Prove that for any square matrix A, the eigenvalues of AAᵀ and AᵀA are the same (excluding zero).

Step 1: Define Eigenvalues and Eigenvectors

An eigenvalue $\lambda$ of a matrix $M$ is a scalar such that there exists a non-zero vector $v$ (called an eigenvector) satisfying the equation $Mv = \lambda v$. We want to show that the non-zero eigenvalues of $AA^T$ and $A^TA$ are identical.

Step 2: Consider an Eigenvalue of $A^TA$

Let $\lambda$ be a non-zero eigenvalue of $A^TA$ with a corresponding eigenvector $v$. This means $A^TAv = \lambda v$, where $v \neq 0$.

Step 3: Multiply by A

Multiply both sides of the equation $A^TAv = \lambda v$ by $A$ from the left. This gives $A(A^TAv) = A(\lambda v)$, which simplifies to $(AA^T)(Av) = \lambda (Av)$.

Step 4: Show $Av$ is an Eigenvector of $AA^T$

For $Av$ to be an eigenvector of $AA^T$, it must be non-zero. Since $\lambda \neq 0$ and $A^TAv = \lambda v$, if $Av = 0$, then $A^T(0) = \lambda v$, which implies $0 = \lambda v$. Since $\lambda \neq 0$, this would mean $v = 0$, contradicting our assumption that $v$ is a non-zero eigenvector. Therefore, $Av \neq 0$.

Step 5: Conclusion for one direction

Since $Av \neq 0$, the equation $(AA^T)(Av) = \lambda (Av)$ shows that $\lambda$ is an eigenvalue of $AA^T$ with $Av$ as its corresponding eigenvector. This proves that every non-zero eigenvalue of $A^TA$ is also an eigenvalue of $AA^T$.

Step 6: Symmetry of the Proof

The proof is symmetric. Let $\mu$ be a non-zero eigenvalue of $AA^T$ with eigenvector $w$. Then $AA^Tw = \mu w$. Multiplying by $A^T$ from the left gives $A^T(AA^Tw) = A^T(\mu w)$, which simplifies to $(A^TA)(A^Tw) = \mu (A^Tw)$. Similar to the previous steps, we can show that $A^Tw \neq 0$. Thus, every non-zero eigenvalue of $AA^T$ is also an eigenvalue of $A^TA$.

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