Show that rank($A^{n+1}$) = rank($A^n$)

Question

Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.

Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.

Answer 1

$\newcommand{\rg}{\operatorname{range}}$ Obviously for every $m$, $\rg A^{m+1}\subset\rg A^{m}$, so if $d_m=\dim\rg A^m$, $d_{m+1}\le d_m$. If $d_{m+1}=d_m$ for some $m$, then $\rg A^{m+1}=\rg A^{m}$ and therefore $\rg A^m=\rg A^{m+1}=\rg A^{m+2}=\dotsb{}$. That is, the sequence $d_0,d_1,\dots$ becomes constant once it stops descending.
Because $d_0= n$, the sequence must stop descending within $n$ terms.

Edit: For the problem you phrased in the comment, $\rg A^{m+1}=\{AA^{m}y:y\in \mathbb C^n\}=\{Ax:x=A^my\in\rg A^m\}=\{Ax:x\in\rg A^m\}$,
Therefore $\rg A^{m}=\rg A^{m+1}\implies$
$\rg A^{m+1}=\{Ax:x\in\rg A^m\}=\{Ax:x\in\rg A^{m+1}\}=\rg A^{m+2}$.

Answer 2

Hint

You can prove that for $k \ge 0$ $$\mathrm{rank}(A^{k+2}) - \mathrm{rank}(A^{k+1}) \le \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^{k})$$

Therefore, $$\mathrm{rank}(A^{n+1}) < \mathrm{rank}(A^{n})$$ would imply the contradiction $\mathrm{rank}(A) \gt n$.

Answer 3

Every matrix corresponds to a linear map. Suppose that $A=\mathcal{M}(T)$, where $T\in \mathcal{L}(V)$ and $dim (V)=n$. Using Theorem 8.4 in Axler, $$Null(T^n)=Null(T^{n+1}).$$

According to the Fundamental Theorem of Linear Maps, $$n=dim Null(T^n)+ dim Range(T^n)$$ $$n=dim Null(T^{n+1})+dim Range(T^{n+1})$$

$dim Range(T^n)=dim Range(T^{n+1})$ implies that $rank(A^n)=rank(A^{n+1})$.

Answer 4

All does depend on $n$. So this is a nice case for complete induction over n.

n=1: A=a real or complex and nonzero. $Rank(A)=1=Rank(A^{n=1})=n=1=rank(a^2)=rank(A^2)=rank(A^{n+1})$

For $n$ natural the hyptheses is $true$.

For $n+1$ A change exactly in one row or column to the case $n$. This row or column can be either but not linear dependent on the other making up the A for $n$. The implies implicitly the at least one element in the column or row is nonzero exactly in the dimension added to A for $n$.

We can now uses some of the equivalent definitions for the $rank$ of a square matrix. With restrict to generality the added row or column has only one nonzero element. This acts as a factor for example in the determinate development or is a new eigenvalue or the matrix A for $n+1$. So the determinante is nonzero at least in that development because we have a nonzero value and the knowledge that the determinant of our A for $n$ is nonzero and $rank(A)=n$.

The main idea for the induction step is ring of matrice rank or Rankconserse by multiplication of a nonsingular matrix in general a $A$ itself specially. Matrices with nonzero $rank$ preserve the rank under multiplication. The multiplication under consideration is commutative because we only multiply A. That is another indicator for our hypotheses for $n+1$. Eigenvalues and Schur decomposition are closely related. One of the matrices in the Schur decomposition is an upper triangle matrice. So increasing the dimension from $n$ to $n+1$ simply adds another last if the last row and column in a unite vector with only a value in the new dimension.

The Schur decomposition is equivalent to that the matrix $∈ℂ^{+1×+1}$ has the property relying on the matrix $∈ℂ^{×}$. The matrice of $rank$ from a group and can transformed into each other under conservation of the $rank$. And the prove is done.