Relationship between Nilpotent Matrix and Matrix with all zero diagonal factors.

Solution 1:

Let $A$ be a $n\times n$ matrix for some $n\times n$, over some algebraically closed field. The following holds:

$$A \text{ is nilpotent }\iff A\text {'s only eigenvalue is }0.$$

Question 1: Does $A$ being nilpotent imply its diagonal entries are all $0$ ?

According to the above characterization of nilpotency, absolutely not.

Take for instance the matrix $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}$. It's easy to check that $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}^2=0_{2\times 2}$.

Question 2: Is any matrix with only $0$'s on the diagonal entries necessarily nilpotent?

Again, no. There are matrices in these conditions which don't even have $0$ has an eigenvalue. For instance $\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$.

More can be said, but it depends on your knowledge whether it is worth saying or not. As far as I know it all comes down to a matrix's Jordan Normal Form:

Any nilpotent matrix $n\times n$ is similar to some block diagonal matrix $$ {\begin{bmatrix} \color{blue}{J_1} & 0 &\dots &\dots & 0\\ 0 & \color{blue}{J_2} & 0 & \dots &0\\ \vdots & \ddots & \ddots & \ddots &\vdots\\ \vdots & \ddots & \ddots &\ddots & 0\\ 0 & \dots & \dots & 0 & \color{blue}{ J_k}\\ \end{bmatrix}}_{n\times n},$$ for some $k\in \Bbb N$. Where, for each $i\in \{1\ldots ,k\},\,J_i$ is a $m_i\times m_i$ matrix, for some $m_i\in \Bbb N$, that looks like

$$\begin{bmatrix}0& 1 &&& \\ & 0 & 1 &\huge 0& \\ & & \ddots & \ddots &\\ &\huge 0 && 0 &1 \\ &&& & 0 \\ \end{bmatrix}_{m_i\times m_i}.$$

Solution 2:

We have this general result:

For a linear transformation $f$ such that $\mathrm{tr}(f)=0$ there's a basis $\mathcal B$ in which the matrix of $f$ is with all zero diagonal factors.

The proof of this result is in these points:

• $f$ isn't an homothetie so there's $x$ such that $x$ and $f(x)$ are linearly independant
• We consider $F$ a complement space of $\mathrm{span}(x)$ containing $f(x)$ and we define the projection $p$ on $F$ with kernel $\mathrm{span}(x)$ so the linear transformation $v:=p\circ f_{|F}$ of $F$ verify $\mathrm{tr}(v)=0$
• We prove the result by induction using the last point.

Solution 3:

By diagonal factors, do you mean diagonal elements of the matrix? If so, it's wrong.

See Wikipedia for more examples, but here is one:

$$A = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$$

$$P = \left( \begin{array}{cc} 3 & 1 \\ 5 & 2 \end{array} \right)$$

Then let $B=P^{-1} A P$,

$$B = \left( \begin{array}{cc} 10 & 4 \\ -25 & -10 \end{array} \right)$$

$A$ and $B$ are nilpotent, and $A^2=B^2=0$.

The converse is also wrong, the following matrix $C$ has zero diagonal entries, but it's not nilpotent, and actually $C^2=I$.

$$C = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$