Find an eigenvector of M corresponding to eigenvalue k

Question

It is given that $M - kI$= \begin{bmatrix}a_{11}&a_{12}\\1&\pi\end{bmatrix}

And I know that for $M$ (which is 2 x 2 matrix), there are two unique eigenvalues such that one of them is $k$.

FYI, I don't have any other info on $a_{11}$ and $a_{12}$. Thanks!

Answer 1

Hint: Since $k$ is an eigenvalue of $M$, $M - kI$ is singular. So that, we have $a_{12} = a_{11}\pi$ and $a_{11}$ is arbitrary. Now, can you find an eigenvector?

Answer 2

Hint If $k$ is an eigenvalue, the rows of $M-kI$ are linearly dependent. Since $M-kI$ is a $2 \times 2$ matrix, this means they are proportional.

Alternate solution Let $v\begin{bmatrix} a \\ b \end{bmatrix}$ be the corresponding eigenvector. Then $$\begin{bmatrix}a_{11}&a_{12}\\1& \pi \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}=(M-kI)v=Mv-kV=0$$

Therefore, $$a_{11}a+a_{12}b=0 \\ a + \pi b=0 $$

Just solve the second equation.