Solution: Let $B = A - 10000I_6$, where $I_6$ is the identity matrix of order $6$. That is, \begin{align*} B = \begin{bmatrix} 1 & 3 & 5 & 7 & 9 & 11 \\ 1 & 3 & 5 & 7 & 9 & 11 \\ 1 & 3 & 5 & 7 & 9 & 11 \\ 1 & 3 & 5 & 7 & 9 & 11 \\ 1 & 3 & 5 & 7 & 9 & 11 \\ 1 & 3 & 5 & 7 & 9 & 11 \end{bmatrix}. \end{align*} It is clear that the matrix $B$ is a rank-1 matrix since all rows are identical. Therefore, $B$ has one non-zero eigenvalue and the rest are zero.

To find the non-zero eigenvalue, we can use the fact that the sum of the eigenvalues of a matrix is equal to the trace of the matrix. The trace of $B$ is \[ \operatorname{tr}(B) = 1 + 3 + 5 + 7 + 9 + 11 = 36. \] Thus, the non-zero eigenvalue of $B$ is 36, and the remaining eigenvalues are 0. Now we know that the eigenvalues of $A = B + 10000I_6$, are $\{\lambda + 10000: \lambda \text{ is an eigenvalue of $B$ } \}$ (see problem of the day (26-01-2025)). Thus, the eigenvalues of $A$ will be \[ 10036, 10000, 10000, 10000, 10000, 10000. \] We know that the eigenvectors of $A$ are the same as the eigenvectors of $B$ (see problem of the day (26-01-2025)).

Observe that \[ B \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 36 \\ 36 \\ 36 \\ 36 \\ 36 \\ 36 \end{bmatrix}. \] Thus, corresponding to the eigenvalue 36, the eigenvector is \[ \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}. \] Hence, the eigenvector of $A$ corresponding to the eigenvalue 10036 is \[ \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}. \]

Now let us find the eigenvectors of $B$ corresponding to the eigenvalue $0$. We need to solve $B \mathbf{x} = 0$, which is same as finding the null space of $B$. Applying the elementary row operations on $B$ gives \[ \begin{bmatrix} 1 & 3 & 5 & 7 & 9 & 11 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}. \] Thus, the null space of $B$ is spanned by the vector (which are also eigenvectors of $B$ corresponding to the eigenvalue 0) \[ \left\{ \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -5 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -7 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -9 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -11 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} . \] Therefore, the eigenvectors of $A$ corresponding to the eigenvalue $10000$ are \[ \left\{ \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -5 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -7 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -9 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -11 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} . \]