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$X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$, $2n\times 2n$ matrix

Then

a) A, B,C,D are nilpotent ⇔ X is nilpotent

b)If X is diagonalisable so is A,B,C,D.

c)min polynomial of X divides the lcm of minimal polynomials of A, B, C,D.

d)If A,B, C,D are diagonalisable so is X.

I am sure that a) is true if $X^k=\begin{pmatrix}A&B\\C&D\end{pmatrix}^k=\begin{pmatrix}A^k&B^k\\C^k&D^k\end{pmatrix}$ holds, thank you for help.I need to find out which of the above statemements are true/false

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What you have for $a)$ is not true, for example consider the $2x2$ all one's matrix, and $k=2$. –  Ryan Sullivant Jun 22 '13 at 6:04
    
For $d)$ consider the $2x2$ matrix with $A = C = D = 0$ and $B = 1$. Then each of $A,B,C,D$ are diagonal, but is $X$? (What is its minimal polynomial?) –  Ryan Sullivant Jun 22 '13 at 6:10

1 Answer 1

For a), c), d) you can take the counterexample $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. (with $n=1$).

For b) I tried a bit with Wolfram Alpha. Here is a counterexample (although I have no good intuition why it works):

$$\begin{pmatrix}1&0&0&1\\1&1&0&0\\2&1&1&0\\0&2&0&1\end{pmatrix}.$$

Here $n=2$ and $A,B,C$ are not diagonalisable. But the whole matrix is. As you can find out with Wolfram alpha.

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@user1551 $\operatorname{lcm}(X,X,X,X-1)=X(X-1)$ and $X^2\nmid X(X-1)$. –  Julian Kuelshammer Jun 22 '13 at 11:30
    
You're right, I forgot the L in LCM. +1 –  user1551 Jun 22 '13 at 11:32

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