A nilpotent matrix is diagonalisable $\iff$ it is the zero matrix.

Question

A nilpotent matrix is diagonalisable $\iff$ it is the zero matrix

For any $A$, $0=A=D^{-1}AD$ which is trivial $\rightarrow$ it's diagonalisable.

However, how do I show $\Rightarrow$ implication?

Answer 1

If a nilpotent matrix $A$ is diagonalisable, then $A=P D P^{-1}$ for $P$ invertible and $D$ diagonal.

Now $0=A^n = P D^n P^{-1}$ implies $D=0$ and so $A=0$.

Answer 2

Since $A$ is nilpotent then for some $p\in \Bbb N$ $A^p=0$ hence if $\lambda $ is an eigenvalue of $A$ and $x$ is an associated eigenvector then $0=A^p x=\lambda^px$ so with $x\ne0$ we get $\lambda^p=0$ and so $\lambda=0$. Hence $0$ is the only eigenvalue of $A$ which means that $A$ is similar to zero matrix and then $A=0$. The other implication is trivial. Now we can conclude the desired result.

Answer 3

Since $A$ is nilpotent the minimalpolynom is $t^k$ for a $k\in\mathbb{N}$. The minimalpolynom of a diagonalisable matrix is a product of distinct linear factors. So it must be $t$ for a diagonalisable nilpotent matrix and so $A=0$, because every matrix is a zero of its minimalpolynom.

Answer 4

I add another proof though not as simple as others:

A matrix is diagonalizable iff its minimal polynomial splits into different linear factors. Since $\;x^n\;$ is the characteristic pol. of any nilpotent matrix of order $\;n\;$, the matrix is diagonalizable iff its minimal polynomial is $\;x\;$ and this happens iff the matrix itself is the zero matrix.