Let $T : R^7 \to R^7$ be a linear transformation such that $T^2 = 0$

Question

Let $T : R^7 \to R^7$ be a linear transformation such that $T^2 = 0$ then what can we tell about the rank of $T$?

I cannot conclude from here. Help Needed.

Answer 1

If $T^2 = 0$, then $\text {im}(T) \subseteq \text{ker}(T)$, so in particular, the rank of $T$ is not more than the nullity of $T$. Since the sum of rank and nullity is 7, we conclude that the rank of $T$ is 3 or smaller.

Here's an example of the matrix of a rank 3 transformation $T$ with the property that $T^2=0$:

$$ \left[\begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] $$

This matrix is block diagonal with three 2x2 diagonal blocks $N$ for which $N^2=0$.

Answer 2

Since $T(\mathbb{R}^7)\subset \text{ker}(T)$ we conclude that $$\text{rank}(T)\le \dim\text{ker}(T)=7-\text{rank}(T)$$ Therefore, $\text{rank}(T)\le 3$.