Which of the following is/are the eigenvector(s) for the matrix given below? $$\begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix}$$

A non-zero vector $v$ is an eigenvector of matrix $A$ if there exists a scalar $\lambda$ such that $Av = \lambda v$.

Let $A = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix}$.

Let us test each option:

• Option A:$v = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$

  $$Av = \begin{pmatrix} -9(-1) - 6(1) - 2(0) - 4(1) \\ -8(-1) - 6(1) - 3(0) - 1(1) \\ 20(-1) + 15(1) + 8(0) + 5(1) \\ 32(-1) + 21(1) + 7(0) + 12(1) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = 1 \cdot v$$

  Since $Av = 1 \cdot v$, this is an eigenvector with eigenvalue $\lambda = 1$.

• Option B:$v = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}$

  $$Av = \begin{pmatrix} -9(1) - 2(-1) \\ -8(1) - 3(-1) \\ 20(1) + 8(-1) \\ 32(1) + 7(-1) \end{pmatrix} = \begin{pmatrix} -7 \\ -5 \\ 12 \\ 25 \end{pmatrix} \neq \lambda v$$

  This is not a scalar multiple of $v$.

• Option C:$v = \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix}$

  $$Av = \begin{pmatrix} -9(-1) - 2(2) - 4(2) \\ -8(-1) - 3(2) - 1(2) \\ 20(-1) + 8(2) + 5(2) \\ 32(-1) + 7(2) + 12(2) \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 6 \\ 6 \end{pmatrix} = 3 \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix} = 3v$$

  Since $Av = 3v$, this is an eigenvector with eigenvalue $\lambda = 3$.

• Option D:$v = \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix}$

  $$Av = \begin{pmatrix} -6(1) - 2(-3) \\ -6(1) - 3(-3) \\ 15(1) + 8(-3) \\ 21(1) + 7(-3) \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -9 \\ 0 \end{pmatrix} = 3 \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix} = 3v$$

  Since $Av = 3v$, this is an eigenvector with eigenvalue $\lambda = 3$.

Hence, the correct options are A, C, and D.