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

Question

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.

score 3 · Accepted Answer · answered Mar 09 '18 at 16:18

3

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 $Tv = 0$.)

answered Mar 09 '18 at 16:18

Arthur

199,419

thank you... should have seen this before. – Iconoclast Mar 09 '18 at 16:57

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

1 Answers1

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