Question

Check whether the matrices A and B are diagonalisable. Diagonalise those matrices
which are diagonalisable. (11)
i) A =


−2 −5 −1
3 6 1
−2 −3 1

 ii) B =


−1 −3 0
2 4 0
−1 −1 2

.

Answer 1

Answer on Question #79869 – Math – Linear Algebra

Question

Check whether the matrices $A$ and $B$ are diagonalizable. Diagonalize those matrices which are diagonalizable. (11)

i)

$$A = \begin{pmatrix} -2 & -5 & -1 \\ 3 & 6 & 1 \\ -2 & -3 & 1 \end{pmatrix}$$

ii)

$$B = \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix}$$

Solution

i)

$$A = \begin{pmatrix} -2 & -5 & -1 \\ 3 & 6 & 1 \\ -2 & -3 & 1 \end{pmatrix}$$

Characteristic equation:

$$\begin{aligned} & -2 - \lambda - 5 - 1 \\ & 3 \quad 6 - \lambda \quad 1 \\ & -2 & -3 \quad 1 - \lambda \\ & = (-2 - \lambda)(\lambda^2 - 7\lambda + 9) + 25 - 15\lambda - 3 + 2\lambda = -\lambda^3 - 2\lambda^2 + 7\lambda^2 + 14\lambda - 9\lambda - 18 - 13\lambda + 22 = \\ & = -(\lambda^3 - 5\lambda^2 + 8\lambda - 4) = -(\lambda - 1)(\lambda - 2)^2 \end{aligned}$$

Eigenvalues are $\lambda_1 = 1, \lambda_2 = 2$. There is one eigenvector corresponding to $\lambda_1$. There can be one or two eigenvectors corresponding to $\lambda_2$. We need to define which of these cases holds now.

$$\begin{pmatrix} -4 & -5 & -1 \\ 3 & 4 & 1 \\ -2 & -3 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\begin{cases} -4x_1 - 5x_2 - x_3 = 0 \\ 3x_1 + 4x_2 + x_3 = 0 \\ -2x_1 - 3x_2 - x_3 = 0 \end{cases}$$ $$\begin{cases} x_3 = -4x_1 - 5x_2 \\ -x_1 - x_2 = 0 \\ 2x_1 + 2x_2 = 0 \end{cases}$$ $$\begin{cases} x_2 = -x_1 \\ x_3 = -4x_1 - 5x_2 \end{cases}$$

$(x_1, x_2, x_3) = (t, -t, t) = t(1, -1, 1)$ is linear space of dimension 1 on the vector $(1, -1, 1)$. Thus, $(1, -1, 1)$ is the one eigenvector corresponding to $\lambda_2 = 2$. So there will be two eigenvectors in total.

Since the number 2 of eigenvectors is less than dimension 3 of the matrix, then the matrix is not diagonalizable.

ii)

$$B = \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix}$$

Characteristic equation:

$$\begin{array}{l} \left| \begin{array}{ccc} -1 - \lambda & -3 & 0 \\ 2 & 4 - \lambda & 0 \\ -1 & -1 & 2 - \lambda \end{array} \right| = (-1 - \lambda)(4 - \lambda)(2 - \lambda) + 3 \cdot 2(2 - \lambda) = -\lambda^3 + 5\lambda^2 - 2\lambda - 8 + 12 - 6\lambda = \\ = -(\lambda^3 - 5\lambda^2 + 8\lambda - 4) = -(\lambda - 1)(\lambda - 2)^2 \end{array}$$

Eigenvalues are $\lambda_1 = 1, \lambda_2 = 2$. There is one eigenvector corresponding to $\lambda_1 = 1$. There can be one or two eigenvectors corresponding to $\lambda_2 = 2$. We need to define which of these cases holds now.

$$\begin{pmatrix} -3 & -3 & 0 \\ 2 & 2 & 0 \\ -1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\begin{cases} -3x_1 - 3x_2 = 0 \\ 2x_1 + 2x_2 = 0 \\ -x_1 - x_2 = 0 \end{cases}$$ $$\{x_2 = -x_1$$

$(x_1, x_2, x_3) = (t, -t, z) = t(1, -1, 0) + z(0, 0, 1)$ is linear space of dimension 2 on vectors $(1, -1, 0)$ and $(0, 0, 1)$.

The vectors $(1, -1, 0), (0, 0, 1)$ are two eigenvectors corresponding to $\lambda_2 = 2$. So there will be three eigenvectors in total, that's why the matrix is diagonalizable.

Find the remaining eigenvector corresponding to $\lambda_1 = 1$:

$$\begin{pmatrix} -2 & -3 & 0 \\ 2 & 3 & 0 \\ -1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$$ $$\begin{cases} -2x_1 - 3x_2 = 0 \\ 2x_1 + 3x_2 = 0 \\ -x_1 - x_2 + x_3 = 0 \end{cases}$$ $$\begin{cases} x_2 = -\dfrac{2}{3}x_1 \\ x_3 = x_1 + x_2 \end{cases}$$ $$\begin{cases} x_2 = -\dfrac{2}{3}x_1 \\ x_3 = \dfrac{1}{3}x_1 \end{cases}$$

The solution, the eigenvector is $(3, -2, 1)$.

Diagonalization of the matrix $B$.

$$B = \begin{pmatrix} 3 & 1 & 0 \\ -2 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 & 0 \\ -2 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}^{-1}$$

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