Given the Matrix $A \in \mathbb{R}^{2\times2}$ with $Spec(A) = \{300,333\}$ I have the Block matrix
$M = \begin{bmatrix}A&A\\A&A\end{bmatrix} \in \mathbb{R}^{4\times4}$ for which I have to determine the Trace and Jordan normal form.
My attempt (not formally written but roughly the plan) :
- Considering that $M$ consist of four blocks the same Matrix $A$ and therefore of equally partitioned blocks we could easily deduce that $tr(M) = tr(A) + tr(A)$. Because a matrix's trace is the sum of its eigenvalues we have that $tr(A) = 300 + 333 = 633 \Rightarrow tr(M) = 633 + 633 = 1266$
- We clearly have at least one row identical to another row which means the $det(M) = 0$ and because the determinant is the product of its eigenvalues we could deduce that at least one of the eigenvalues is $0$.
- If I conpute the characteristic polynomial mechanically I get:$\mathcal{p}_M=-\frac{λ^2\left(-λ^3d-λ^2bc+2λ^2d^2+3λ^2ad+2λabc-6λad^2-2λa^2d+6λbcd+4b^2c^2+4a^2d^2-8abcd\right)}{bc-d\left(a-λ\right)}$ for $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ which is very ineffective but I can at least deduce that the algebraic multiplicity of the eigenvalue $0$ is at least $2$ which tells me that the sum of the sizes of all Jordan blocks corresponding to the eigenvalue $0$ is at least $2$.
I am unsure how to proceed past this point. Any reminder about any theorems I'm clearly not remembering would be greatly appreciated!
Best Answer
We already know that $A$ has zero as an eigenvalue. Moreover, because the rank of $M$ is $2$, the geometric multiplicty of the zero-eigenvalue is $4 - 2 = 2$. We just need to find the other eigenvalues.
One approach is to try to go directly for the eigenvectors of $M$. Suppose that $v,w \in \Bbb R^2$, and let $(v,w)$ denote the corresponding "block vector" in $\Bbb R^4$. In order for $v,w$ to be an eigenvector of $M$ associated with $\lambda \neq 0$, we need to have $$ \pmatrix{A & A\\ A & A} \pmatrix{v\\w} = \lambda \pmatrix{v\\w} \implies \begin{cases}Av + Aw = \lambda v\\ Av + Aw = \lambda w\end{cases}. $$ We see that $\lambda v = Av + Aw = \lambda w$, and because $\lambda \neq 0$ it must be the case that $v = w$. With that substitution, both equations simply become $$ Av + Av = \lambda v \implies Av = \frac{\lambda}{2}v. $$ In other words, there is an eigenvalue $\mu$ (non-zero) of $A$ for which $\mu = \frac{\lambda}{2} \implies \lambda = 2 \mu$.
We know that $A$ has the non-zero eigenvalues $300,333$. It follows that $M$ has the eigenvalues $2 \cdot 300 = 600$ and $2 \cdot 333 = 666$.
Now, $M$ has a two dimensional eigenspace associated with zero and two other eigenvalues, which means that the dimensions of the eigenspaces adds up to $4$. It follows that $M$ is diagonalizable.