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 $A^n=O$.

Hint.

Hint for (a)

1. $(\Rightarrow)$ Consider $A \mathbf{x}=\lambda \mathbf{x}$, where $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an eigenvector corresponding to $\lambda$.
2. $(\Leftarrow)$ 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.

The post Nilpotent Matrix and Eigenvalues of the Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that

(1) $$\det(A)=\prod_{i=1}^n \lambda_i$$

(2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$

Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.

Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
We give two different proofs.

Plan 1.

1. Use the definition of eigenvalues (the characteristic polynomial).
2. Compare coefficients.

Plan 2.

1. Make $A$ upper triangular matrix or in the Jordan normal/canonical form.
2. Use the property of determinants and traces.

Proof. [Method 1]

(1) Recall that eigenvalues are roots of the characteristic polynomial $p(\lambda)=\det(A-\lambda I_n)$.
It follows that we have
\begin{align*}
&\det(A-\lambda I_n) \\
&=\begin{vmatrix}
a_{1 1}- \lambda & a_{1 2} & \cdots & a_{1,n} \\
a_{2 1} & a_{2 2} -\lambda & \cdots & a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{m 2} & \cdots & a_{n n}-\lambda
\end{vmatrix} =\prod_{i=1}^n (\lambda_i-\lambda). \tag{*}
\end{align*}

Letting $\lambda=0$, we see that $\det(A)=\prod_{i=1}^n \lambda_i$ and this completes the proof of part (a).

(2) Compare the coefficients of $\lambda^{n-1}$ of the both sides of (*).
The coefficient of $\lambda^{n-1}$ of the determinant on the left side of (*) is

$$(-1)^{n-1}(a_{11}+a_{22}+\cdots a_{n n})=(-1)^{n-1}\tr(A).$$
The coefficient of $\lambda^{n-1}$ of the determinant on the right side of (*) is
$$(-1)^{n-1}\sum_{i=1}^n \lambda_i.$$
Thus we have $\tr(A)=\sum_{i=1}^n \lambda_i$.

Proof. [Method 2]

Observe that there exists an $n \times n$ invertible matrix $P$ such that
\[P^{-1} A P= \begin{bmatrix}
\lambda_1 & * & \cdots & * \\
0 & \lambda_2 & \cdots & * \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n \tag{**}
\end{bmatrix}.
\] This is an upper triangular matrix and diagonal entries are eigenvalues.
(If this is not familiar to you, then study a “triangularizable matrix” or “Jordan normal/canonical form”.)

(1) Since the determinant of an upper triangular matrix is the product of diagonal entries, we have
\begin{align*}
\prod_{i=1}^n \lambda_i & =\det(P^{-1} A P)=\det(P^{-1}) \det(A) \det(P) \\
&= \det(P)^{-1}\det(A) \det(P)=\det(A),
\end{align*}
where we used the multiplicative property of the determinant.

(2) We take the trace of both sides of (**) and get
\begin{align*}
\sum_{i=1}^n \lambda_i =\tr(P^{-1}AP) =\tr(A).
\end{align*}
(Here for the last equality we used the property of the trace that $\tr(AB)=\tr(BA)$ for any $n\times n$ matrices $A$ and $B$.)
Thus we obtained the result $\tr(A)=\sum_{i=1}^n \lambda_i$.

Comment.

The proof of (1) in the first method is simple, but that of (2) requires a bit observation, especially when we find the coefficient of the left-hand side.

The proof of (2) in the second method is simpler although you need to know about the Jordan normal/canonical form.

These two formulas relate the determinant and the trace, and the eigenvalue of a matrix in a very simple way.

The post Determinant/Trace and Eigenvalues of a Matrix first appeared on Problems in Mathematics.