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I have attempted to solve this problem and I reach a certain point (finding eigenvalues) before getting stuck.

Let $$ A= \left[ {\begin{array}{cc} 2 & 0 & k \\ 0 & 3 & 0\\ 0 & 0 & k \\ \end{array} } \right] $$

To find the values for which this matrix is diagonalizable, I first found the characteristic polynomial, which I calculated to be: $(2- \lambda)(3- \lambda)(k - \lambda)$

I calculated this by solving for the determinant of $A - \lambda I$.

The answer is apparently for all $k \neq 2$, but I am unsure of how to prove this. Could someone provide any insight on this? Thanks! For self-study.

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First note that if $k\neq 2,3$, then all eigenvalues are different, and hence the matrix is diagonalizable.

In case when $k = 2$ or $k=3$, i.e. you have an eigenvalue with multiplicity more than $1$, you need to calculate dimension of the eigenspace. If all the dimensions of eigenspaces are equal to multiplicities of corresponding eigenvalues, then the matrix is diagonalizable, otherwise not.

Can you continue from here?

Edit: Let $\lambda$ be an eigenvalue of $A$ and $E_\lambda$ its corresponding eigenspace. Let $p_A$ be the characteristic polynomial of $A$.

Since $\lambda$ is an eigenvalue, $p_A(\lambda) = 0$ and thus, there exists positive integer $n$ such that $p_A(x) = (x-\lambda)^nq(x)$ and $q(\lambda)\neq 0$. Define $a_\lambda$ as this $n$ in the corresponding factorization of $p_A$. We call this algebraic multiplicity of eigenvalue $\lambda$.

On the other hand, we are also interested in number $g_\lambda = \dim E_\lambda = \dim\ker(A-\lambda I)$ which is called geometric multiplicity of eigenvalue $\lambda$.

In general, $1\leq g_\lambda \leq a_\lambda$ and $A$ is diagonalizable if and only if $a_\lambda = g_\lambda$ for all eigenvalues $\lambda$ of $A$.

In your case, you have characteristic polynomial $p_A(x) = (x-2)(x-3)(x-k)$. If $k\neq 2,3$, then $a_2 = a_3 = a_k = 1$ and since $1\leq g_\lambda \leq a_\lambda$, we have $g_2=g_3 = g_k = 1$, as well. That is, all geometric and algebraic multiplicities are equal and the matrix is diagonalizable.

In the case $k = 3$, we have $a_2 = 1$, $a_3 = 2$, so again, $a_2 = g_2 = 1$, but we need to check whether $g_3 = 1$ or $g_3 = 2$. To find $g_3$ solve the homogeneous linear system $(A-3I)v = 0$. You will find out that solution space is of dimension $2$ (its obvious as soon as you write it), and hence $a_3 = g_3 = 2$, so $A$ is diagonalizable.

In the case $k = 2$, similar to the previous case, we have $a_3 = g_3 = 1$, but $a_2 = 2$ so we need to solve homogeneous linear system $(A-2I)v = 0$, but in this case you will find out that $g_2 = 1 \neq a_2$, and thus $A$ is not diagonalizable.

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  • $\begingroup$ This makes sense! The only trouble I'm having is understanding the determination of multiplicities. I believe that's where I'm making errors. If it's not too much trouble would you mind adding that calculation to your answer? Thank you! $\endgroup$ – rubyquartz Jul 24 '17 at 17:29
  • $\begingroup$ @user7090180, please see my edit. $\endgroup$ – Ennar Jul 24 '17 at 17:48
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A square matrix over a field $\mathbb F$ is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors over $\mathbb F$. In your case, when $\lambda=2$, you only need to check whether the polynomial $(x-2)(x-3)$ annihilates $A$. If so, $A$ is diagonalisable, otherwise the minimal polynomial of $A$ has to be $(x-2)^2(x-3)$ and hence $A$ is not diagonalisable.

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