Matrices in $\,\mathbb R^{3\times 3}$ having only zero as eigenvalue

Question

Does there exist a matrix $A$ in $\Bbb{R}^{3\times 3}$ that has only $0$ as eigenvalue?

As far as I think, an upper diagonal matrix with only zeros on the diagonal is such an example.

And if I'm correct, is such a matrix diagonalizable?

Answer 1

Yes you are correct: by taking $$A=\begin{bmatrix} 0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$ you have $sp(A)=\{0\}$. Anyway it isn’t diagonalizable, it is already in Jordan canonical form.

In general if $B\in\mathbb R^{n\times n}$ so that $B$ has only eigen-value $0$ ($sp(B)=\{0\}$) then $B$ is diagonalizable $\iff B=0$ (denoting with $q_B$ the minimal polinomial of $B$): $B$ is diagonalizable $\iff q_B$ is product of linear coprime factors, and because $B$ has only $0$ as eigen-value it must be $q_B(t)=t\iff B=0$.

Answer 2

You're right with your thinking: an upper-triangular matrix with only zeros on the diagonal has $0$ as the only eigenvalue, for example, the zero matrix. Your second statement is incorrect; consider the matrix \begin{align} A&= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align} This matrix has $0$ as the only eigenvalue, however, it is not diagonalizable. One way to see this is to notice that $A$ is already a Jordan matrix, but not all the Jordan blocks are $1 \times 1$.

Alternatively, it's easy to verify that $\ker(A)$ (which is the eigenspace corresponding to the eigenvalue $0$) is $2$-dimensional, and in fact $\{e_1, e_3\}$ is a basis for $\ker(A)$. i.e we only have $2$ linearly independent eigenvectors of $A$, whereas for $A$ to be diagonalizable, we need $3$ linearly independent eigenvectors.