Let $\lambda \in \mathbb R^*$, $n \ge 2$.
Let $A \in M_n(\mathbb R)$, $f$ the linear map defined by $A$.
Let $E_\lambda (A^TA) = \ker (A^TA - I_n)$ and $E_\lambda (A^TA) = \ker (AA^T - I_n)$.
Show that $\dim E_\lambda (A^TA) = \dim E_\lambda (AA^T)$.
I started by showing that $\dim E_\lambda (A^TA) \le \dim E_\lambda (AA^T)$:
Let $\tilde{f}$ be the restriction of $f$ on $E_\lambda (A^TA)$. We can prove that $\tilde{f}$ is injective:
Let $x, x'$ such that $\tilde{f}(x) = \tilde{f}(x')$, then $$Ax = Ax'$$ thus $$A^TAx = A^TAx'$$ $i.e$ $$\lambda x = \lambda x'$$ as $\lambda \ne 0$, $x=x'$ wich shows that $\tilde{f}$ is injective.
We can also prove that $\operatorname{Im} \tilde{f} \subset E_\lambda (AA^T)$:
Let $y \in \operatorname{Im} \tilde{f}$, there exists $x \in E_\lambda (A^TA)$ such that $y = Ax$, hence $$AA^T y = AA^T Ax = \lambda Ax = \lambda y.$$ Now, using Rank-Nullity theorem, we can write $$\dim E_\lambda (A^TA) = \dim \operatorname{Im} \tilde{f} \le \dim E_\lambda (AA^T).$$
To finish, I would have to prove that $$\dim E_\lambda (A^TA) \ge \dim E_\lambda (AA^T).$$
I can't figure out how to do it, any help would be greatly appreciated.
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $\mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $\mathbb{R}$.