2 = 5x_2}$

$\bigg \{ \array{8x_1 - 4x_2 = 0 \\ -4x_1 + 2x_2 = 0}$

Both these equations give $x_2 = 2x_1$.

Note that for a $2 \times 2$ system we do not actually need to introduce the parameter $\alpha$ as we did in the previous example. We can simply choose a convenient numerical value for either of the components $x_1$ or $x_2$ of the eigenvector. So here we can let $x_1 = 1$, say, giving $x_2 = 2$.

Thus an eigenvector corresponding to the eigenvalue $\lambda_1 = 5$ is $\boldsymbol{x}_1 = \pmatrix{1 \\ 2}$.

Case 2: To find an eigenvector $\boldsymbol{x}_2$ corresponding to eigenvalue $\lambda_2 = 15$ we solve,

$A \;\boldsymbol{x}_2 = \lambda_2 \; \boldsymbol{x}_2$

$\pmatrix{\;13 & -4 \\ -4 & \;\;\;7}\pmatrix{x_1 \\ x_2} = 15\pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{13x_1 - 4x_2 = 15x_1 \\ -4x_1 + 7x_2 = 15x_2}$

$\bigg \{ \array{-2x_1 - 4x_2 = 0 \\ -4x_1 - 8x_2 = 0}$

Both these equations give $x_1 = -2x_2$. Let $x_2 = 1$, say, giving $x_1 = -2$.

Then an eigenvector corresponding to the eigenvalue $\lambda_2 = 15$ is $\boldsymbol{x}_2 = \pmatrix{-2 \\ 1}$.

In summary, we therefore have the eigenvalue/eigenvector pairs,

$\lambda_1 = 5, \; \boldsymbol{x}_1 = \pmatrix{1 \\ 2}$; $\lambda_2 = 15, \; \boldsymbol{x}_2 = \pmatrix{-2 \\ \;\;\;1}$.

Find the eigenvalues and eigenvectors of the matrix, $A = \pmatrix{\;\;\;3 & -6 \\ -2 & \;\;\;4}$.

Solution

To find the eigenvalues we need to solve

$\text{det}(A - \lambda I) = \bigg | \matrix{3 - \lambda & -6 \\ -2 & 4 - \lambda} \bigg | = 0$

$(3 - \lambda)(4 - \lambda) - 12 = 0$

$\lambda^2 - 7\lambda = 0$

$\lambda(\lambda - 7) = 0$

$\lambda_1 = 0, \; \lambda_2 = 7$

We now find the eigenvectors:

Case 1: To find an eigenvector $\boldsymbol{x}_1$ corresponding to eigenvalue $\lambda_1 = 0$ we solve, $A\,\boldsymbol{x}_1 = 0\,\boldsymbol{x}_1$.

$\pmatrix{\;\;\;3 & -6 \\ -2 & \;\;\; 4}\pmatrix{x_1 \\ x_2} = \pmatrix{0 \\ 0}$

$\bigg \{ \array{3x_1 - 6x_2 = 0 \\ -2x_1 + 4x_2 = 0}$.

Both these equations give $x_1 = 2x_2$. Let $x_2 = 1$, say, giving $x_1 = 2$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = 0$ is $\boldsymbol{x}_1 = \pmatrix{2 \\ 1}$.

Case 2: To find an eigenvector $\boldsymbol{x}_2$ corresponding to eigenvalue $\lambda_2 = 7$, solve $A\,\boldsymbol{x}_2 = 7\,\boldsymbol{x}_2 $.

$\pmatrix{\;\;\;3 & -6 \\ -2 & \;\;\; 4}\pmatrix{x_1 \\ x_2} = 7\pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{3x_1 - 6x_2 = 7x_1 \\ -2x_1 + 4x_2 = 7x_2}$.

$\bigg \{ \array{-4x_1 - 6x_2 = 0 \\ -2x_1 - 3x_2 = 0}$.

Both these equations give $x_1 = - {\large\frac{3}{2}}x_2$. Let $x_2 = 2$, say, giving $x_1 = - 3$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_2 = 7$ is $\boldsymbol{x}_2 = \pmatrix{-3 \\ \;\;\;2}$.

To summarise we have:

$\lambda_1 = 0, \; \boldsymbol{x}_1 = \pmatrix{2 \\ 1}$; $\lambda_2 = 7, \; \boldsymbol{x}_2 = \pmatrix{-3 \\ \;\;\;2}$.

Find the eigenvalues and eigenvectors of the matrix, $A = \pmatrix{3 & 0 \\ 0 & 3}$.

Solution

The matrix $A$ is a diagonal matrix so that its eigenvalues are the entries on the main diagonal. Hence, $A$ has one repeated eigenvalue, i.e. $\lambda = 3$. We now try to find two linearly independent eigenvectors for this eigenvalue. Solve $A\pmb{x} = 3\pmb{x}$ , i.e.

$\pmatrix{3 & 0 \\ 0 & 3}\pmatrix{x_1 \\x_2} = 3\pmatrix{x_1 \\x_2}$

$\Bigg \{ \matrix{3x_1 + 0x_2 = 3x_1 \\ 0x_1 + 3x_2 = 3x_2}$

$\Bigg \{ \matrix{0x_1 + 0x_2 = 0 \\ 0x_1 + 0x_2 = 0}$

Neither of these equations place any restrictions on $x_1$ or $x_2$ and so they can take any values we choose, say $x_1 = \alpha$ and $x_2 = \beta$ . An eigenvector corresponding to the repeated eigenvalue $\lambda = 3$ will therefore have the form, $\pmb{x} = \pmatrix{\alpha \\ \beta}$. We can now obtain two linearly independent eigenvectors through suitable choices for $\alpha$ and $\beta$. The most obvious choices are:

$\alpha = 1$, $\beta = 0$ giving the eigenvector, $\pmb{x}_1 = \pmatrix{1 \\ 0}$ and

$\alpha = 0$, $\beta = 1$ giving the eigenvector, $\pmb{x}_2 = \pmatrix{0 \\ 1}$.

We have therefore been able to find two linearly independent eigenvectors for the repeated eigenvalue $\lambda = 3$.

Find the eigenvalues and eigenvectors of the matrix, $A = \pmatrix{3 & -9 \\ 1 & \;\;\,9}$.

Solution

In Example 4 we found that the matrix has one repeated eigenvalue, i.e. $\lambda = 6$. We now try to find two linearly independent eigenvectors for this eigenvalue. Solve $A\pmb{x} = 6\pmb{x}$, i.e.

$\pmatrix{3 & -9 \\ 1 & \;\;\,9}\pmatrix{x_1 \\x_2} = 6\pmatrix{x_1 \\x_2}$

$\Bigg \{ \matrix{3x_1 - 9x_2 = 6x_1 \\ \;x_1 + 9x_2 = 6x_2}$

$\Bigg \{ \matrix{-3x_1 - 9x_2 = 0 \\ \;\;\;\,x_1 + 3x_2 = 0}$

Both these equations give $x_1 = -3x_2$. Let $x_2 = \alpha$, $(\alpha \neq 0 )$ giving $x_1 = -3\alpha$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda = 6$ will be of the form $\pmb{x} = \pmatrix{-3 \\ \;\;\;1}\alpha$.

No matter what choice is made for $\alpha$ the resulting eigenvectors will be scalar multiples of each other. We have therefore been unable to find two linearly independent eigenvectors for the eigenvalue $\lambda = 6$.

Find the eigenvectors of the matrix, $A = \pmatrix{0 & -1 \\ 1 & \;\;\;0}$.

Solution

In Example 6(i) we found that $A$ had complex eigenvalues, $\lambda_1 = j$ and $\lambda_2 = - j$.

We now calculate the eigenvectors of the matrix $A$

Case 1: To find an eigenvector $\boldsymbol{x}_1$ corresponding to eigenvalue $\lambda_1 = j$ we solve

$A\,\boldsymbol{x}_1 = \lambda_1\,\boldsymbol{x}_1$.

$\pmatrix{0 & -1 \\ 1 & \;\;\; 0}\pmatrix{x_1 \\ x_2} = j\pmatrix{x_1 \\ x_2}$

$\bigg \{ \array{-x_2 = j\,x_1 \\\;\; x_1 = j\,x_2}$.

If, for example, we multiply the first equation by $j$ both equations give $x_1 = j\,x_2$.

Let $x_2 = 1$, say, then $x_1 = j$.

An eigenvector corresponding to the eigenvalue $\lambda_1 = j$ will then be $\boldsymbol{x}_1 = \pmatrix{j \\ 1}$.

Case 2: To find an eigenvector $\pmb{x_2}$ corresponding to eigenvalue $\lambda_2 = - j$ simply take the complex conjugates of the entries of $\boldsymbol{x}_1$ giving, $\boldsymbol{x}_2 = \pmatrix{-j \\ \;\;\;1}$.

To summarise we have:

$\lambda_1 = j, \; \boldsymbol{x}_1 = \pmatrix{j \\ 1}$; $\lambda_2 = -j, \; \boldsymbol{x}_2 = \pmatrix{-j \\ \;\;\;1}$.

Determine the eigenvalues and eigenvectors of the matrix, $A = \pmatrix{\;\;\;1 & 0 & 2 \\ -7 & 2 & 4 \\ \;\;\;8 & 0 & 1}$

Solution

To calculate the eigenvalues we need to solve

\begin{align} \text{det}(A - \lambda I) = \left| \begin{matrix} 1-\lambda & 0 & 2 \\ -7 & 2-\lambda & 4 \\ 8 & 0 & 1-\lambda \end{matrix} \right| = 0 \end{align}.

Here we will expand down Column 2 as it is the row/column with most zeroes.

$(2 - \lambda) \bigg | \matrix{1-\lambda & 2 \\ 8 & 1-\lambda} \bigg | = 0$

$(2 - \lambda)[(1 - \lambda)(1-\lambda)-16] = 0$

$(2-\lambda)[\lambda^2 - 2\lambda - 15] = 0$

$(2 - \lambda)(\lambda - 5)(\lambda + 3) = 0$

Hence, $\lambda_1 = 2, \; \lambda_2 = 5$ and $\lambda_3 = -3$ are the eigenvalues of the matrix $A$.

We now calculate the eigenvectors corresponding to the eigenvalues.

Case 1: For an eigenvector $\boldsymbol{x}_1$, corresponding to eigenvalue $\lambda_1 = 2$, we solve $A\, \boldsymbol{x}_1 = 2\,\boldsymbol{x}_1$, i.e.

$\pmatrix{\;\;\;1 & 0 & 2 \\ -7 & 2 & 4 \\ \;\;\;8 & 0 & 1}\pmatrix{x_1 \\ x_2 \\ x_3} = 2 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Bigg \{ \matrix{\;x_1 \;\;\;\;\;\; &+& 2x_3 = 2x_1 \\ -7x_1 + 2x_2 &+& 4x_3 = 2x_2 \\ 8x_1 \;\;\;\;\;\; &+& \;\;x_3 = 2x_3}$

$\Bigg \{ \matrix{\;-x_1\;\;\; & + & 2x_3 = 0 \\ -7x_1 & + & 4x_3 = 0 \\ \;\;\,8x_1 & - & \;\;x_3 = 0}$.

From the three equations the only possibility is that $x_1 = x_3 = 0$ . We can choose $x_2$ to have any value, $\alpha \neq 0$. Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = 2$ is of the form $ \boldsymbol{x}_1 = (0, \; \alpha, \; 0)^T $. For example, letting $\alpha = 1$ gives $\boldsymbol{x}_1 = (0, \; 1, \; 0)^T$.

Case 2: For an eigenvector $\boldsymbol{x}_2$, corresponding to eigenvalue $\lambda_2 = 5$, we solve $A\,\boldsymbol{x}_2 = 5\,\boldsymbol{x}_2$, i.e.

$\pmatrix{\;\;\;1 & 0 & 2 \\ -7 & 2 & 4 \\ \;\;\;8 & 0 & 1}\pmatrix{x_1 \\ x_2 \\ x_3} = 5 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Bigg \{ \matrix{\;x_1 \;\;\;\;\;\; &+& 2x_3 = 5x_1 \\ -7x_1 + 2x_2 &+& 4x_3 = 5x_2 \\ 8x_1 \;\;\;\;\;\; &+& \;\;x_3 = 5x_3}$

$\Bigg \{ \matrix{-4x_1 \;\;\;\;\;\;\;\;\;\;& + & 2x_3 = 0 \;.........(1) \\ -7x_1 - 3x_2 & + & 4x_3 = 0\; .........(2) \\ \;\;\;8x_1 \;\;\;\;\;\;\;\;\;\; & - & 4x_3 = 0 \; ..........(3)}$.

Equations (1) and (3) both say that $x_3 = 2x_1$. Set $x_1 = \alpha \;\,(\alpha \neq 0)$ to obtain $x_3 = 2\alpha$. Substituting these values in Eq. (2) gives, $-7\alpha - 3x_2 + 8\alpha = 0 \Rightarrow 3x_2 = \alpha \Rightarrow x_2 = \alpha / 3$. Hence, an eigenvector corresponding to the eigenvalue $\lambda_2 = 5$ is of the form $\boldsymbol{x}_2 = (\alpha, \; \alpha/3, \; 2\alpha)^T$. Choosing $\alpha = 3$ gives $\boldsymbol{x}_2 = (3, \; 1, \; 6)^T$.

Case 3: For an eigenvector $\boldsymbol{x}_3$, corresponding to eigenvalue $\lambda_3 = - 3$, solve $A\,\boldsymbol{x}_3 = - 3\, \boldsymbol{x}_3$, i.e.

$\pmatrix{\;\;\;1 & 0 & 2 \\ -7 & 2 & 4 \\ \;\;\;8 & 0 & 1}\pmatrix{x_1 \\ x_2 \\ x_3} = -3 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Bigg \{ \matrix{\;x_1 \;\;\;\;\;\; &+& 2x_3 = -3x_1 \\ -7x_1 + 2x_2 &+& 4x_3 = -3x_2 \\ 8x_1 \;\;\;\;\;\; &+& \;\;x_3 = -3x_3}$

$\Bigg \{ \matrix{4x_1\;\;\;\;\;\;\; & + & 2x_3 = 0 \; .........(1) \\ -7x_1 + 5x_2 & + & 4x_3 = 0\; .........(2) \\ 8x_1 \;\;\;\;\;\;\;& + & 4x_3 = 0\; .........(3)}$.

Equations (1) and (3) both say that $x_3 = -2x_1$. Set $x_1 = \alpha \;\, (\alpha \neq 0)$ to obtain $x_3 = -2\alpha$. Substituting these values in Eq. (2) gives, $-7\alpha + 5x_2 - 8\alpha = 0 \Rightarrow 5x_2 = 15\alpha \Rightarrow x_2 = 3\alpha$. Hence, an eigenvector corresponding to the eigenvalue $\lambda_3 = -3$ is of the form $\boldsymbol{x}_3 = (\alpha, \; 3\alpha, \; -2\alpha)^T$. Choosing $\alpha = 1$ gives $\boldsymbol{x}_3 = (1, \; 3, \; -2)^T$.

In summary, we therefore have,

$\lambda_1 = 2, \; \boldsymbol{x}_1 = \pmatrix{0 \\ 1 \\ 0}$; $\lambda_2 = 5, \; \boldsymbol{x}_2 = \pmatrix{3 \\ 1 \\ 6}$;$\lambda_3 = -3, \; \boldsymbol{x}_3 = \pmatrix{\;\;\;1 \\ \;\;\;3 \\ -2}$.

(i). Calculate the eigenvalues and eigenvectors of the matrix

$A = \pmatrix{-4 & -6 \\ \,\,\,\,3 & \,\,\,\,5}$.

(ii). Determine an invertible matrix $P$ and a diagonal matrix $D$ such that such that $P^{-1}A\,P=D$ .

Solution

(i). First we find the eigenvalues by solving,

$\text{det}(A - \lambda I) = \bigg | \matrix{-4-\lambda & -6 \\3 & 5-\lambda} \bigg | = 0$

$\Rightarrow (-4 - \lambda)(5 - \lambda) + 18 = 0$

$\Rightarrow \lambda^2 - \lambda - 2 = 0$

$\Rightarrow (\lambda - 2)(\lambda + 1) = 0$

$\Rightarrow \lambda_1 = 2,\; \lambda_2 = -1.$

As the eigenvalues of the matrix $A$ are real and distinct $A$ is diagonalisable.

We now calculate the eigenvectors corresponding to each of the eigenvalues.

For $\lambda_1 = 2$ we have $A\pmb{x}_1 = \lambda_1 \pmb{x}_1$ and so we need to solve,

$\pmatrix{-4 & -6 \\ \;\,\,3 & \;\,\,5}\pmatrix{x_1 \\ x_2} = 2.\pmatrix{x_1 \\ x_2}$

$\Rightarrow \bigg \{ \matrix{-4x_1 - 6x_2 = 2x_1 \\ \;\;3x_1 + 5x_2 = 2x_2}$

$\Rightarrow \bigg \{ \matrix{-6x_1 - 6x_2 = 0 \\ \;\;\,3x_1 + 3x_2 = 0.}$

Both these equations give $x_1 = -x_2$ and if we let $x_2 = 1$ then $x_1 = -1$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = 2$ is $\pmb{x}_1 = \pmatrix{-1 \\ \;\;\,1}$ .

For $\lambda_2 =\; ‒ 1$ we have $A\pmb{x}_2 = \lambda_2 \pmb{x}_2$ and so we need to solve,

$\pmatrix{-4 & -6 \\ \;\;\,3 & \;\;\,5}\pmatrix{x_1 \\ x_2} = -1.\pmatrix{x_1 \\ x_2}$

$\Rightarrow \bigg \{ \matrix{-4x_1 - 6x_2 = -x_1 \\ \;\;\,3x_1 + 5x_2 = -x_2}$

$\Rightarrow \bigg \{ \matrix{-3x_1 - 6x_2 = 0 \\ \;\;\,3x_1 + 6x_2 = 0}$

Both these equations give $x_1 = -2x_2$ and if we let $x_2 = 1$ then $x_1 = -2$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_2 = - 1$ is $\pmb{x}_2 = \pmatrix{-2 \\ \;\;\,1}$.

The eigenvalues and eigenvectors of $A$ are therefore,

$\lambda_1 = 2$, $\pmb{x}_1 = \pmatrix{-1 \\ \;\;\,1}$;$\lambda_2 = -1$,$\pmb{x}_2 = \pmatrix{-2 \\ \;\;\, 1}$.

(ii). Using the above results we can define the invertible matrix,

$P = \pmatrix{-1 & -2 \\ \;\;\,1 & \;\;\,1}$.

Now form the diagonal matrix $D$ by writing the eigenvalues on the main diagonal of $D$ in the same order the corresponding eigenvectors appear in $P$, i.e.

$D = \pmatrix{2 & \;\;\,0 \\ 0 & -1}$

We now check that $P^{-1} A P = D$ as required.

Since we have a $2 \times 2$ matrix we easily find that $P^{-1} = \pmatrix{\;\;\,1 & \;\;\,2 \\ -1 & -1}$ and so $P^{-1} A P = \pmatrix{\;\;\,1 & \;\;\,2 \\-1 & -1}\pmatrix{-4 & -6 \\ \;\;\,3 & \;\;\,5}\pmatrix{-1 & -2 \\ \;\;\,1 & \;\;\, 1} = \pmatrix{2 & 4 \\ 1 & 1}\pmatrix{-1 & -2 \\ \;\;\,1 & \;\;\,1} = \pmatrix{2 & \;\;\,0 \\ 0 & -1} = D$ as required.

(i). Determine the eigenvalues and eigenvectors of the matrix,

$A = \pmatrix{\;-5 & 2 & 1 \\ \;-8 & 3 & 2 \\ -16 & 2 & 6}$.

(ii). Write down an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1} A P = D$ .

Solution

The eigenvalues are calculated using,

$\text{det}(A - \lambda I) = \left | \matrix{-5-\lambda & 2 & 1 \\ -8 & 3-\lambda & 2 \\ -16 & 2 & 6-\lambda} \right | = 0$

$\Rightarrow (-5-\lambda) \left|\matrix{3 - \lambda & 2 \\ 2 & 6-\lambda}\right| - 2 \left | \matrix{-8 & 2 \\ -16 & 6-\lambda} \right| + 1 \left | \matrix{-8 & 3-\lambda \\ -16 & 2} \right| = 0$

$\Rightarrow (-5-\lambda)[(3-\lambda)(6-\lambda) - 4] - 2 [-8(6-\lambda) + 32] + 1 [-16 + 16(3 - \lambda)] = 0$

$\Rightarrow (-5-\lambda)[\lambda^2 - 9\lambda + 14] - 2 [8\lambda - 16] + [32 - 16\lambda] = 0$

$\Rightarrow (-5-\lambda)(\lambda-2)(\lambda-7) - 2[8(\lambda-2)] + 16(2 - \lambda) = 0$

$\Rightarrow (-5-\lambda)(\lambda-2)(\lambda-7) - 16(\lambda-2) - 16(\lambda-2) = 0$

$\Rightarrow (\lambda-2)(-5-\lambda)(\lambda-7) - 32(\lambda-2) = 0$

$\Rightarrow (\lambda-2)[(-5-\lambda)(\lambda-7)-32] = 0$

$\Rightarrow (\lambda-2)[-\lambda^2 + 2\lambda+ 3] = 0$

$\Rightarrow -(\lambda-2)[\lambda^2 - 2\lambda - 3] = 0$

$\Rightarrow -(\lambda-2)(\lambda-3)(\lambda+1) = 0$

$\Rightarrow \lambda_1 = 2, \;\lambda_2 = 3, \; \lambda_3 = -1$.

We now calculate the eigenvectors corresponding to each of the eigenvalues.

Case 1: For an eigenvector $\pmb{x}_1$, corresponding to eigenvalue $\lambda_1 = 2$ , we solve $A\pmb{x}_1 = 2\pmb{x}_1$ , i.e.

$\pmatrix{-5 & 2 & 1 \\ -8 & 3 & 2 \\ -16 & 2 & 6}\pmatrix{x_1 \\x_2 \\x_3} = 2 \pmatrix{x_1\\x_2\\x_3}$

$\Bigg\{ \array{-5x_1 + 2x_2 + x_3 = 2x_1 \\ -8x_1 + 3x_2 + 2x_3 = 2x_2 \\ -16x_1 + 2x_2 + 6x_3 = 2x_3}$

$\Bigg \{ \array{-7x_1 + 2x_2 + x_3 = 0 \dots\dots(1) \\-8x_1 + x_2 + 2x_3 = 0 \dots\dots(2) \\ -16x_1 + 2x_2 + 4x_3 = 0 \dots\dots(3)}$.

Now, (1) ‒ (3) gives $9x_1 - 3x_3 = 0 \Rightarrow x_3 = 3x_1$.

Then (2) gives $-8x_1 + x_2 + 6x_1 = 0 \Rightarrow -2x_1 + x_2 = 0 \Rightarrow x_2 = 2x_1$.

Let $x_1 = 1$ then $x_2 = 2$ and $x_3 = 3$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = 2$ is $\pmb{x}_1 = \pmatrix{1 \\ 2 \\ 3}$.

Case 2: For an eigenvector $\pmb{x}_2$ , corresponding to eigenvalue $\lambda_2 = 3$ , we solve $A \pmb{x}_2 = 3 \pmb{x}_2$, i.e.

$\pmatrix{-5 & 2 & 1 \\ -8 & 3 & 2 \\ -16 & 2 & 6} \pmatrix{x_1 \\ x_2 \\x_3} = 3\pmatrix{x_1 \\ x_2 \\x_3}$

$\Bigg \{ \array{-5x_1 + 2x_2 + x_3 = 3x_1 \\ -8x_1 + 3x_2 + 2x_3 = 3x_2 \\ -16x_1 + 2x_2 + 6x_3 = 3x_3}$

$\Bigg \{ \array{-8x_1 + 2x_2 + x_3 = 0 \dots\dots(1)\;\; \\-8x_1\;\;\;\;\;\;\;\;\;\;\;+ 2x_3 = 0 \dots\dots(2)\,\\-16x_1 + 2x_2 + 3x_3 = 0 \dots\dots(3)\;\;}$.

From Equation (2) we have that $x_3 = 4x_1$.

Also, (3) $- 2 \times$ (1) gives, $-2x_2 + x_3 = 0 \Rightarrow x_3 = 2x_2$.

Let $x_1 = 1$ then $x_3 = 4$ and $x_2 = 2$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_2 = 3$ is $\pmb{x}_2 = \pmatrix{1\\2\\4}$.

Case 3: For an eigenvector $\pmb{x}_3$, corresponding to eigenvalue $\lambda_3 = -1$, solve $A \pmb{x}_3 = ‒1 \pmb{x}_3 $, i.e.

$\pmatrix{-5 & 2 & 1 \\ -8 & 3 & 2 \\ -16 & 2 & 6} \pmatrix{x_1 \\ x_2 \\ x_3} = -1 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Bigg \{ \array{-5x_1 + 2x_2 + x_3 = -x_1 \\ -8x_1 + 3x_2 + 2x_3 = -x_2 \\ -16x_1 + 2x_2 + 6x_3 = -x_3}$

$\Bigg \{ \array{-4x_1 + 2x_2 + x_3 = 0 \dots\dots(1) \\ -8x_1 + 4x_2 + 2x_3 = 0 \dots\dots(2) \\ -16x_1 + 2x_2 + 7x_3 = 0 \dots\dots(3)}$.

Now, (1) ‒ (3) gives $12x_1 - 6x_3 = 0 \Rightarrow x_3 = 2x_1$.

Substitute in (2) to give, $-8x_1 + 4x_2 + 4x_1 = 0 \Rightarrow -4x_1 + 4x_2 = 0 \Rightarrow x_1 = x_2$.

Let $x_1 = 1$ then $x_2 = 1$ and $x_3 = 2$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_3 = -1$ is $\pmb{x}_3 = \pmatrix{1\\1\\2}$.

In summary, the eigenvalue/eigenvectors pairs are

$\lambda_1 = 2, \; \pmb{x}_1 = \pmatrix{1 \\ 2 \\ 3}$; $\lambda_2 = 3, \; \pmb{x}_2 = \pmatrix{1 \\ 2 \\ 4}$;$\lambda_3 = -1, \; \pmb{x}_3 = \pmatrix{1 \\ 1 \\ 2}$

(ii). As the eigenvalues of $A$ are distinct $A$ is diagonalisable and we can define

$P = \pmatrix{1 & 1 & 1 \\ 2 & 2 & 1 \\ 3 & 4 & 2}$ and $D = \pmatrix{2 & 0 & \;\;\,0 \\0 & 3 & \;\;\,0 \\ 0 & 0 & -1}$ so that $P^{-1} A P = D$.

As a check we could calculate $P^{-1}$ and form the matrix product $P^{-1} A P$ , but we would prefer not to have to determine $P^{-1}$ for the $3 \times 3$ matrix. Alternatively, we note that $P^{-1}A P = D$ is equivalent to writing $AP = PD$ and we show this latter relationship holds.

Here

$AP = PD = \pmatrix{2 & \;3 & -1 \\ 4 & \;6 & -1 \\6 & 12 & -2}$

thereby confirming that our calculations are correct.

In Example 11 (Section 2.2.2) we saw that the matrix $A = \pmatrix{3 & -9 \\ 1 & \;\;\;9}$ has a repeated eigenvalue, $\lambda = 6$ and that all possible eigenvectors are scalar multiples of $\pmb{x} = \pmatrix{-3 \\\;\;\;1}$. It is therefore impossible to find two linearly independent eigenvectors. A consequence of this is that the matrix $P$ will have columns that are scalar multiples of each other meaning that the determinant of $P$ is zero so that the inverse does not exist. Hence, $A$ cannot be diagonalised.

For the matrix $A = \pmatrix{1 & -3 & 3 \\ 0 & -5 & 6 \\ 0 & -3 & 4}$ we can show that the characteristic equation is

$(\lambda + 2)(\lambda - 1)^2 = 0$.

Hence, the eigenvalues are $\lambda_1 = -2$, $\lambda_2 = \lambda_3 = 1$.

We now calculate the eigenvectors noting that the eigenvalue $\lambda = 1$ is repeated.

Case 1: For eigenvector $\pmb{x}_1$, corresponding to eigenvalue $\lambda_1 = -2$, solve $A\pmb{x}_1 = -2\pmb{x}_1 $, i.e.

$\pmatrix{1 & -3 & 3 \\ 0 & -5 & 6 \\ 0 & -3 & 4}\pmatrix{x_1 \\ x_2 \\ x_3} = -2 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Rightarrow \Bigg \{ \array{x_1 - 3x_2 + 3x_3 = -2x_1 \\ \;\;\;-5x_2 \;+ 6x_3 = -2x_2 \\ \;\;\;-3x_2 \;+ 4x_3 = -2x_3}$

$\Rightarrow \Bigg \{ \array{3x_1 - 3x_2 + 3x_3 = 0 \dots\dots(1) \\ \;\;\;\;\;\;-3x_2 \;+ 6x_3 = 0 \dots \dots(2) \\ \;\;\;\;\;\;-3x_2 \;+ 6x_3 = 0\dots\dots(3)}$

Equations (2) and (3) both give $x_2 = 2x_3$.

Substitute in Eq. (1) giving,

$3x_1 - 6x_3 + 3x_3 = 0 \Rightarrow x_1 = x_3$.

Let $x_3 = 1 \Rightarrow x_2 = 2$, $x_1 = 1$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = -2$ is, $\pmb{x}_1 = \pmatrix{1 \\ 2 \\ 1}$.

Case 2: For the repeated eigenvalue $\lambda_1 = 1$ we try to find two linearly independent eigenvectors. Solve $A\pmb{x} = 1.\pmb{x}$, i.e.,

$\pmatrix{1 & -3 & 3 \\ 0 & -5 & 6 \\ 0 & -3 & 4}\pmatrix{x_1 \\ x_2 \\ x_3} = 1 \pmatrix{x_1 \\ x_2 \\ x_3}$

$\Bigg \{ \array{x_1 - 3x_2 + 3x_3 = x_1 \\ \;\;\;\;-5x_2 \;+ 6x_3 = x_2 \\ \;\;\;\;-3x_2 \;+ 4x_3 = x_3}$

$\Bigg \{ \array{-3x_2 + 3x_3 = 0 \dots\dots(1) \\ -6x_2 \;+ 6x_3 = 0 \dots \dots(2) \\ -3x_2 + 3x_3 = 0\dots\dots(3)}$

All three equations give $x_2 = x_3$ and so we can set $x_2 = x_3 = \alpha$, $(\alpha \neq 0 )$.

As the equations are all independent of $x_1$ it can take any value, say $x_1 = \beta$.

An eigenvector corresponding to the repeated eigenvalue $\lambda = 1$ will therefore have the form, $\pmb{x} = \pmatrix{\beta \\ \alpha \\ \alpha}$. We can now obtain two linearly independent eigenvectors through suitable choices for $\alpha$ and $\beta$. The most obvious choices are:

$\alpha = 1$, $\beta = 0$ giving the eigenvector, $\pmb{x}_2 = \pmatrix{0 \\ 1 \\ 1}$ and

$\alpha = 0$, $\beta = 1$ giving the eigenvector, $\pmb{x}_3 = \pmatrix{1 \\ 0 \\ 0}$.

In summary, we have

$\lambda_1 = -2, \pmb{x}_1 = \pmatrix{1 \\ 2 \\ 1}$; $\lambda_2 = 1, \pmb{x}_2 = \pmatrix{0 \\ 1 \\ 1}$;$\lambda_3 = 1, \pmb{x}_3 = \pmatrix{1 \\ 0 \\ 0}$.

We have therefore found three linearly independent eigenvectors and so $A$ is diagonalisable with .

$P = \pmatrix{1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & 0}$ and $D = \pmatrix{-2 & 0 & 0 \\ \;\;0 & 1 & 0 \\ \;\;0 & 0 & 1}$ so that $P^{-1}A P = D$.

Determine the general solution of the coupled system of ordinary differential equations,

$\dot{x}_1 = -5x_1 + 2x_2 + x_3$

$\dot{x}_2 = -8x_1 + 3x_2 + 2x_3$

$\dot{x}_3 = -16x_1 + 2x_2 + 6x_3$.

Solution

Step 1: Write the system as a matrix equation.

Let $\dot{\pmb{x}} = \pmatrix{\dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3}, \; A = \pmatrix{-5 & 2 & 1 \\ -8 & 3 & 2 \\ -16 & 2 & 6}$ and $\pmb{x} = \pmatrix{x_1 \\ x_2 \\ x_3}$ so that the system may be written as the matrix equation, $\dot{\pmb{x}} = A \pmb{x}$, i.e. $\pmatrix{\dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3} = \pmatrix{-5 & 2 & 1 \\ -8 & 3 & 2 \\ -16 & 2 & 6}\pmatrix{x_1 \\ x_2 \\ x_3}$.

Step 2: Calculate the eigenvalues and eigenvectors of the coefficient matrix $A$.
The eigenvalues and eigenvectors of the matrix $A$ were previously found in Example 15 to be,

$\lambda_1 = 2$, $\pmb{x}_1 = \pmatrix{1 \\ 2 \\ 3}$; $\lambda_2 = 3$, $\pmb{x}_2 = \pmatrix{1 \\ 2 \\ 4}$ $\lambda_3 = -1$, $\pmb{x}_3 = \pmatrix{1 \\ 1 \\ 2}$.

Step 3: Diagonalise the matrix $A$.
As the eigenvalues of $A$ are real and distinct we know that $A$ is diagonalisable with

$P = \pmatrix{1 & 1 & 1 \\ 2 & 2 & 1 \\ 3 & 4 & 2}$    and    $P^{-1}AP = \pmatrix{2 & 0 & \;\;\;0 \\ 0 & 3 & \;\;\;0 \\ 0 & 0 & -1} = D$.

Step 4: Determine the general solution of the system.

We saw above that by letting $\pmb{x} = P \pmb{u}$ the coupled system $( \dot{\pmb{x}} = A\pmb{x} )$ can be expressed in uncoupled form as $\dot{\pmb{u}} = D\pmb{u}$ which in this case gives

$\pmatrix{\dot{u}_1 \\ \dot{u}_2 \\ \dot{u}_3} = \pmatrix{2 & 0 & \;\;0 \\ 0 & 3 & \;\;0 \\ 0 & 0 & -1}\pmatrix{u_1 \\ u_2 \\ u_ 3}$.

Expanding this expression gives the three uncoupled differential equations

${\Large{\frac{du_1}{dt}}} = 2u_1$, ${\Large{\frac{du_2}{dt}}} = 3u_2$ and ${\Large{\frac{du_3}{dt}}} = -u_3$

The solutions of these separable first order differential equations (refer to notes on ODEs) are

$u_1 = C_1e^{2t}$, $u_2 = C_2e^{3t}$    and    $u_3 = C_3e^{-t}$

where $C_1$, $C_2$ and $C_3$ are arbitrary constants.

The solution, in terms of the original variables, $x_1(t)$, $x_2(t)$ and $x_3(t)$ can now be found.

Since $\pmb{x} = P\pmb{u} $ we have, $\pmatrix{x_1 \\ x_2 \\ x_3} = \pmatrix{1 & 1 & 1 \\ 2 & 2 & 1 \\ 3 & 4 & 2}\pmatrix{u_1 \\ u_2 \\ u_3}$.

Multiplying out the above gives

$x_1(t) = u_1 + u_2 + u_3 = C_1e^{2t} + C_2e^{3t} + C_3e^{-t}$

$x_2(t) = 2u_1 + 2u_2 + u_3 = 2C_1e^{2t} + 2C_2e^{3t} + C_3e^{-t}$

$x_3(t) = 3u_1 + 4u_2 + 2u_3 = 3C_1e^{2t} + 4C_2e^{3t} + 2C_3e^{-t}$.

A neater way of writing the general solution (GS) is

$\class{outbox}{\pmatrix{x_1 \\ x_2 \\ x_3} = C_1\pmatrix{1 \\ 2 \\ 3}e^{2t} + C_2 \pmatrix{1 \\ 2 \\ 4}e^{3t} + C_3\pmatrix{1 \\ 1 \\ 2}e^{-t}}$.

Can you see how the general solution is connected to the eigenvalues and eigenvectors?

The eigenvalues and eigenvectors appear is the GS as follows

$\pmb{x}(t) = C_1 \pmb{x}_1e^{\lambda_1t} + C_2\pmb{x}_2e^{\lambda_2t} + C_3\pmb{x}_3e^{\lambda_3t}$.

Note: If initial conditions were specified we could determine the unknown constants $C_1$, $C_2$ and $C_3$ to obtain a particular solution.

Extending the above, the general solution of a linear system $\dot{\pmb{x}} = A\pmb{x}$, where the $n \times n$ coefficient matrix $A$ is diagonalisable, is given by

$\class{outbox}{\pmb{x}(t) = C_1 \pmb{x}_1e^{\lambda_1t} + C_2\pmb{x}_2e^{\lambda_2t} + C_3\pmb{x}_3e^{\lambda_3t} + \dots + C_n \pmb{x}_ne^{\lambda_nt}}$

where $\pmb{x}_i$ is the eigenvector associated with the real eigenvalue $\lambda_i$.

A system of two coupled ordinary differential equations is given by

$\dot{x}_1 = -2x_1 + 5x_2$

$\dot{x}_2 = \;\;x_1 + 2x_2$.

Determine the particular solutions for $x_1(t)$ and $x_2(t)$ that satisfy the initial conditions, $x_1(0) = -1$, $x_2(0) = 5$.

Solution

Step 1: Write the system in matrix form, $\dot{\pmb{x}} = A\pmb{x}$, i.e.

$\pmatrix{\dot{x}_1 \\ \dot{x}_2} = \pmatrix{-2 & 5 \\ \;\;\;1 & 2}\pmatrix{x_1 \\ x_2}$

Step 2: Determine the eigenvalues by solving,

$\det(A - \lambda I) = \Bigg | \matrix{-2-\lambda & 5 \\ 1 & 2-\lambda} \Bigg| = 0$

$\Rightarrow (-2 -\lambda)(2 - \lambda) - 5 = 0$

$\Rightarrow \lambda^2 - 9 = 0$

$\Rightarrow (\lambda - 3)(\lambda + 3) = 0$

$\Rightarrow \lambda_1 = 3$, $\lambda_2 = -3$.

As the eigenvalues of the matrix $A$ are real and distinct $A$ is diagonalisable.

Step 3: Calculate the eigenvectors corresponding to each of the eigenvalues.

For $\lambda_1 = 3$ we have $A \pmb{x}_1 = \lambda_1 \pmb{x}_1$ and so we need to solve,

$\pmatrix{-2 & 5 \\ \;\;\;1 & 2}\pmatrix{x_1 \\ x_2} = 3.\pmatrix{x_1 \\ x_2}$

$\Rightarrow \Bigg \{ \array{-2x_1 + 5x_2 = 3x_1 \\ \;\;\;\;x_1 + 2x_2 = 3x_2}$

$\Rightarrow \Bigg \{ \array{-5x_1 + 5x_2 = 0 \\ x_1 - x_2 = 0}$

Both these equations give $x_1 = x_2$ and if we let $x_2 = 1$ then $x_1 = 1$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_1 = 3$ is $\pmb{x}_1 = \pmatrix{1 \\ 1}$.

For $\lambda_2 = -3$ we have $A \pmb{x}_2 = \lambda_2 \pmb{x}_2$ and so we need to solve,

$\pmatrix{-2 & 5 \\ \;\;\;1 & 2}\pmatrix{x_1 \\ x_2} = -3.\pmatrix{x_1 \\ x_2}$

$\Rightarrow \Bigg \{ \array{-2x_1 + 5x_2 = -3x_1 \\ \;\;\;\;x_1 + 2x_2 = -3x_2}$

$\Rightarrow \Bigg \{ \array{x_1 + 5x_2 = 0 \\ x_1 + 5x_2 = 0}$

Both these equations give $x_1 = -5x_2$ and if we let $x_2 = 1$ then $x_1 = -5$.

Hence, an eigenvector corresponding to the eigenvalue $\lambda_2 = -3$ is $\pmb{x}_2 = \pmatrix{-5 \\ \;\;\;1}$.

The eigenvalues and eigenvectors of $A$ are therefore,

$\lambda_1 = 3$, $\pmb{x}_1 = \pmatrix{1 \\ 1}$;$\lambda_2 = -3$,$\pmb{x}_2 = \pmatrix{-5 \\ \;\;\;1}$.

Step 4: Diagonalise the matrix $A$.

Using the above results we can define the invertible matrix,

$P = \pmatrix{1 & -5 \\ 1 & \;\;\;1}$.

Now form the diagonal matrix $D$ by writing the eigenvalues on the main diagonal of $D$ in the same order the corresponding eigenvectors appear in $P$, i.e.

$P^{-1}AP = D = \pmatrix{3 & \;\;\;0 \\ 0 & -3}$.

Step 5: Determine the general solution of the system.

Let $\pmb{x} = P\pmb{u}$. Differentiating both sides with respect to $t$, gives, $\dot{\pmb{x}} = P \dot{\pmb{u}}$ (since $P$ is a constant matrix). The coupled system $\dot{\pmb{x}} = A \pmb{x}$ can then be written as

$ P\dot{\pmb{u}} = AP\pmb{u} \;\;\;\;\; \Rightarrow \dot{\pmb{u}} = P^{-1}AP \pmb{u} \;\;\;\;\; \Rightarrow \dot{\pmb{u}} = D \pmb{u}$.

Hence,

$\pmatrix{\dot{u}_1 \\ \dot{u}_2} = \pmatrix{3 & \;\;\;0 \\ 0 & -3}\pmatrix{u_1 \\ u_2}$.

Expanding this expression gives the two uncoupled first order separable differential equations

${\Large{\frac{du_1}{dt}}} = 3u_1$ and ${\Large{\frac{du_2}{dt}}} = -3u_2$

The solutions of these differential equations (refer to notes on ODEs) are

$u_1 = C_1e^{3t}$ and $u_2 = C_2e^{-3t}$

where $C_1$ and $C_2$ are arbitrary constants

The solution, in terms of the original variables, $x_1(t)$ and $x_2(t)$ can now be found.

Since $\pmb{x} = P\pmb{u}$ we have, $\pmatrix{x_1 \\ x_2} = \pmatrix{1 & -5 \\ 1 & \;\;\;1}\pmatrix{u_1 \\ u_2}$.

Expanding this expression, gives

$x_1 = u_1 - 5u_2$

$x_2 = u_1 + u_2$.

Now use $u_1 = C_1e^{3t}$ and $u_2 = C_2e^{-3t}$ to write down the general solution

$\class{outbox}{\array{x_1(t) = C_1e^{3t} - 5C_2e^{-3t} \\ \\ x_2(t) = C_1e^{3t} + C_2e^{-3t}}}$.

Step 6: Determine the particular solution that satisfies,$x_1(0) = -1$, $x_2(0) = 5$. Substituting these values into the general solution gives,

$x_1(0) = -1 \Rightarrow C_1 - 5C_2 = -1$

$x_2(0) = 5 \Rightarrow C_1 + C_2 = 5$.

Solve as simultaneous equations to obtain, $C_1 = 4$ and $C_2 = 1$.

The particular solution (PS) is therefore,

$\class{outbox}{\array{x_1(t) = 4e^{3t} - 5e^{-3t} \\ \\ x_2(t) = 4e^{3t} + e^{-3t}}}$.

Step 7 (OPTIONAL): Check the answers for $x_1(t)$ and $x_2(t)$.

The PS gave that $x_1(t) = 4e^{3t} - 5e^{-3t}$ and so differentiating gives,

$\dot{x}_1 = 12e^{3t} + 15e^{-3t}$. (1)

The original coupled system gave that

$\dot{x}_1 = -2x_1 + 5x_2$.

Substituting the expressions for $x_1(t)$ and $x_2(t)$ from the PS we have,

$\dot{x}_1 = -2(4e^{3t} - 5e^{-3t}) + 5(4e^{3t} + e^{-3t})$

$\Rightarrow \dot{x}_1 = 12e^{3t} + 15e^{-3t}$. (2)

Expressions ( 1 ) and ( 2 ) are identical thereby verifying our answer for $x_1(t)$.

Now verify the answer for $x_2(t)$.

The PS gave that $x_2(t) = 4e^{3t} + e^{-3t}$ and so differentiating gives,

$\dot{x}_2 = 12e^{3t} - 3e^{-3t}$. (3)

The coupled system gave that

$\dot{x}_2 = x_1 + 2x_2$.

Substituting the expressions for $x_1(t)$ and $x_2(t)$ from the PS we have,

$\dot{x}_2 = (4e^{3t} - 5e^{-3t}) + 2(4e^{3t} + e^{-3t})$

$\Rightarrow \dot{x}_2 = 12e^{3t} - 3e^{-3t}$. (4)

Expressions ( 3 ) and ( 4 ) are identical thereby verifying our answer for $x_2(t)$.

Determine the general solution of the coupled system

$\dot{x}_1 = x_1 - 3x_2 + 3x_3$

$\dot{x}_2 = \;\;\;\;\,-5x_2 + 6x_3$

$\dot{x}_3 = \;\;\;\;\,-3x_2 + 4x_3$.

Solution

In matrix form the system is given by $\dot{\pmb{x}} = A\pmb{x}$, i.e.

$\pmatrix{\dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3} = \pmatrix{1 & -3 & 3 \\ 0 & -5 & 6 \\ 0 & -3 & 4}\pmatrix{x_1 \\ x_2 \\ x_3}$.

In Example 17 we found that the matrix has the following eigenvalues and eigenvectors where the eigenvalue $\lambda = 1$ is repeated

$\lambda_1 = -2$, $\pmb{x}_1 = \pmatrix{1 \\ 2 \\ 1};$ $\lambda_2 = 1$, $\pmb{x}_2 = \pmatrix{0 \\ 1 \\ 1}$;$\lambda_3 = 1$, $\pmb{x}_3 = \pmatrix{1 \\ 0 \\ 0}$;

As we were able to find three linearly independent eigenvectors the matrix, $A$, is diagonalisable and the general solution of the system is given by

$x(t) = C_1\pmb{x}_1e^{\lambda_1t} + C_2\pmb{x}_2e^{\lambda_2t} + C_3\pmb{x}_3e^{\lambda_3t}$

Substituting the eigenvalues and eigenvectors gives

$\class{outbox}{x(t) = C_1\pmatrix{1 \\ 2 \\ 1}e^{-2t} + C_2 \pmatrix{0 \\ 1 \\ 1}e^{t} + C_3 \pmatrix{1 \\ 0 \\ 0}e^t}$

which expands to

$\class{outbox}{\array{x_1(t) = C_1e^{-2t} + C_3e^t \\ \\ x_2(t) = 2C_1e^{-2t} + C_2e^t \\ \\ x_3(t) = C_1e^{-2t} + C_2e^t.}}$

Determine the general solution of the coupled system

$\ddot{x}_1 = -2x_1 + x_2$

$\ddot{x}_2 = x_1 - 2x_2$.

Solution

Step 1: Write the system in matrix form, $\ddot{x} = A\pmb{x}$, i.e.

$\pmatrix{\ddot{x}_1 \\ \ddot{x}_2} = \pmatrix{-2 & \;\;\;1 \\ \;\;\;1 & -2}\pmatrix{x_1 \\ x_2}$.

Step 2: The eigenvalues and eigenvectors of $A$ are found to be,

$\lambda_1 = -3$, $\pmb{x}_1 = \pmatrix{\;\;\;1 \\ -1}$;$\lambda_2 = -1$, $\pmb{x}_2 = \pmatrix{1 \\ 1}$.

Since $A$ has distinct eigenvalues it can be diagonalised.

Step 3: Diagonalise the matrix $A$.

Using the above results, $P = \pmatrix{\;\;\;1 & 1 \\ -1 & 1}$ and $P^{-1}AP = \pmatrix{-3 & \;\;\;0 \\ \;\;\; 0 & -1} = D$.

Step 4: Determine the general solution of the system.

Let $\pmb{x} = P\pmb{u}$. Differentiating twice gives, $\ddot{\pmb{x}} = P\ddot{\pmb{u}}$ (since $P$ is a constant matrix). The coupled system $\ddot{\pmb{x}} = A\pmb{x}$ can then be written as

$P\ddot{\pmb{u}} = AP\pmb{u} \;\;\;\;\;\Rightarrow \ddot{\pmb{u}} = P^{-1}AP\pmb{u} \;\;\;\;\;\Rightarrow \ddot{\pmb{u}} = D \pmb{u}$.

Hence,

$\pmatrix{\ddot{u}_1 \\ \ddot{u}_2} = \pmatrix{-3 & \;\;\; 0 \\ \;\;\; 0 & -1}\pmatrix{u_1 \\ u_2}$.

Expanding this expression gives the two uncoupled second order differential equations

${\Large{\frac{d^2u_1}{dt^2}}} = -3u_1$    and    ${\Large{\frac{d^2u_2}{dt^2}}} = -u_2$.

The solutions of the differential equations

${\Large{\frac{d^2u_1}{dt^2}}} + 3u_1 = 0$    and    ${\Large{\frac{d^2u_2}{dt^2}}} + u_2 = 0$

are

$u_1(t) = A_1 \sin(\sqrt{3}t) + B_1\cos(\sqrt{3}t)$ and

$u_2(t) = A_2\sin(t) + B_2\cos(t)$

where $A_1$, $B_1$, $A_2$ and $B_2$ are arbitrary constants.

Note: In each case the roots of the auxiliary equation are complex with real part zero so solutions are in terms of the trig functions - refer to notes on ODEs.

The solution in terms of the original variables, $x_1(t)$ and $x_2(t)$ can now be found.

Since $\pmb{x} = P\pmb{u}$ we have, $\pmatrix{x_1 \\ x_2} = \pmatrix{\;\;\; 1 & 1 \\ -1 & 1}\pmatrix{u_1 \\ u_2}$.

Expanding this matrix equation gives

$x_1 =\;\;\, u_1 + u_2$

$x_2 = -u_1 + u_2$.

From earlier, $u_1 = A_1 \sin(\sqrt{3}t) + B_1\cos(\sqrt{3}t)$ and $u_2 = A_2\sin(t) + B_2\cos(t)$. Hence, on substitution, the general solution is

$\class{outbox}{\array{x_1(t) = A_1\sin(\sqrt{3}t) + B_1\cos(\sqrt{3}t) + A_2\sin(t) + B_2\cos(t)\;\;\;\;\;\; \\ \\ x_2(t) = -A_1\sin(\sqrt{3}t) - B_1\cos(\sqrt{3}t) + A_2\sin(t) + B_2\cos(t).}}$

A neater way of writing out the general solution is

$\class{outbox}{\pmatrix{x_1 \\ x_2} = \big[A_1\sin(\sqrt{3}t) + B_1\cos(\sqrt{3}t)\big]\pmatrix{\;\;\; 1 \\ -1} + \big [A_2\sin(t) + B_2\cos(t)\big]\pmatrix{1 \\ 1}}$.

Can you see how the general solution is connected to the eigenvalues and eigenvectors?

The general solution of a linear system $\ddot{\pmb{x}} = A \pmb{x}$ where the $n \times n$ matrix $A$ has $n$ distinct negative eigenvalues $-\omega_1^2$, $-\omega_2^2$, $-\omega_3^2 , \dots , -\omega_n^2$, with associated real eigenvectors $\pmb{x}_1, \pmb{x}_2, \pmb{x}_3, \dots , \pmb{x}_n$, is given by

$$\class{outbox}{\pmb{x}(t) = \sum_{i=1}^{n}\big[A_i\sin(\omega_it) + B_i\cos(\omega_it)\big]\pmb{x_i}}$$

Determine the particular solution of the coupled system

$\ddot{x}_1 = -2x_1 + x_2$

$\ddot{x}_2 = x_1 - 2x_2$

subject to the initial conditions $x_1(0) = 0$, $x_2(0) = 0$, $\dot{x}_1(0) = 1$, $\dot{x}_2(0) = 3$.

Solution

The general solution of the system was obtained in Example 21 to be

$x_1(t) = A_1\sin(\sqrt{3}t) + B_1\cos(\sqrt{3}t) + A_2\sin(t) + B_2\cos(t)$

$x_2(t) = -A_1\sin(\sqrt{3}t) - B_1\cos(\sqrt{3}t) + A_2\sin(t) + B_2\cos(t)$.

First apply the conditions, $x_1(0) = 0, x_2(0) = 0$:

$x_1(0) = 0 : A_1\sin(0) + B_1\cos(0) + A_2\sin(0) + B_2\cos(0) = 0 \Rightarrow B_1 + B_2 = 0$

$x_2(0) = 0 : -A_1\sin(0) - B_1\cos(0) + A_2\sin(0) + B_2\cos(0) = 0 \Rightarrow -B_1 + B_2 = 0$.

Solving these simultaneous equations gives $B_1 = B_2 = 0$.

Hence, with $B_1 = B_2 = 0$ the general solution becomes,

$x_1(t) = A_1\sin(\sqrt{3}t) + A_2\sin(t)$

$x_2(t) = -A_1\sin(\sqrt{3}t) + A_2\sin(t)$.

To apply the remaining conditions $\dot{x}_1(0) = 1$ and $\dot{x}_2(0) = 3$ we must first differentiate $x_1(t)$ and $x_2(t)$:

$\dot{x}_1(t) = \sqrt{3}A_1\cos(\sqrt{3}t) + A_2\cos(t) \Rightarrow \sqrt{3}A_1 + A_2 = 1$

$x_2(t) = - \sqrt{3}A_1\cos(\sqrt{3}t) + A_2\cos(t) \Rightarrow -\sqrt{3}A_1 + A_2 = 3$.

Solving these simultaneous equations gives, $A_1 = {\Large{\frac{1}{\sqrt{3}}}}$ and $A_2 = 2$.

We now have values $A_1 = {\Large{\frac{1}{\sqrt{3}}}}$, $A_2 = 2$, $B_1 = 0$ and $B_2 = 0$ which are substituted in the general solution to obtain the particular solution:

$\class{outbox}{\array{x_1(t) = {\Large{\frac{1}{\sqrt{3}}}}\sin(\sqrt{3}t) + 2\sin(t) \\ \\ x_2(t) = -{\Large{\frac{1}{\sqrt{3}}}}\sin(\sqrt{3}t) + 2\sin(t).}}$

A neater way of writing out the particular solution is,

$\class{outbox}{\pmatrix{x_1 \\ x_2} = {\Large{\frac{1}{\sqrt{3}}}}\sin(\sqrt{3}t)\pmatrix{\;\;\;1 \\ -1} + 2\sin(t)\pmatrix{1 \\ 1}.}$

A system of two coupled linear ordinary differential equations is given by,

$\ddot{x}_1 = -7x_1 + 12x_2$

$\ddot{x}_2 = -4x_1 + 7x_2.$

(i). express the system in matrix form, $\dot{\pmb{x}} = A \pmb{x}$.

(ii). determine the eigenvalues and eigenvectors of the coefficient matrix, $A$.

(iii). write down an invertible matrix $P$ such that $P^{-1}AP = D$ where $D$ is a diagonal matrix. Write down the matrix $D$.

(iv). determine the general solution of the system.

Solution

(i). In matrix form the system is,

$\pmatrix{\ddot{x}_1 \\ \ddot{x}_2} = \pmatrix{-7 & 12 \\ -4 & \;\;7}\pmatrix{x_1 \\ x_2}$.

(ii). Exercise: Show that the matrix $A$ has the following eigenvalues and eigenvectors,

$\lambda_1 = -1, \;\;\;\;\;\pmb{x}_1 = \pmatrix{2 \\ 1}; \;\;\;\;\;\;\;\;\lambda_2 = 1, \;\;\;\;\;\;\pmb{x}_2 = \pmatrix{3 \\ 2}$.

(iii). $\;\;P = \pmatrix{2 & 3 \\ 1 & 2}, \;\;\;\;\;\;\;\;\; P^{-1}AP = D = \pmatrix{-1 & 0 \\ \;\;\;0 & 1}$.

(iv). Determine the general solution of the system.

Let $\pmb{x} = P\pmb{u}$. Differentiating twice gives, $\ddot{\pmb{x}} = P\ddot{\pmb{u}}$ (since $P$ is a constant matrix).

The coupled system $\ddot{\pmb{x}} = A\pmb{x}$ can then be written as

$P \ddot{\pmb{u}} = AP\pmb{u} \;\;\;\;\;\; \Rightarrow \ddot{\pmb{u}} = P^{-1}AP \pmb{u} \;\;\;\;\;\; \ddot{\pmb{u}} = D \pmb{u}$.

Hence,

$\pmatrix{\ddot{u}_1 \\ \ddot{u}_2} = \pmatrix{-1 & 0 \\ \;\;\; 0 & 1}\pmatrix{u_1 \\ u_2}$

Expanding this expression gives the two uncoupled second order differential equations,

$\ddot{u}_1 + u_1 = 0$    and    $\ddot{u}_2 - u_2 = 0$.

Solving these equations gives,

$u_1(t) = A_1\sin(t) + B_1\cos(t)$ and

$u_2(t) = A_2e^t + B_2e^{-t}$

where $A_1$, $B_1$, $A_2$ and $B_2$ are arbitrary constants.

The solution in terms of the original variables, $x_1(t)$ and $x_2(t)$ can now be found.

Since $\pmb{x} = P \pmb{u}$ we have, $\pmatrix{x_1 \\ x_2} = \pmatrix{2 & 3 \\ 1 & 2}\pmatrix{u_1 \\ u_2}$.

Expanding this matrix equation gives

$x_1 = 2u_1 + 3u_2$

$x_2 = \;\;u_1 + 2u_2$.

From earlier, $u_1(t) = A_1\sin(t) + B_1\cos(t)$ and $u_2(t) = A_2e^t + B_2e^{-t}$. Hence, on substitution, the general solution is

$\class{outbox}{\array{x_1(t) = 2A_1\sin(t) + 2B_1\cos(t) + 3A_2e^t + 3B_2e^{-t} \\ \\ x_2(t) = A_1\sin(t) + B_1\cos(t) + 2A_2e^t + 2B_2e^{-t}.}}$