Questions: From the following, select all in which v is an eigenvector of A. There may be more than one correct answer. A. A=[9 8 -8 -20; 6 7 -10 -10; 4 4 -5 -8; 4 4 -2 -11], v=[5; 2; 2; 3] B. A=[-1 4 0 4; 0 1 0 3; 0 2 -1 2; 0 0 0 -2], v=[0; 1; -1; -1] C. A=[1 2 2 2; 2 1 -2 -2; 2 2 1 4; 0 0 0 -3], v=[0; 1; 1; -1] D. None of the above.

From the following, select all in which v is an eigenvector of A. There may be more than one correct answer.
A. A=[9 8 -8 -20; 6 7 -10 -10; 4 4 -5 -8; 4 4 -2 -11], v=[5; 2; 2; 3]
B. A=[-1 4 0 4; 0 1 0 3; 0 2 -1 2; 0 0 0 -2], v=[0; 1; -1; -1]
C. A=[1 2 2 2; 2 1 -2 -2; 2 2 1 4; 0 0 0 -3], v=[0; 1; 1; -1]
D. None of the above.

Transcript text: From the following, select all in which $\mathbf{v}$ is an eigenvector of $A$. There may be more than one correct answer. A. $A=\left[\begin{array}{cccc}9 & 8 & -8 & -20 \\ 6 & 7 & -10 & -10 \\ 4 & 4 & -5 & -8 \\ 4 & 4 & -2 & -11\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l}5 \\ 2 \\ 2 \\ 3\end{array}\right]$ B. $A=\left[\begin{array}{cccc}-1 & 4 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 2 & -1 & 2 \\ 0 & 0 & 0 & -2\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{c}0 \\ 1 \\ -1 \\ -1\end{array}\right]$ C. $A=\left[\begin{array}{cccc}1 & 2 & 2 & 2 \\ 2 & 1 & -2 & -2 \\ 2 & 2 & 1 & 4 \\ 0 & 0 & 0 & -3\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{c}0 \\ 1 \\ 1 \\ -1\end{array}\right]$ D. None of the above.

Solution

Solution Steps

To determine if a vector $\mathbf{v}$ is an eigenvector of a matrix $A$, we need to check if there exists a scalar $\lambda$ such that $A\mathbf{v} = \lambda \mathbf{v}$. This involves computing the product $A\mathbf{v}$ and checking if the result is a scalar multiple of $\mathbf{v}$.

Step 1: Check Eigenvector for A1 and v1

To determine if $\mathbf{v}_1 = \begin{bmatrix} 5 \\ 2 \\ 2 \\ 3 \end{bmatrix}$ is an eigenvector of the matrix $A_1 = \begin{bmatrix} 9 & 8 & -8 & -20 \\ 6 & 7 & -10 & -10 \\ 4 & 4 & -5 & -8 \\ 4 & 4 & -2 & -11 \end{bmatrix}$, we compute $A_1 \mathbf{v}_1$:

\[ A_1 \mathbf{v}_1 = \begin{bmatrix} 9 & 8 & -8 & -20 \\ 6 & 7 & -10 & -10 \\ 4 & 4 & -5 & -8 \\ 4 & 4 & -2 & -11 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \]

Since $A_1 \mathbf{v}_1 = 1 \cdot \mathbf{v}_1$, $\mathbf{v}_1$ is indeed an eigenvector of $A_1$ with eigenvalue $\lambda = 1$.

Step 2: Check Eigenvector for A2 and v2

Next, we check if $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \\ -1 \end{bmatrix}$ is an eigenvector of the matrix $A_2 = \begin{bmatrix} -1 & 4 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 2 & -1 & 2 \\ 0 & 0 & 0 & -2 \end{bmatrix}$:

\[ A_2 \mathbf{v}_2 = \begin{bmatrix} -1 & 4 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 2 & -1 & 2 \\ 0 & 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix} \]

Since $A_2 \mathbf{v}_2$ does not equal $\lambda \mathbf{v}_2$ for any scalar $\lambda$, $\mathbf{v}_2$ is not an eigenvector of $A_2$.

Step 3: Check Eigenvector for A3 and v3

Finally, we check if $\mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ -1 \end{bmatrix}$ is an eigenvector of the matrix $A_3 = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 1 & -2 & -2 \\ 2 & 2 & 1 & 4 \\ 0 & 0 & 0 & -3 \end{bmatrix}$:

\[ A_3 \mathbf{v}_3 = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 1 & -2 & -2 \\ 2 & 2 & 1 & 4 \\ 0 & 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ 0 \\ 3 \end{bmatrix} \]

Since $A_3 \mathbf{v}_3$ does not equal $\lambda \mathbf{v}_3$ for any scalar $\lambda$, $\mathbf{v}_3$ is not an eigenvector of $A_3$.