On exact sequences of finite-dimensional inner product spaces

Question

Consider the exact sequence $U \stackrel{f}{\to} V\stackrel{g}{\to}W$ of finite-dimensional inner product spaces. Show that the sequence $W\stackrel{g^*}{\to}V\stackrel{f^*}{\to}U$ is exact. In other words I want to show that $\text{im}(g^*)=\ker (f^*)$. It seems to me that the names of the maps $g^*$ and $f^*$ are suggesting that they're the dual maps or adjoints, but I'm not sure what to do with that.

I'm really struggling with this problem so any help is greatly appreciated.

Answer 1

Yes, it seems that $f^*$ and $g^*$ are the adjoint maps of $f$ and $g$. In other words, $f^*$ is defined such that for all $u \in U$ and $v \in V$, we have $\langle f(u) , v\rangle = \langle u, f^*(v)\rangle$.

One approach is to first show that $\operatorname{im}(g^*) \subseteq \ker(f^*)$, then use the dimensions of $\operatorname{im}(g^*)$ and $\ker(f^*)$ to deduce that $\operatorname{im}(g^*) = \ker(f^*)$. For the first part of this approach, note that it generally holds for composable linear maps $\alpha,\beta$ over inner product spaces that

• $\operatorname{im}(\beta) \subseteq \ker(\alpha)$ iff $\alpha \circ \beta = 0$,
• $(\alpha \circ \beta)^* = \beta^* \circ \alpha^*$.

Thus, we have $$ f^* \circ g^* = (g \circ f)^* = 0^* = 0, $$ which means that $\operatorname{im}(g^*) \subseteq \ker(f^*)$. From there, use the rank-nullity theorem to deduce that $\dim \operatorname{im}(g^*) = \dim \ker(f^*)$.

Answer 2

Another way is to use the following:

If $T \colon E \to F$ is a linear map between finite dimensional inner product spaces, then

1. $\newcommand{\im}{\operatorname{im}} \ker(T^*) = \im(T)^\perp$, and
2. $\im(T^*) = \ker(T)^\perp$;

where $(\_)^\perp$ denotes the orthogonal complement.

Proof: Given $y \in F$, \begin{align} y \in \ker(T^*) & \iff T^*(y)=0 \\ & \iff \forall x \in E, \ \langle x,T^*(y) \rangle =0 \\ & \iff \forall x \in E, \ \langle T(x),y \rangle =0 \\ & \iff y \in \im(T)^\perp. \end{align} This proves 1. Taking $(\_)^\perp$ on both sides of 1 we obtain $\im(T) = \ker(T^*)^\perp$ (since $(W^\perp)^\perp=W$ for a finite dimensional subspace $W$), and replacing $T$ with $T^*$ we obtain 2.