Please help me understand this proof of why $(T^{-1})^\ast = (T^\ast)^{-1}$

Question

Let $V,W$ be finite-dimensional inner product spaces over $\mathbb C$. Let $T \in \mathcal L(V,W)$ be an invertible linear transformation. Prove that $(T^{-1})^\ast = (T^\ast)^{-1}$.

Here is the proof my professor gave of this: by definition, $\langle Tx,y \rangle = \langle x,T^\ast y\rangle$ for all $x \in V, y \in W$. Then, we set $x = T^{-1}z$. I don't really understand why we do this step, this is where I have trouble.

Applying this definition, we get

$$ \langle z,y \rangle = \langle TT^{-1}z,y\rangle = \langle T^{-1}z, T^\ast y \rangle $$

Then, by definition of adjoint,

$$ \langle T^{-1}z,T^\ast y \rangle = \langle z, (T^{-1})^\ast T^\ast y \rangle $$

From this we can see:

$$ \langle z, (T^\ast)^{-1} T^\ast y \rangle = \langle z,y \rangle = \langle z, (T^{-1})^\ast T^\ast y \rangle $$

Then, by the Riesz Representation Theorem,

$$ (T^\ast)^{-1} T^\ast y = (T^{-1})^\ast T^\ast y $$

So we can conclude

$$ (T^{-1})^\ast = (T^\ast)^{-1} $$

Can someone help me make sense of this proof? I really don't understand why we need to set $x$ to some other vector, or how this helps us prove the statement.

Answer 1

To verify directly, $$ TT^{-1}=I=T^{-1}T \\ (TT^{-1})^*=I=(T^{-1}T)^* \\ (T^{-1})^*T^*=I=T^*(T^{-1})^* $$ The last line implies that $T^*$ has an inverse and that inverse is $(T^{-1})^*$. In other words, $$ (T^*)^{-1}=(T^{-1})^*. $$

Answer 2

This feels like the theorem that shows that $T$ being invertible implies $T^*$ is invertible, but in that case there are a couple steps that beg the question without further explanation.

This is how I'd prove the same:

Let $x,y\in W$. Then $$\langle x, (T^{-1})^*T^*y\rangle = \langle T^{-1}x, T^*y\rangle = \langle TT^{-1}x,y\rangle = \langle x,y\rangle = \langle x, Iy\rangle,$$ so $$\langle x, ((T^{-1})^*T^*-I)y\rangle = \langle x, (T^{-1})^*T^*y\rangle - \langle x,Iy\rangle = 0.$$ In particular, by setting $x = ((T^{-1})^*T^*-I)y$, we see $x=0$, and since this is true for all $y$, we see $(T^{-1})^*T^*=I.$

Similarly, suppose $x,y \in V$. Then $$\langle x, T^*(T^{-1})^*y\rangle = \langle Tx, (T^{-1})^*y\rangle = \langle T^{-1}Tx, y\rangle = \langle x,y\rangle = \langle x, Iy\rangle.$$ Then, setting $x = (T^*(T^{-1})^*-I)y$ shows $x = 0$, and since this is true for all $y$, we see $T^*(T^{-1})^*=I.$

We conclude $(T^*)^{-1} = (T^{-1})^*$.