For which value of $k$ is the following matrix diagonalizable?

Question

\begin{pmatrix}7&k\\ \:0&7\end{pmatrix}

I could not figure out how to derive the eigenvalues and eigenvectors of the matrix above because of the letter $k$. How am I supposed to deal with a value in terms of $k$? And how would I be able to find out if the matrix is diagonalizable or not through that?

Answer 1

Let $\mathcal A$ be the matrix \begin{pmatrix}7&k\\ \:0&7\end{pmatrix}

The characteristic polynomial of $\mathcal A$ is $p(\lambda)=(7-\lambda)^2$.

Observe that $ (\mathcal A - 7\mathcal Id)(x,y) = (0,0) \iff ky=0$. Clearly if $k=0$ $\mathcal A$ is diagonalizable. If $k \ne 0$ then $ky = 0 \iff y=0$. What can you conclude?

Answer 2

When $k \neq 0:$

A column vector that is not sent to zero by $$ \left( \begin{array}{cc} 0 & k \\ 0 & 0 \end{array} \right) $$ is $$ \left( \begin{array}{c} 0 \\ 1 \end{array} \right), $$ and $$ \left( \begin{array}{cc} 0 & k \\ 0 & 0 \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} k \\ 0 \end{array} \right) $$ Putting the columns in reverse order, we get $$ P = \left( \begin{array}{cc} k & 0 \\ 0 & 1 \end{array} \right) $$ with $$ P^{-1} = \left( \begin{array}{cc} \frac{1}{k} & 0 \\ 0 & 1 \end{array} \right) $$ after which $$ \left( \begin{array}{cc} \frac{1}{k} & 0 \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} 7 & k \\ 0 & 7 \end{array} \right) \left( \begin{array}{cc} k & 0 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} 7 & 1 \\ 0 & 7 \end{array} \right) $$

The point being that, as soon as $k \neq 0,$ the specific value of $k$ is not that important. The last matrix is the Jordan form of your original.