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Suppose $A, B, C$ are square matrices of size n>1 and let $M = \begin{bmatrix} A & B \\ 0_n & C \end{bmatrix}$

If $A, C$ are diagonalizable, is $M$ diagonalizable?

Also, what if we are given that $M$ is diagonalizable, can we conclude that A and C are diagonalizable?

I just seem clueless right now, but I have been thinking of using the fact that if a matrix is diagonalizable, then its minimal polynomial is just a product of linear expressions, say $\prod_i(x-\lambda_i)$ where the $\lambda_i$'s are distinct.

I also tried looking for the proper $P$ so that I can write $P^{-1}MP$ as a diagonal matrix and that did not work.

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  • $\begingroup$ Your favourite example of a non-diagonalisable $2\times2$ matrix should in fact be of this form $\endgroup$ – Marc van Leeuwen Oct 24 '16 at 16:03
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Not for the first, try with $$M=\left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right) $$

The second part is true. Suppose $M$ diagonalizable but that $A$ or $C$ is not. If $A$ is not diagonalizable there is an eigenvalue $\lambda$ of $A$ whose algebraic multiplicity is strictly larger than the geometric. This implies that there is $X\neq 0$ so that $(A-\lambda)X\neq 0$ but $(A-\lambda)^2X =0$. Then with $V=\left(\begin{matrix} X \\ 0 \end{matrix} \right)$ we have $(M-\lambda)V\neq 0$ and $(M-\lambda)^2V =0$ so $M$ was not diagonalizable. If $C$ is not then there is $Y^T$ so that $Y^T(C-\lambda)\neq 0$ but $Y(C-\lambda)^2 =0$. Then with $W^T=( 0^T\ \ Y^T)$ we have $W^T(M-\lambda)\neq 0$ and $W^T(M-\lambda)^2 =0$ so $M$ was not diagonalizable.

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  • $\begingroup$ What do you mean? Let's assume $n>1$. $\endgroup$ – user198504 Oct 16 '16 at 10:26
  • $\begingroup$ Thanks! But I do not get the part wherein you said that $(A - \lambda)^2X = 0$. Can you expound a bit? $\endgroup$ – user198504 Oct 16 '16 at 11:25
  • $\begingroup$ Have added a bit, but you may think of the usual matrix $[0 \; 1 ; 0 \; 0]$ and the eigenvalue 0 which has geometric multiplicity 1 but algebraic 2. The vector $X=[0 ; 1]$ then corresponds to my argument. $\endgroup$ – H. H. Rugh Oct 16 '16 at 11:35
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The first part is clearly not true, as $(\begin{smallmatrix}0&1\\0&0\end{smallmatrix})$ shows.

For the second part, this is related to the fact that the restriction of a diagonalisable operator to a stable subspace is diagonalisable. This in fact directly gives the result for $A$, which describes the restriction of $M$ to the (invariant) subspace$~W$ spanned by the first $n$ standard basis vectors. That argument does not apply as directly to $B$, but by transposition symmetry it is clear that it must be diagonalisable as well.

In fact the mentioned fact is proved most easily using an argument that also directly shows $B$ is diagonalisable. It can be shown using the property that if $\Lambda=\{\lambda_1,\ldots,\lambda_k\}$ is a finite set (no repetitions among the $\lambda_i$), then an operator $\phi$ is diagonalisable with eigenvalues contained in$~\Lambda$ if and only if the polynomial $P=\prod_{\lambda\in\Lambda}(X-\lambda)$ annihilates$~\phi$, in other words if $P[\phi]=(\phi-\lambda_1I)\circ\cdots\circ(\phi-\lambda_kI)$ is the zero operator. This is applied with $\Lambda$ the set of eigenvalues of$~M$, and the fact that $P[M]$ has diagonal blocks $P[A]$ and $P[B]$ for any polynomial$~P$. The mentioned property is easy to show: the composed operator clearly kills each eigenspace for $\lambda_i$, and on the other hand the dimension of its kernel cannot exceed the sum of the dimensions of the kernels of the individual $(\phi-\lambda_iI)$, which is the sum of the dimensions, and dimension of the direct sum, of the eigenspaces for the $\lambda_i$.

Concretely, you get that the sets of eigenvalues of $A$ and of $B$ are contained in that of$~M$, that the eigenspaces of $A$ are obtained by intersecting the eigenspaces of$~M$ with the invariant subspace$~W$, and those of$~B$ by projecting those eigenspace parallel to$~W$ to a complementary subspace.

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