# Show that the eigenvalues of $T^*T$ and $TT^*$ are the same for a linear operator $T$ on finite dimensional space. [duplicate]

the following question has me stumped: Show that for a linear operator $T$ defined on a finite-dimensional space, the eigenvalues of $T^*T$ and $TT^*$ are the same. Its given in a tutorial chapter on finite-frame theory without proof.

Hint: If $\lambda$ is an eigenvalue of $T^*T$, and $v$ is a corresponding eigenvector, what happens if you apply $TT^*$ to $Tv$? (Caution: some care is needed if $\lambda = 0$ and $Tv = 0$.)