Eigenvalues of Adjoint Transformation

Question

\begin{array} { l } { \text { Let } T : E \rightarrow F \text { be a linear transformation between finite dimension vector spaces.} } \\ { \text { Show that } T ^ { * } T \text { and } T T^ { * } \text { share the same non zero eigenvalues (T* is the adjoint of T). } } \end{array}.

Answer 1

If $T^*Tu=\lambda u$ with $u\ne0$ and $\lambda\ne0$, then $v=Tu$ is also no zero and:

$$T(T^*Tu)=TT^*(Tu) = \lambda(Tu)~\Rightarrow TT^*v=\lambda v\,.$$

So $v=Tu$ is an eigenvector of $TT^*$ with the same eigenvalue $\lambda$, i.e. $\lambda$ is an eigenvalue of both $TT^*$ and $T^*T$. Note that $T^*$ does not need to be the adjoin of $T$, it could be any other linear transformation from $F\to E$.