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\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?

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    $\begingroup$ You should be able to find the eigenvalues, k will not be involved for this. Then, compare algebraic multiplicity with geometric multiplicity. $\endgroup$ – Joppy Oct 26 '17 at 0:36
  • $\begingroup$ I see. But how will I be able to find the value of k by comparing the algebraic and geometric multiplicities? $\endgroup$ – sktsasus Oct 26 '17 at 0:39
  • $\begingroup$ First start with the eigenvalues of this matrix. $\endgroup$ – imranfat Oct 26 '17 at 0:42
  • $\begingroup$ Just try and tell us what you get. $\endgroup$ – Mr. T Oct 26 '17 at 0:42
  • $\begingroup$ 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! $\endgroup$ – sktsasus Oct 26 '17 at 0:53
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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?

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

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