# Every Diagonalizable Nilpotent Matrix is the Zero Matrix ## Problem 504

Prove that if $A$ is a diagonalizable nilpotent matrix, then $A$ is the zero matrix $O$. Add to solve later

### Definition (Nilpotent Matrix)

A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$.

## Proof.

### Main Part

Since $A$ is diagonalizable, there is a nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.

As we show below, the only eigenvalue of any nilpotent matrix is $0$.
Thus, $S^{-1}AS$ is the zero matrix.
Hence $A=SOS^{-1}=O$.

### The only eigenvalue of each nilpotent matrix is $0$

It remains to show that the fact we used above: the only eigenvalue of the nilpotent matrix $A$ is $0$.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$.
That is,
$A\mathbf{v}=\lambda \mathbf{v}, \tag{*}$

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.

Then we use the relation (*) inductively and obtain
\begin{align*}
A^k\mathbf{v}&=A^{k-1}A\mathbf{v}\\
&=\lambda A^{k-1}\mathbf{v} && \text{by (*)}\\
&=\lambda A^{k-2}A\mathbf{v}\\
&=\lambda^2 A^{k-2}\mathbf{v} && \text{by (*)}\\
&=\dots =\lambda^k \mathbf{v}.
\end{align*}

Hence we have
$\mathbf{0}=O\mathbf{v}=A^k\mathbf{v}=\lambda^k \mathbf{v}.$

Note that the eigenvector $\mathbf{v}$ is a nonzero vector by definition.
Thus, we must have $\lambda^k=0$, hence $\lambda=0$.
This proves that the only eigenvalue of the nilpotent matrix $A$ is $0$, and this completes the proof.

### Another Proof of the Fact

Even though the fact proved above is true regardless of diagonalizability of $A$, we can make use that $A$ is diagonalizable to prove the fact as follows.

Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of the $n\times n$ nilpotent matrix $A$.
Then we have
$S^{-1}AS=D,$ where
$D:=\begin{bmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 &\lambda_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n \end{bmatrix}.$

Then we have
$\require{cancel}$
\begin{align*}
D^k&=(S^{-1}AS)^k\\
&=(S^{-1}A\cancel{S})(\cancel{S}^{-1}A\cancel{S})\cdots (\cancel{S}^{-1}AS)\\
&=S^{-1}A^kS\\
&=S^{-1}OS=O.
\end{align*}

Since
$D^k=\begin{bmatrix} \lambda_1^k & 0 & \dots & 0 \\ 0 &\lambda_2^k & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n^k \end{bmatrix},$ it yields that $\lambda_i^=0$, and hence $\lambda_i=0$ for $i=1, \dots, n$.

## Related Question.

The converse of the above fact is also true: if the only eigenvalue of $A$ is $0$, then $A$ is a nilpotent matrix.

See the post ↴
Nilpotent Matrix and Eigenvalues of the Matrix
for a proof of this fact. Add to solve later

### 1 Response

1. 07/10/2017

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