Canonical isomorphism between spaces of endomorphisms

Question

In my course notes, given $E$ a $K$-vector space, the canonical isomorphism ($E^{*}$ is the dual space of $E$) between spaces of endomorphisms:

$$ \Phi : End(E) \to End(E^{*}) $$

is constructed via the transposition, i.e. it maps $u \mapsto ({}^{t}u : f \mapsto f \circ u)$. Then it says when $E = K^{n}$, from the canonical isomorphism: $$ \psi : E \to E^{*} $$

which maps $x \mapsto (\phi_{x} : y \mapsto \langle x,y \rangle)$, there is a composed isomorphism:

$$ M_{n}(K) \to End(E) \overset{\Phi}{\to} End(E^{*}) \overset{\psi}{\to} End(E) \to M_{n}(K) $$

which is nothing but the matrix transposition.

I still do not understand how the isomorphism $End(E^{*}) \overset{\psi}{\to} End(E)$ can be derived from the isomorphism $\psi: E \to E^{*}$.

Answer 1

Take $f \in \operatorname{End}(E^*)$ to $\psi^{-1} \circ f \circ \psi \in \operatorname{End}(E)$. It's an isomorphism since the inverse is given by $g \mapsto \psi \circ g \circ \psi^{-1}$.

To add some intuition, an isomorphism should be able to transport any relevant structure back and forth between its domain and target. You can think of it as relabeling the elements of the domain by elements of the target. So if you have an endomorphism $g$ of $X$ that takes $u$ to $v$ and an isomorphism $\psi : X \to Y$, then you get an endomorphism of $Y$ that takes $\psi(u)$ to $\psi(v)$. If we want to write this directly in terms of elements of $Y$, that is, know what it sends $y \in Y$ to, we first find the element $u \in X$ for which $\psi(u) = y$, equivalently $u = \psi^{-1}(y)$. Then $v = g(u) = g(\psi^{-1}(y))$ and finally $\psi(v) = \psi(g(\psi^{-1}(y)))$. So on $Y$ we get exactly the map $\psi \circ g \circ \psi^{-1}$. If you start with an endomorphism $f$ of $Y$ then you switch the roles of $\psi$ and $\psi^{-1}$ so $f$ is taken to $\psi^{-1} \circ f \circ \psi \in \operatorname{End}(X)$.

Answer 2

Just to elaborate the answer of @ronno (and take note for myself), one way to understand the morphism $End(E^{*}) \overset{\psi}{\to} End(E)$ is to look at the adjoint via scalar product and the transposition.

For $u \in L(E)$, the adjoint $u^{*}$ satisfies $\langle u(x) \mid y \rangle = \langle x \mid u^{*}(y) \rangle $ and the transposition ${}^{t}u$ satisfies $\langle u(x), f \rangle = \langle x, {}^{t}u(f) \rangle$ for all $x,y \in E$ and $f \in L(E)$. There is also the canonical isomorphism $\psi: E \to E^{*}$ which maps $x \mapsto \langle x \mid \_ \rangle$.

Now consider

$$ \begin{align*} \langle u(x) \mid y \rangle & = \langle u(x), \psi(y) \rangle = \langle x, {}^{t}u \circ \psi (y) \rangle & \\ & = \langle x \mid u^{*}(y) = \langle x, \psi \circ u^{*}(y) & \end{align*} $$

so ${}^{t}u \circ \psi (y) = \psi \circ u^{*}(y)$, or $u^{*} = \psi^{-1} \circ {}^{t}u \circ \psi $ and we have the derived isomorphism

$$ \begin{align*} End(E^{*}) & \overset{\psi}{\to} End(E) \\ {}^{t}u & \mapsto u^{*} \end{align*} $$