Given a square matrix A of order n, prove $\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$

Question

Given $A\in F^{n \times n}$ prove:

$$\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$$

$\operatorname{rank}(A^{n+1}) \leq \operatorname{rank}(A^n)$ is easy, just from:

How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?

But how can I prove the other direction? or should I do it otherwise?

Answer 1

Coming back to this question after a few years, I've found a simpler proof, using only basic linear algebra knowledge.

First, if $\operatorname{rank}(A)=n$, use the facts:

• Matrix is full rank iff it is invertible
• Product of invertible matrices is invertible

so $\operatorname{rank}(A^{k})=n$ for any natural $k$.

Otherwise, use induction to show the following:

if $rank(T^k) = rank(T^{k+1})$ for some positive integer $k$, then $rank(T^k) = rank(T^m)$ for all positive integer $m \geq k$.

Finally, we have to show that if $n \gt \operatorname{rank}(A)$, then $rank(A^k) = rank(A^{k+1})$ for some $k\le n$. $$ rank(A^k) = \dim(\operatorname{im}(A^k)) $$$$ \operatorname{im}(A) \supseteq \operatorname{im}(A^2) \supseteq \operatorname{im}(A^3) \supseteq \dots $$

$$ n \gt \operatorname{rank}(A) \ge \operatorname{rank}(A^2) \ge \operatorname{rank}(A^3) \ge \dots \ge \operatorname{rank}(A^n) \ge \operatorname{rank}(A^{n+1}) \ge 0 $$ There are n possible values ($0,\dots,n-1$) for n+1 ranks, so there are at least two ranks that are equal.

Answer 2

Note that we can assume the field is algebraically closed, as the rank of the matrix does not change if we look at it as being over a larger field.

Now the matrix is similar to an upper triangular matrix. We can assume that it has a block form consisting of an upper triangular $m\times m$ matrix with only non-zero elements on the diagonal, and a block consisting of a strictly upper triangular $(n-m)\times (n-m)$ matrix. Now both the $n$'th and the $n+1$'st power of such a matrix will simply consist of some $m\times m$ upper triangular block with only non-zero elements on the diagonal (as we kill off the strictly upper triangular block when the power is at least $n-m$). This shows that these two powers have the same rank (namely $m$).

Answer 3

Using Fitting's Lemma, one can give another version of the fine argument of @Tobias.

The sequence $$ \ker(A) \subseteq \ker(A^2) \subseteq \ker(A^3) \subseteq \dots $$ is ascending, and the sequence $$ \operatorname{im}(A) \supseteq \operatorname{im}(A^2) \supseteq \operatorname{im}(A^3) \supseteq \dots $$ is descending. Choose the smallest $m$ such that $$ \ker(A^m) = \ker(A^{m+i}), \qquad \operatorname{im}(A^m) = \operatorname{im}(A^{m+i}) $$ for all $i \ge 0$. Note that if $\ker(A^m) = \ker(A^{m+1})$, then $\ker(A^m) = \ker(A^{m+i})$ for all $i \ge 0$. In particular $m \le n$.

Now Fitting's Lemma states that $$ F^n = \ker(A^m) \oplus \operatorname{im}(A^m), $$ and $A$ is nilpotent on the first summand, and invertible on the second one.

Then for any $k \ge m$ (actually, I believe, exactly for these values of $k$) we will have $$\operatorname{rank}(A^k) = \operatorname{rank}(A^{k+1}).$$

Answer 4

Here's my solution?

If $\operatorname{rank}(A)=n$, which means $A$ is invertable, then $\operatorname{rank}(A^n)=\operatorname{rank}(A^{n+1})=n$.

Now let's consider the case where $0 \le\operatorname{rank}(A)<n$.

From $$ \operatorname{rank}(AB) \le \min\{\operatorname{rank}(A), \operatorname{rank}(B)\} $$ we get $$ \operatorname{rank}(A)\ge \operatorname{rank}(A^2)\ge \cdots \ge \operatorname{rank}(A^n)\ge \operatorname{rank}(A^{n+1}). $$ Since $0\le \operatorname{rank}(A)<n$, there exists $i \in \{1,2, \ldots ,n\}$ such that $\operatorname{rank}(A^i)=\operatorname{rank}(A^{i+1})$.

Let $N(A)$ be the nullspace of $A$, then $$ \operatorname{dim}N(A^{i})=n-\operatorname{rank}(A^{i}) =n-\operatorname{rank}(A^{i+1})=\operatorname{dim}N(A^{i+1}) $$ and also $N(A^{i}) \subseteq N(A^{i+1})$, thus $N(A^{i})=N(A^{i+1})$.

Use this conclusion, for all $x \in N(A^{i+2})$, $$ A^{i+2}x=0=A^{i+1}(Ax)=A^{i}(Ax)=A^{i+1}x. $$ Hence, $N(A^{i+2}) \subseteq N(A^{i+1})$. Also $N(A^{i+1}) \subseteq N(A^{i+2})$, then we have $N(A^{i+1})=N(A^{i+2})$.

By induction, we get $$ N(A^{i})=N(A^{i+1})=\cdots=N(A^{n})=N(A^{n+1}) $$ so $\operatorname{rank}(A^{n})=n-\operatorname{dim}N(A^{n}) =n-\operatorname{dim}N(A^{n+1})=\operatorname{rank}(A^{n+1})$.

Answer 5

A more straightforward proof than the existing ones.

If $\operatorname{rank}(A^{n+1})<\operatorname{rank}(A^n)$, then $$\ker(A^n)\subsetneq\ker(A^{n+1})$$ there exists a vector $v$, such that $A^{n+1}v=0$ but $A^{n}v\not=0$.

Now we show that $v, Av, \cdots, A^nv$ are linearly independent, but $F^n$ cannot have more than $n$ linearly independent vectors, a contradiction.

If $\sum_{i=0}^n a_iA^i v=0$, then apply $A^n$ on both sides, we have $a_0A^nv=0$, therefore $a_0=0$.

Similarly, apply $A^{n-1}$ on both sides, we have $a_1=0$, etc. So $a_0=\cdots = a_n=0$.