# 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 $O$.

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### 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.

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