Eigenvalues and Eigenvectors

Yes, with eigenvalue $4$

$A(1, 1) = (3 + 1,\ 2 + 2) = (4, 4) = 4(1, 1)$.

$4$ and $1$

$(3 - \lambda)(2 - \lambda) - 2 = \lambda^2 - 5\lambda + 4 = (\lambda - 4)(\lambda - 1)$.

$(1, -2)$, or any nonzero multiple

$A - I = \begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix}$ gives $2x_1 + x_2 = 0$, so $x_2 = -2x_1$. Check: $A(1, -2) = (1, -2)$.

$5$ and $-2$

A diagonal matrix is triangular, so its eigenvalues are its diagonal entries, with eigenvectors $\mathbf{e}_1$ and $\mathbf{e}_2$.

$(2, 6, 3)$

$A - 5I = \begin{bmatrix} -3 & 1 & 0\\ 0 & -2 & 4\\ 0 & 0 & 0 \end{bmatrix}$. With $x_3 = 3$, the second row gives $x_2 = 6$ and the first gives $x_1 = 2$. Check: $A(2, 6, 3) = (10, 30, 15)$.

Its characteristic equation is $\lambda^2 + 1 = 0$.

$\det\begin{bmatrix} -\lambda & -1\\ 1 & -\lambda \end{bmatrix} = \lambda^2 + 1$, which is positive for every real $\lambda$.

$9\mathbf{v}$ and $3^{10}\mathbf{v}$

$A^2\mathbf{v} = A(3\mathbf{v}) = 3A\mathbf{v} = 9\mathbf{v}$, and each further factor of $A$ multiplies by another $3$.

$\tfrac{1}{\lambda}$, with the same eigenvector

Apply $A^{-1}$ to both sides of $A\mathbf{v} = \lambda\mathbf{v}$: $\mathbf{v} = \lambda A^{-1}\mathbf{v}$. Since $A$ is invertible, $\lambda \ne 0$, so $A^{-1}\mathbf{v} = \tfrac{1}{\lambda}\mathbf{v}$.

No

An eigenvalue of $0$ means $A\mathbf{v} = \mathbf{0}$ for some nonzero $\mathbf{v}$, so the null space is not trivial. Equivalently, $\det A$ is the product of the eigenvalues, which is $0$.

Do the reported eigenvalues add to the diagonal sum of the original?

Row operations change eigenvalues. The true ones are $\tfrac{5 \pm \sqrt{33}}{2} \approx 5.37$ and $-0.37$.

$\det(A - \lambda I) = (1 - \lambda)(4 - \lambda) - 6 = \lambda^2 - 5\lambda - 2$, whose roots are $\tfrac{5 \pm \sqrt{33}}{2}$. The student’s values add to $-1$, not to the diagonal sum $5$, which already shows they are wrong.