Let $X$ and $\lambda$ respectively be the eigenvectors and eigenvalues of $A.$ i.e., $AX = \lambda X.$
$$\begin{aligned} AX- \lambda X & = O \\ AX - \lambda I X &= O \\ (A- \lambda I ) X &= O \end{aligned}$$
From here, we suppose that $X$ is not a zero vector. Then the matrix $A - \lambda I$ does not have an inverse matrix, which indicates that the determinant of $A- \lambda I$ is zero as follows:
$\text{det}(A - \lambda I) = 0$
$\implies \begin{vmatrix}
6 - \lambda & 5\\
5 & 6 - \lambda
\end{vmatrix} = 0$
$\implies (6-\lambda)^{2} - 25 = 0$
$\implies 36 + \lambda^{2} - 12 \lambda - 25 = 0$
$\implies \lambda^{2} - 12 \lambda + 11 = 0$
$\implies \lambda = 1,11$
Now let's find the eigenvectors.
First, when $\lambda = 1 ,$ the eigenvector $X = \begin{pmatrix} X_1 \\ X_2 \end{pmatrix}$ can be obtained as follows:
$(A - I)X = O$
$\implies \begin{pmatrix}
5 & 5 \\
5 & 5
\end{pmatrix} \cdot \begin{pmatrix} X_{1} \\ X_{2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
Perform the operation $R_{2} \rightarrow R_{2} - R_{1},$ then we get
$\implies \begin{pmatrix}
5 & 5 \\
0 & 0
\end{pmatrix} \cdot \begin{pmatrix} X_{1} \\ X_{2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
- Rank of the coefficient matrix $r = 1$
- Number of variables (unknowns) $n = 2$
Here, $r <n \implies 1 < 2$
Infinite number of solutions possible, and $n-r = 2 - 1 = 1$ variable (unknown) will be given some value and others will be evaluated accordingly.
Let $X_{2} = k,$ then $5X_{1} + 5X_{2} = 0$
$\implies X_{1} + X_{2} = 0$
$\implies X_{1} + k = 0$
$\implies X_{1} = -k$
$\therefore$ The eigen vector correspong to the eigen value $\lambda = 1$ is $ \begin{pmatrix} X_{1} \\ X_{2} \end{pmatrix} = \begin{pmatrix} -k \\ k \end{pmatrix} = k \begin{pmatrix} -1 \\ 1 \end{pmatrix}$
Second, when $\lambda = 11,$ the eigenvector $Y = \begin{pmatrix} Y_1 \\ Y_2 \end{pmatrix}$ can be obtained as follows:
$(A - 11I)X = O$
$\implies \begin{pmatrix}
-5 & 5 \\
5 & -5
\end{pmatrix} \cdot \begin{pmatrix} Y_{1} \\ Y_{2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
Perform the operation $R_{2} \rightarrow R_{2} + R_{1},$ then we get
$\implies \begin{pmatrix}
-5 & 5 \\
0 & 0
\end{pmatrix} \cdot \begin{pmatrix} Y_{1} \\ Y_{2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
- Rank of the coefficient matrix $r = 1$
- Number of variables (unknowns) $n = 2$
Here, $r <n \implies 1 < 2$
Infinite number of solutions possible, and $n-r = 2 - 1 = 1$ variable (unknown) will be given some value and others will be evaluated accordingly.
Let $X_{2} = k,$ then $-5X_{1} + 5X_{2} = 0$
$\implies -X_{1} + X_{2} = 0$
$\implies -X_{1} + k = 0$
$\implies X_{1} = k$
$\therefore$ The eigen vector correspong to the eigen value $\lambda = 11$ is $ \begin{pmatrix} Y_{1} \\ Y_{2} \end{pmatrix} = \begin{pmatrix} k \\ k \end{pmatrix} = k \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
So, the correct answer is $A;C.$
$\textbf{PS:}$ If the eigenvalues of A are distinct, it turns out that the eigenvectors are linearly independent; but, if any of the eigenvalues are repeated, further investigation may be necessary.
