Inequality Regarding Ranks of Matrices

Linear Algebra exam problems and solutions at University of California, Berkeley

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)

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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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