A problem on eigenvalues of a matrix.

Question

The imaginary parts of the eigenvalues of the matrix A.

$$ \begin{bmatrix} 3 & 2 & 5 \\ 2 & -3 & 6 \\ 0 & 0 & -3\\ \end{bmatrix} $$ are

(A)$0,0,0$

(B)$1,-1,0$

(C)$2,-2,0$

(D)$3,-3,0$

My input: In this I tried to calculate eigenvalues like we do in normal subtracting $\lambda$ from diagonal entries and solving cubic equation after taking determinant of matrix A. But this question came in $1$ marks . I think there is some fact which is being used here. Please tell me. Thank you!

Answer 1

The eigenvalue of $-3$ (which is real) can be read off directly.

Hence it suffices to study the eigenvalue of the upper left $2$ by $2$ matrix. Of which it is symmetric, which means the eigenvalue is real.

Edit:

Notice that the standard way of computing eigenvalues of matrix $A$ is to compute the charactheristic polynomial, $\det(A-\lambda I ) = 0$. Now, notice that $A$ is a block-triangular matrix, $A = \begin{bmatrix} B & C \\ 0 & D \end{bmatrix}$.

We have \begin{align} \det(A-\lambda I)&= \det \begin{pmatrix} B-\lambda I & C \\ 0 & D-\lambda I \end{pmatrix}\\ &= \det(B-\lambda I) \det(D - \lambda I) \end{align}

here $D$ is $-3$ and hence $\lambda = -3$ is a solution. In fact for this particular example, since $A$ is a $3$ by $3$ matrix and $D$ is a scalar, just perform Laplace expansion along the last row to obtain the result.

Answer 2

The answer is (A).

You don't really have to do any calculation. But first observe that $(0, 0, 1)$ is an eigenvector with corresponding eigenvalue $-3$.

For the other two eigenvalues, all you have to do is observe that the corresponding eigenvectors have the form $a= (a_1, a_2, 0)$ and $b = (b_1, b_2, 0)$. This means the eigenvector corresponds to the eigenvector of the upper left symmetric square matrix. Since it's symmetric then the eigenvalues must be real.

Answer 3

Take the last row to compute the determinant: $(-3-\lambda)((-3-\lambda) (3-\lambda)-4)=0 \implies (-3-\lambda) =0 \text { or }(-3-\lambda)(3-\lambda) -4=0 \implies \lambda =-3 \text { or } \lambda ^2-13=0 \implies \lambda =-3 \text { or } \lambda =\pm \sqrt {13}$.

So there are three real eigenvalues...

Therefore the answer is A...