Question

QUESTION 3 Let A = [ 4 0 1 −2 1 0 −2 0 1 ](its 3by3 matrix).

3.1. Find the eigenvalues of A.

3.2. For each eigenvalue of A find the corresponding eigenvectors.

3.3. Is A diagonalizable? Explain. If it is diagonalizable, then find P such that 𝐷 = 𝑃 −1𝐴𝑃 is a diagonal matrix. Do not compute 𝑃 −1(show all step!) . 

Answer 1

$A = \begin{bmatrix} 4 & 0 & 1 \\ −2 & 1 & 0 \\ −2 & 0 & 1 \end{bmatrix}$

3.1. The characteristic polynomial of this matrix is

$P_{A}(\lambda)=\det (A-\lambda I) =\begin{vmatrix} 4-\lambda & 0 & 1 \\ −2 & 1-\lambda & 0 \\ −2 & 0 & 1-\lambda \end{vmatrix}=$

$(1-\lambda)\begin{vmatrix} 4-\lambda & 1 \\−2 & 1-\lambda \end{vmatrix}=(1-\lambda)(\lambda^2-5\lambda+6)=$

$=(1-\lambda)(\lambda-2)(\lambda-3)$

The eigenvalues of $A$ are the roots of the polynomial $P_A(\lambda)$:

$\lambda_1=1$, $\lambda_2=2$, $\lambda_3=3$.

3.2. The eigenvector $X_1$ corresponding to the eigenvalue $\lambda_1=1$ is the solution to the equation $(A-\lambda_1I)X=0$.

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

which implies $x_1=0$, $x_3=0$, and hence, $X_1=\begin{bmatrix} 0 \\1 \\0 \end{bmatrix}$.

The eigenvector $X_2$ corresponding to the eigenvalue $\lambda_2=2$ is the solution to the equation $(A-\lambda_2I)X=0$.

$\begin{bmatrix} 2 & 0 & 1 \\ −2 & -1 & 0 \\ −2 & 0 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 2x_1+x_3 \\-2 x_1-x_2 \\ -2x_1-x_3 \end{bmatrix}=\begin{bmatrix} 0 \\0 \\0 \end{bmatrix}$

which implies $x_1=t$, $x_2=-2t$, $x_3=-2t$, and hence, $X_2=\begin{bmatrix} 1 \\-2 \\-2 \end{bmatrix}$.

The eigenvector $X_3$ corresponding to the eigenvalue $\lambda_3=3$ is the solution to the equation $(A-\lambda_3I)X=0$.

$\begin{bmatrix} 1 & 0 & 1 \\ −2 & -2 & 0 \\ −2 & 0 & -2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} x_1+x_3 \\-2 x_1-2x_2 \\ -2x_1-2x_3 \end{bmatrix}=\begin{bmatrix} 0 \\0 \\0 \end{bmatrix}$

which implies $x_1=t$, $x_2=-t$, $x_3=-t$, and hence, $X_3=\begin{bmatrix} 1 \\-1 \\-1 \end{bmatrix}$.

3.3. Consider the matrix $P=(X_1 | X_2|X_3)$ (i.e. $X_1, X_2 ,X_3$ are the columns of this matrix).

Then

$AP=(AX_1|AX_2|AX_3)=(\lambda_1X_1|\lambda_2X_2|\lambda_3X_3)=PD$, where

$D=\begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ is a diagonal matrix.

Therefore, $D=P^{-1}AP$, which means that the matrix $A$ is diagonalizable.

Answer. The eigenvalues of $A$ are $\lambda_1=1$, $\lambda_2=2$, $\lambda_3=3$.

The eigenvalues of $A$ are $X_1=\begin{bmatrix} 0 \\1 \\0 \end{bmatrix}$ , $X_2=\begin{bmatrix} 1 \\-2 \\-2 \end{bmatrix}$, $X_3=\begin{bmatrix} 1 \\-1 \\-1 \end{bmatrix}$.

The matrix $P$ equals to $(X_1 | X_2|X_3)=\begin{bmatrix} 0 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & -2 & -1 \end{bmatrix}$.