# Inequality Regarding Ranks of Matrices

## Problem 58

Let $A$ be an $n \times n$ matrix over a field $K$. Prove that
$\rk(A^2)-\rk(A^3)\leq \rk(A)-\rk(A^2),$ where $\rk(B)$ denotes the rank of a matrix $B$.

(University of California, Berkeley, Qualifying Exam)

Contents

## Hint.

Regard the matrix as a linear transformation $A: K^n\to K^n$.
Then consider the restrictions
$A|_{\im(A)}:\im(A) \to K^n$ and $A|_{\im(A^2)}:\im(A^2) \to K^n$.

Use the rank-nullity theorem:
If $F: V\to W$ is a linear transformation from a vector space of dimension $n$, then we have

$\rk(F)+\nullity(F)=n.$ Or equivalently
$\dim(\im(F))+\dim(\ker(F))=\dim(V).$

Another way to prove the problem is to use the quotient spaces $AK^n/A^2K^n$ and $A^2K^n/A^3K^n$.

We will give two proofs.

## Proof 1.

We identify the matrix $A$ with the linear transformation from $K^n$ to itself whose matrix representation is $A$ with a fixed basis for $K^n$.

We first restrict the linear transformation $A:K^n \to K^n$ to the image of $A$ and obtain the linear transformation
$A|_{\im(A)}:\im(A)\to K^n.$ Note that the image of $A|_{\im(A)}$ is $\im(A^2)$.
Thus by the rank-nullity theorem we have
$\rk(A^2)+\dim(\ker(A|_{\im(A)})=\rk(A). \tag{1}$

Similarly, we consider the restricted linear transformation $A|_{\im(A^2)}:\im(A^2)\to K^n$.
The image of $A|_{\im(A^2)}$ is $\im(A^3)$ and by the rank-nullity theorem we have
$\rk(A^3)+\dim(\ker(A|_{\im(A^2)})=\rk(A^2).\tag{2}$

Since $\im(A^2) \subset \im(A)$, we have
$\ker(A|_{\im(A^2)}) \subset \ker(A|_{\im(A)}),$ and thus
$\dim(\ker(A|_{\im(A^2)})) \leq \dim(\ker(A|_{\im(A)})). \tag{3}$ Combining (1), (2), and (3), we obtain
$\rk(A^2)-\rk(A^3)\leq \rk(A)-\rk(A^2),$ as required.

## Proof 2.

Here is another proof that uses quotient spaces.
Consider the quotient vector space $AK^n/A^2K^n$. The dimension of this space is $\dim(AK^n/A^2K^n)=\rk(A)-\rk(A^2).$

Also consider the quotient vector space $A^2K^n/A^3K^n$ whose dimension is $\dim(A^2K^n/A^3K^n)=\rk(A^2)-\rk(A^3).$ The matrix multiplication by $A$ induces a surjective homomorphism from $AK^n/A^2k^n$ to $A^2K^n/A^3K^n$.
Therefore
$\dim(AK^n/A^2K^n) \geq \dim(A^2K^n/A^3K^n)$ and we obtain
$\rk(A^2)-\rk(A^3)\leq \rk(A)-\rk(A^2)$

## Comment.

The second proof used the fancier word “quotient space”, hence might be terse.
But in both proofs, the essential part is the rank-nullity theorem (or homomorphism theorem).

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