Determine all matrices, $A$, that satisfy the condition that $\operatorname{rank}(A^k) = \operatorname{rank}(A)$ for each $k \geq 1$.

Question

My intution is that the condition is true if and only if $A$ is invertible or $A$ is a projection, i.e $A^2 = A$. It is trivial to show that if $A$ is invertible or $A$ is a projection then the condition is true but I couldn't show the other implication.

Answer 1

Let's say $A$ is an $n \times n$ matrix with entries in $\mathbb{k}$ satisfying your property.

Now, the question linked in comments shows that your condition is equivalent to just $\operatorname{rk} A^2 = \operatorname{rk} A$. Since $\operatorname{ker} A \subseteq \operatorname{ker} A^2$, this is the same as $\operatorname{ker} A = \operatorname{ker} A^2$. That means that if $x \in \operatorname{im} A$ and $x \not = 0$ then $A x \not = 0$. In turn, this means that $\operatorname{im} A \cap \operatorname{ker A} = \{0\}$.

These simplification steps form a chain of equivalences, so one answer to your question is exactly those matrices $A$ such that $\operatorname{im} A \cap \operatorname{ker} A = \{0\}$.

Let's figure out how to construct them all. Pick a basis $(v_1, \ldots, v_m)$ of $\operatorname{im} A$ and a basis $(w_{m + 1}, \ldots, w_n)$ of $\operatorname{ker} A$. Then, since $\operatorname{im} A \cap \operatorname{ker} A = \{0\}$, we must have that $(v_1, \ldots, v_m, w_{m + 1}, \ldots, w_n)$ is a basis of all of $\mathbb{k}^n$. Let $S$ be the (invertible) matrix with this basis as its columns, i.e. $S = [v_1\ \ldots\ w_n]$. Now, since the last $n - m$ columns of $S$ are vectors in the kernel of $A$, the matrix $S A S^{-1}$ breaks up into block matrices as $$ \begin{bmatrix} X & 0\\ 0 & 0 \end{bmatrix}\tag{$*$} $$ with $X$ an $m \times m$ matrix and the zeroes representing zero block matrices. Because $\operatorname{im} A \cap \operatorname{ker A} = \{0\}$, the matrix $X$ must have kernel $\{0\}$. In other words, $X$ must be invertible.

Conversely, if $S$ is any invertible matrix at all and $B$ is any matrix of the form $(*)$ with $X$ an invertible $m \times m$ matrix then the matrix $$ (S B S^{-1})^2 = S B^2 S^{-1} = S\begin{bmatrix} X^2 & 0\\ 0 & 0 \end{bmatrix}S^{-1} $$ has rank $m$. Therefore $A = S B S^{-1}$ satisfies your condition.

Putting this all together, we have shown that $A$ satisfies your condition if and only if there is an invertible $n \times n$ matrix $S$ and an invertible $m \times m$ matrix $X$ such that $$ A = S \begin{bmatrix} X & 0\\ 0 & 0 \end{bmatrix} S^{-1}. $$

Answer 2

Looking at the Jordan form, a matrix is similar to a block-diagonal matrix where each block is a Jordan block. The blocks corresponding to nonzero eigenvalues do it lose rank when taking powers. The only way for the matrix to lose rank when taking powers is to have Jordan blocks of size $>1$ for the eigenvalue zero.

In other words the $n\times n$ matrices that keep their rank for all powers are of them form $$ A=S\,\begin{bmatrix} B &0\\0&0_m\end{bmatrix}\,S^{-1}, $$ where $B$ is a $k\times k$ invertible matrix and $k+m=n$. This includes both the invertible matrices and the idempotents, but also many other matrices.