Question

How to obtain the eigenvalues and eigenvectors of the matrix: M=
[2 3 0
3 2 0
0 0 1]

Answer 1

Answer on Question #84642 – Math – Linear Algebra

Question

How to obtain the eigenvalues and eigenvectors of the matrix: $M = \begin{pmatrix} 2 & 3 & 0 \\ 3 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Solution

In such problems, we first find the eigenvalues of the matrix.

Finding eigenvalues

To do this, we find the values of $\lambda$ which satisfy the characteristic equation of the matrix $M$, namely those values of $\lambda$ for which

$\operatorname{det}(M - \lambda I) = 0$

where $I$ is the $3 \times 3$ identity matrix

$$I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Form the matrix $M - \lambda I$:

$$M - \lambda I = \begin{pmatrix} 2 & 3 & 0 \\ 3 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} = \begin{pmatrix} 2 - \lambda & 3 & 0 \\ 3 & 2 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{pmatrix}$$

Calculate $\operatorname{det}(M - \lambda I)$

$$\operatorname{det}(M - \lambda I) = \begin{vmatrix} 2 - \lambda & 3 & 0 \\ 3 & 2 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{vmatrix} = (1 - \lambda) \begin{vmatrix} 2 - \lambda & 3 \\ 3 & 2 - \lambda \end{vmatrix} = (1 - \lambda) ((2 - \lambda)^2 - 3^2) = (1 - \lambda) ((2 - \lambda)^2 - 3^2) = (1 - \lambda) ((2 - \lambda)^2 - 3^2) = (1 - \lambda) ((2 - \lambda) + 3) = - (1 - \lambda) (1 + \lambda) (5 - \lambda)$$

The solutions of the equation $\operatorname{det}(M - \lambda I) = - (1 - \lambda) (1 + \lambda) (5 - \lambda) = 0$ are

$$\lambda_1 = -1, \quad \lambda_2 = 1, \quad \lambda_1 = 5$$

Finding eigenvectors

We can find the eigenvectors by Gaussian Elimination.

STEP 1: For each eigenvalue $\lambda$, we have $(M - \lambda I)X = 0$ where $X$ is the eigenvector associated with eigenvalue $\lambda$.

STEP 2: Find $X$ by Gaussian elimination. That is, convert the augmented matrix.

$$(M - \lambda I, 0)$$

to row echelon form and solve the resulting linear system by back substitution.

Case 1: $\lambda = -1$. We have to find vectors $X$ which satisfy $(M - \lambda I)X = 0$.

First, form the matrix $M - (-1) \cdot I = M + I$

$$M + I = \left( \begin{array}{ccc} 2 + 1 & 3 & 0 \\ 3 & 2 + 1 & 0 \\ 0 & 0 & 1 + 1 \end{array} \right) = \left( \begin{array}{ccc} 3 & 3 & 0 \\ 3 & 3 & 0 \\ 0 & 0 & 2 \end{array} \right)$$

Construct the augmented matrix $(M - \lambda I, 0)$ and convert it to row echelon form

$$\left( \begin{array}{ccc} 3 & 3 & 0 \\ 3 & 3 & 0 \\ 0 & 0 & 2 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array}$$

Rewriting this augmented matrix as a linear system gives

$$\left\{ \begin{array}{l} x _ {1} + x _ {2} = 0, \\ x _ {3} = 0 \end{array} \right.$$

So the form of eigenvector $X_{1}$ is given by:

$$X _ {1} = \left( \begin{array}{c} x _ {1} \\ - x _ {1} \end{array} \right) = x _ {1} \left( \begin{array}{c} 1 \\ - 1 \end{array} \right) = C _ {1} \left( \begin{array}{c} 1 \\ - 1 \end{array} \right)$$

for any real number $C_1 \neq 0$

Case 2: $\lambda = 1$

$$M - I = \left( \begin{array}{ccc} 1 & 3 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

Construct the augmented matrix $(M - \lambda I, 0)$ and convert it to row echelon form

$$\left( \begin{array}{ccc} 1 & 3 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 3 & 0 \\ 0 & - 8 & 0 \\ 0 & 0 & 0 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \sim \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right| \sim \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)$$

Rewriting this augmented matrix as a linear system gives

$$\left\{ \begin{array}{l} x _ {1} = 0, \\ x _ {2} = 0 \\ x _ {3} = x _ {3} \end{array} \right.$$

So the form of eigenvector $X_{2}$ is given by:

$$X _ {2} = \left( \begin{array}{c} 0 \\ 0 \\ x _ {3} \end{array} \right) = x _ {3} \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) = C _ {2} \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right)$$

for any real number $C_2 \neq 0$.

Case 3: $\lambda = 5$.

$$M - 5I = \begin{pmatrix} 2 - 5 & 3 & 0 \\ 3 & 2 - 5 & 0 \\ 0 & 0 & 1 - 5 \end{pmatrix} = \begin{pmatrix} -3 & 3 & 0 \\ 3 & -3 & 0 \\ 0 & 0 & -4 \end{pmatrix}$$

Construct the augmented matrix $(M - \lambda I, 0)$ and convert it to row echelon form

$$\begin{pmatrix} -3 & 3 & 0 & 0 \\ 3 & -3 & 0 & 0 \\ 0 & 0 & -4 & 0 \end{pmatrix} \sim \begin{pmatrix} -3 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$

Rewriting this augmented matrix as a linear system gives

$$\begin{cases} x_1 - x_2 = 0 \\ x_3 = 0 \end{cases} \stackrel{=}{=} \begin{cases} x_1 = x_2 \\ x_3 = 0 \end{cases}$$

So the form of eigenvector $X_3$ is given by:

$$X_3 = \begin{pmatrix} x_1 \\ x_1 \\ 0 \end{pmatrix} = x_1 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = C_3 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$$

for any real number $C_3 \neq 0$.

Answer: Eigenvalues of $M$ are $\lambda_1 = -1$, $\lambda_2 = 1$, $\lambda_1 = 5$, eigenvectors of $M$ are $X_1 = C_1 \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$, $X_2 = C_2 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, $X_3 = C_3 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, where $C_1 \neq 0, C_2 \neq 0, C_3 \neq 0$.

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