Matrix with eigenvalues as $0$ and rank as $n-1$

Question

How do I find a matrix $A \in \mathbb{R}^{n * n}$ with all eigenvalues as zero and rank being $n-1$.

Is this possible? Because the number of non-zero eigenvalues is the rank. But this question has cropped up in my assignment.

Any example will do. Thanks!

Answer 1

$A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$

Edit: a more general example is a matrix with $1$'s just above the diagonal entries, and $0$'s elsewhere. So $$A=\begin{bmatrix}0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 1 & ... & 0 & 0\\ ... & ... & ... & ... & ... & ... \\ 0 & 0 & 0 & ... & 0 & 0\end{bmatrix}$$

Answer 2

Consider a $3\times3$ matrix as below:
$A=\begin{bmatrix} 0 & 3 & 0 \\ 0 & 0 & 3\\ 0 & 0 & 0 \end{bmatrix}.$
Here $rank(A)=2$ and $\lambda_1=\lambda_2=\lambda_3=0$.
The above matrix is a nilpotent matrix.

Answer 3

Consider $A\in \mathbb{R}^{n\times n} $

$$a_{ij} = \begin{cases} 1& j=i+1,\\ 0& \text{otherwise} \end{cases}$$