# 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?

• You should be able to find the eigenvalues, k will not be involved for this. Then, compare algebraic multiplicity with geometric multiplicity. – Joppy Oct 26 '17 at 0:36
• I see. But how will I be able to find the value of k by comparing the algebraic and geometric multiplicities? – sktsasus Oct 26 '17 at 0:39
• First start with the eigenvalues of this matrix. – imranfat Oct 26 '17 at 0:42
• Just try and tell us what you get. – Mr. T Oct 26 '17 at 0:42
• OK. So I believe the eigenvalue is 7 with algebraic multiplicity of 2. Is this correct? If so, how can I find the geometric multiplicity and value of k? Thank you! – sktsasus Oct 26 '17 at 0:53

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?

• Oh I see. So the matrix is only diagonalizable for k = 0? – sktsasus Oct 26 '17 at 0:59
• How you conclude that? – Mr. T Oct 26 '17 at 1:01
• Because if k = 0, then y does not equal 0? And for the matrix to be diagonalizable, y must not be 0? – sktsasus Oct 26 '17 at 1:03
• @sktsasus What does the constraint $y=0$ tell you about the geometric multiplicity of the eigenvalue $7$? – Bungo Oct 26 '17 at 1:05
• @sktsasus The geometric multiplicity is 1. (The eigenvectors are exactly the nonzero multiples of $(1\ 0)^T$.) What can you conclude? – Bungo Oct 26 '17 at 1:09

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.