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I'm really new to linear algebra and I want to show that for $A \in \mathbb{R}^{n\times m}$, $AA^T$ and $A^TA$ share the same non-zero eigenvalues with geometric multiplicity $p$. What I have so far is just the part that shows they have the same eigenvalues.

It's clear that if $\lambda \not = 0$ is an eigenvalue of $A^TA$ with eigenvector $\mathbf{x} \not = 0$, then $A\mathbf{x} \not = 0$ is an eigenvector of $A A^T$ relative to $\lambda$.

How can I extend this proof to also count the geometric multiplicity?

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Let's take $\lambda \neq 0$ eigenvalue of $A^TA$ and a basis $\{v_1, \dots , v_p \}$ of the eigenspace relative to $\lambda$.
We know that $\{Av_1, \dots, Av_p \}$ are all eigenvectors relative to $\lambda$ for $AA^T$ and we want to show that they are all linearly indipendent.
Let's take a linear combination $a_1Av_1+\dots+a_pAv_p=0$, we know that $A(a_1v_1+\dots a_pv_p) = 0$ but this is impossible because $\lambda(a_1v_1+\dots a_pv_p)= A^TA(a_1v_1+\dots a_pv_p)= A^T(0)=0$.
So we have that the geometric multiplicity of $\lambda$ in $AA^T$ is greater than the geometric multiplicity of $\lambda$ in $A^TA$.
Can you conclude from here? (Notice the symmetry in the problem)

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The first part you've proved is ok. You've proved the linear map

$A: \mathbb{R}^m\to \mathbb{R}^n$

sends an eigenvector $v$ of $A^tA$ of eigenvalue $\lambda$ to the eigenvector $Av$ of $AA^t$ of eigenvalue $\lambda$. Moreover you can observe

$A(ker(A^tA-\lambda I))\subseteq ker(AA^t-\lambda I)$

and $ker(A^tA-\lambda I)\cap ker(A)=0$

so you've proved the set of eigenvalues of $A^tA$ is contained in the set of eigenvalues of $AA^t$ and

$p_\lambda^{A^tA}=dim (ker(A^tA-\lambda I))=dim (A(ker(A^tA-\lambda I)))\leq dim (ker(AA^t-\lambda I))=p_\lambda^{AA^t}$

Now if you consider the other map

$A^t: \mathbb{R}^n\to \mathbb{R}^m$

you get also the set of eigenvalues of $AA^t$ is contained in the set of eigenvalues of $A^tA$, i.e. the two matrixes has the same eigenvalues, and

$p_\lambda^{AA^t}\leq p_\lambda^{A^tA}$

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Every symmetric matrix is diagonalizable and its eigenvectors are orthogonal to each other (see http://www.maths.nuigalway.ie/~rquinlan/linearalgebra/section2-2.pdf), which immediately implies that the eigenbasis of a symmetric matrix is full rank, and each of its eigenvalues have geometric multiplicity equal to their algebraic multiplicity, i.e. there is at most one unique eigenvector associated to each (repeated) eigenvalue.

You have already concluded that both $P = A^TA$ and $Q = AA^T$ have the same eigenvalues, with their eigenvectors related by $u=Av$, where $u$ is an eigenvector of $Q$ and $v$ an eigenvector of $P$. It is trivial to see that both of these matrices are symmetric, therefore the geometric multiplicities of their eigenvalues are equivalent.

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