Let $U, V,$ and $W$ be finite dimensional vector spaces with inner products. If $A: U \rightarrow V$ and $B: V \rightarrow W$ are linear maps with adjoints $A^{*}$ and $B^{*},$ define the linear map $C: V \rightarrow V$ by $$ C=A A^{*}+B^{*} B $$ If $U \stackrel{A}{\longrightarrow} V \stackrel{B}{\longrightarrow} W$ is exact , then my question is how to show $C: V \rightarrow V$ is invertible?
Here is my thought. We obviously have $W \stackrel{B^*}{\longrightarrow} V \stackrel{A^*}{\longrightarrow} U$ is exact. Then $V=\text{Im}(B^*)\oplus \text{Im}(A)=\text{Ker}(A^*)\oplus\text{Ker}(B)$ thanks to it. Hence,
$C(V)=AA^*(\text{Ker}(B))+B^*B(\text{Ker}(A^*))=AA^*(\text{Im}(A))+B^*B(\text{Im}(B^*)),$ but what should I do next? I still wonder how to solve this problem by the matrix method.
Any help would be appreciated. Thanks a lot!
As you observed, there is a decomposition $V=E\oplus F$ where $E=\ker(B)=\operatorname{im}(A)$ and $F=\ker(A^*)=\operatorname{im}(B^*)$. Note that $AA^*$ vanishes on $F$ and maps $E$ to itself. Moreover, $\ker(AA^*)=\ker(A^*)=F$ (if $AA^*v=0$ then $0=\langle AA^*v,v\rangle=\langle A^*v,A^*v\rangle$ so $A^*v=0$), so $AA^*$ is injective when restricted to $E$. This means that the restriction of $AA^*$ to a map $E\to E$ is invertible. Similarly, $B^*B$ vanishes on $E$ and restricts to an invertible map $F\to F$.
So, writing $C$ as a block matrix with respect to the decomposition $V=E\oplus F$, it is $\begin{pmatrix} AA^* & 0 \\ 0 & B^*B\end{pmatrix}$ where both diagonal entries are invertible (as maps $E\to E$ and $F\to F$). Thus $C$ is invertible.