Show that $\mathrm{rank}(I_n-A^TA)-\mathrm{rank}(I_m-AA^T) \geq n-m.$

Question

For any $m×n$ matrix $A$, where $n>m$ and $\mathrm{rank}(A)=m,$ show that $$\mathrm{rank}(I_n-A^TA)-\mathrm{rank}(I_m-AA^T) \geq n-m$$

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Using the hint by Omnomnomnom, I've known that $AB$ and $BA$ have the same nonzero eigenvalues with the same geometric multiplicity when they are both square.

It seems that I should consider geometric multiplicity of eigenvalue $1$ of $A^TA$ and $AA^T$, but I couldn't see how to make it.

Any help will be appreciated.

Answer 1

Hint: If $A^TAv = v$ for some vector $v$, then $(AA^T)A v = Av$. That is: if $v$ is in the nullspace of $I - A^TA$, then $Av$ is in the nullspace of $I - AA^T$.

Answer 2

By the rank-nullity theorem, we have $$ \dim \ker (I_n -A^t A)=n-\mathrm{rank}(I_n-A^tA),$$ so your question is equivalent to showing that $$\dim \ker (I_n -A^t A)\leq\dim \ker (I_m -A A^t).$$ First notice that (as in the hint above), if $ v \in \ker (I_n -A^t A)$ then $Av\in\ker (I_m -A A^t)$.

Next, let $v_1,\ldots,v_k$ (possibly $k=0$) be a basis of $\ker (I_n -A^t A)$. You need to show that $Av_1,\ldots ,Av_k$ are linearly independent in $\ker (I_n -A A^t)$. This completes the proof.