For which value of $k$ is the following matrix diagonalizable? \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?
 A: 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?
A: 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.
