Nilpotent Matrix and Eigenvalues of the Matrix
Problem 11
An $n\times n$ matrix $A$ is called nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix.
Prove the followings.
(a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
(b) The matrix $A$ is nilpotent if and only if $A^n=O$.
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Hint.
Hint for (a)
- $(\Rightarrow)$ Consider $A \mathbf{x}=\lambda \mathbf{x}$, where $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an eigenvector corresponding to $\lambda$.
- $(\Leftarrow)$ Consider triangulation or Jordan normal/canonical form of $A$. Or use Cayley-Hamilton theorem.
Proof of (a).
$(\Rightarrow)$
Suppose the matrix $A$ is nilpotent. Namely there exists $k \in \N$ such that $A^k=O$. Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be the eigenvector corresponding to the eigenvalue $\lambda$.
Then they satisfy the equality $A\mathbf{x}=\lambda \mathbf{x}$. Multiplying this equality by $A$ on the left, we have
\[A^2\mathbf{x}=\lambda A\mathbf{x}=\lambda^2 \mathbf{x} .\]
Repeatedly multiplying by $A$, we obtain that $A^k \mathbf{x}=\lambda^k \mathbf{x}$. (To prove this statement, use mathematical induction.)
Now since $A^k=O$, we get $\lambda^k \mathbf{x}=0_n$, $n$-dimensional zero vector.
Since $\mathbf{x}$ is an eigenvector and hence nonzero by definition, we obtain that $\lambda^k=0$, and hence $\lambda=0$.
$(\Leftarrow)$
Now we assume that all the eigenvalues of the matrix $A$ are zero.
We prove that $A$ is nilpotent.
There exists an invertible $n\times n$ matrix $P$ such that $P^{-1} A P$ is an upper triangular matrix whose diagonal entries are eigenvalues of $A$.
(This is always possible. Study a triangularizable matrix or Jordan normal/canonical form.)
Hence we have
\[P^{-1} A P= \begin{bmatrix}
0 & * & \cdots & * \\
0 & 0 & \cdots & * \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0
\end{bmatrix}.
\]
Then we have $(P^{-1}AP)^n=O$. This implies that $P^{-1} A^n P=O$ and thus $A^n=POP^{-1}=O$.
Therefore the matrix $A$ is nilpotent.
Another proof of $(\Leftarrow)$ using Cayley-Hamilton theorem
Suppose that all the eigenvalues of the matrix $A$ are zero.
Then the characteristic polynomial of the matrix $A$ is
\[p(t)=\det(A-tI)=\pm t^n.\]
Hence by the Cayley-Hamilton theorem says that
\[p(A)=\pm A^n=O,\]
the zero matrix.
Thus, $A$ is nilpotent.
Note also that this method also proves the part (b).
Proof of (b).
If $A^n=O$, then by definition the matrix $A$ is nilpotent.
On the other hand, suppose $A$ is nilpotent. Then by Part (a), the eigenvalues of $A$ are all zero. Then by the same argument of the proof of part (a) $(\Leftarrow)$, we have $A^n=O$.
Comment.
Part (b) implies the following.
Suppose that you are given $n \times n$ matrix $B$.
You calculate the power $B^n$, and if it is not zero, then the power $B^k$ is never going to be the zero matrix $O$ no matter how large the number $k$ is.
Related Question.
See the post ↴
Every Diagonalizable Nilpotent Matrix is the Zero Matrix
for a proof of this problem.
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