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I'm trying to prove the existence of the linear map $f^*$ in the following theorem about the adjoint:

Let $f: V \to W$ be a linear map, with $V$ finite-dimensional and $V,W$ inner product spaces over the same field. Then there is a unique linear map $f^*: W \to V$ such that $$\langle f(v),w \rangle = \langle v, f^*(w)\rangle$$ for all $v\in V, w \in W$.

Here's how far I got so far.

Try 1:

Fix $w \in W$. Because $V$ is finite-dimensional there exists a unique $x \in V$ for all linear forms $\psi_1 = \langle f(\cdot),w \rangle \in V^*$ --- namely we have the isomorphism $V \to V^*$. Since the linear form $\langle f(\cdot),w \rangle$ is uniquely determined by $w \in W$, define the linear map $f^*$ by $x = f^*(w)$. Let $\psi_2 = \langle \cdot,f^*(w) \rangle \in V^*$ and $\theta = \psi_1 - \psi_2 \in V^*$.

Now consider ker$(\theta)$. If ker$(\theta)$ = $V$ then our claim is proven --- namely $\psi_1 = \psi_2$.

Mistake: Firstly $\psi_1 \in W^*$, and not in $V^*$. And because $W \to W^*$ need not be an isomorphism the argument doens't follow.

EDIT: $\psi_1 \in V^*$ indeed. Namely $\psi_1: v \mapsto \langle f(v),w \rangle$ and $w \in W$ is fixed. However note that $\psi_1$ uses an inner product on $W$.

EDIT: Note that rk$(\theta) = 1$, so we have dim$\big($ker$(\theta)\big) = $ dim$(V) - 1$. Which fails my attempted proof. Namely take dim$(V) = 1$. So that, ker$(\theta)$ = $\{0\}$. And thus for all $0 \neq v \in V$ we have $\psi_1 \neq \psi_2$.

Try 2: the same as Try 1 but with $v = x$.

I'm assuming that the argument can be proven by a less cumbersome method. Also I think that my method is not correct, since I seem to get stuck.

Suggestions or an alternative method are welcome.

EDIT: Here's a summary of Fischer's proof.

Define the linear map $\lambda: W \to V^*$ by $$\lambda(w) = \langle \cdot, w \rangle_W \circ f.$$ Next define the isomorphism $\sigma: V \to V^*$ by $$\sigma(v_0) = \langle \cdot, v_0 \rangle_V.$$ Now, consider the composition $$\varphi: \sigma^{-1} \circ \lambda: W \to V.$$ We then have $$\langle v, \varphi(w) \rangle_V = \langle f(v), w \rangle_W$$ for all $v \in V, w \in W$. We can now define $f^*:= \varphi$. This proves the theorem for the existence part.

For uniqueness, assume that $\psi$ satisfies $\langle f(v),w \rangle_W = \langle v, \psi(w) \rangle_V$. Fixing $w_0 \in W$, we get $$\langle v,\psi(w_0) - f^*(w_0) \rangle_V = 0$$ for all $v \in V$. Choosing $v = \psi(w_0) - f^*(w_0)$ we get $\psi = f^*$, which proves the uniqueness part.

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    $\begingroup$ Your notation seems to imply that both spaces are endowed with the same inner product, is this the case? As a hint may I suggest that you write out the statement you wish $f^*$ to satisfy, and then see if you can construct $f^*$ using the function $f$ and the isomorphisms between $W,W^*$ and $V,V^*$. $\endgroup$ – R R Dec 3 '13 at 19:03
  • $\begingroup$ $V$ and $W$ are not endowed with the same inner product and I have listed the mistake in the question. $\endgroup$ – Mussé Redi Dec 3 '13 at 20:13
  • $\begingroup$ What do you mean with the isomorphisms between $W,W^*$ and $V,V^*$? If you mean Hom$(W,W^*) \to$Hom$(V,V^*)$, then why does there exist such an isomorphim. Note that the isomorphism $W \to W^*$ need not exist. $\endgroup$ – Mussé Redi Dec 4 '13 at 1:30
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Let's take a step back, and look at it from a more abstract perspective.

First, we define a map $\lambda \colon W \to V^\ast$ by

$$\lambda(w) \colon v \mapsto \langle f(v), w\rangle_W,$$

or $\lambda(w) = \langle\,\cdot\,,w\rangle_W \circ f$. Now, looking at it, we notice that $\lambda$ is linear (or, if the scalar field is $\mathbb{C}$, and the inner product linear in the first and antilinear in the second argument, $\lambda$ is antilinear; if the spaces are complex and the inner product is antilinear in the first, and linear in the second argument, consider $\langle w,f(v)\rangle_W$ instead):

$$\begin{align} \lambda(w_1+w_2)(v) &= \langle f(v), w_1 + w_2\rangle_W\\ &= \langle f(v), w_1\rangle_W + \langle f(v),w_2\rangle_W\\ &= \lambda(w_1)(v) + \lambda(w_2)(v)\\ &= \bigl(\lambda(w_1) + \lambda(w_2)\bigr)(v), \end{align}$$

and since $v$ was arbitrary, that means $\lambda(w_1+w_2) = \lambda(w_1) + \lambda(w_2)$. And for a scalar $s$, we have

$$\begin{align} \lambda(s\cdot w)(v) &= \langle f(v), s\cdot w\rangle_W\\ &= s\cdot \langle f(v), w\rangle_W\\ &= s\cdot \lambda(w)(v)\\ &= \bigl(s\cdot \lambda(w)\bigr)(v), \end{align}$$

that is, $\lambda(s\cdot w) = s\cdot \lambda(w)$ (for complex scalars, we would have to conjugate $s$ and $\lambda$ would be antilinear).

Now, with the (anti-) isomorphism $\sigma \colon V \to V^\ast$, given by $\sigma(v) = \langle\,\cdot\,,v\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle w, v\rangle_V$ (or $\sigma(v) = \langle v,\,\cdot\,\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle v,w\rangle_V$ if $V$ is a complex vector space and the inner product is antilinear in the first argument, as is customary in physics), consider the composition

$$\varphi = \sigma^{-1} \circ \lambda \colon W \to V.$$

Then $\varphi$ is linear (as the composition of two linear maps in the real case, and as the composition of two antilinear maps in the complex case), and for $v\in V,\; w \in W$, we have

$$\begin{align} \langle v, \varphi(w)\rangle_V &= \langle v, \sigma^{-1}(\lambda(w))\rangle_V\\ &= \lambda(w)(v)\tag{definition of $\sigma$}\\ &= \langle f(v),w\rangle_W,\tag{definition of $\lambda$} \end{align}$$

and that means we can define $f^\ast := \varphi = \sigma^{-1}\circ \lambda$.

It remains to be seen that there is only one such map $W \to V$.

Suppose that $\psi \colon W \to V$ is a map with $\langle f(v), w\rangle_W = \langle v, \psi(w)\rangle_V$ for all $v\in V,\; w\in W$.

Fix an arbitrary $w_0\in W$. Then for all $v\in V$

$$\begin{align} \langle v,\psi(w_0) - f^\ast(w_0)\rangle_V &= \langle v,\psi(w_0)\rangle_V - \langle v, f^\ast(w_0)\rangle_V\\ &= \langle f(v), w_0\rangle_W - \langle f(v), w_0\rangle_W\\ &= 0. \end{align}$$

In particular, we can choose $v = \psi(w_0) - f^\ast(w_0)$, so we have

$$\langle \psi(w_0) - f^\ast(w_0), \psi(w_0) - f^\ast(w_0)\rangle_V = 0,$$

which means $\psi(w_0) - f^\ast(w_0) = 0$. Since $w_0 \in W$ was arbitrary, it follows that $\psi = f^\ast$, and the uniqueness is also established.

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  • $\begingroup$ How does $\langle v, \sigma^{-1}(\lambda (w)) \rangle_V = \lambda (w)(v)$ follow from the definition of $\sigma$? $\endgroup$ – Mussé Redi Dec 3 '13 at 23:20
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    $\begingroup$ The definition of $\sigma \colon V \to V^\ast$ is $\sigma(v_0)(v) = \langle v, v_0\rangle_V$. And that means $\mu(v) = \sigma(\sigma^{-1}(\mu))(v) = \langle v, \sigma^{-1}(\mu)\rangle_V$ for $\mu \in V^\ast$. $\endgroup$ – Daniel Fischer Dec 3 '13 at 23:24
  • $\begingroup$ How is $\langle \cdot, w \rangle_W \circ f$ a correct notation for $\lambda(w)$? $\endgroup$ – Mussé Redi Dec 3 '13 at 23:56
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    $\begingroup$ Both denote the map $v \mapsto \langle f(v), w\rangle_W$. Of course one may have qualms about a notation $\langle \,\cdot\,,w\rangle_W$, then ignore that part. $\endgroup$ – Daniel Fischer Dec 3 '13 at 23:59
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    $\begingroup$ $\langle v,w\rangle_W$ doesn't even make sense, since $v\in V$. It's the composition of the two maps $f\colon v \mapsto f(v)$, and $\langle\,\cdot\,,w\rangle_W\colon z \mapsto \langle z, w\rangle_W$. The value that $z$ is instantiated with is the result of the application of $f$, namely $f(v)$, so $$\bigl(\langle\,\cdot\,,w\rangle_W\circ f\bigr)(v) = \langle\,\cdot\,,w\rangle_W \bigl(f(v)\bigr) = \langle f(v),w\rangle_W.$$ $\endgroup$ – Daniel Fischer Dec 4 '13 at 0:09
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Fix $w\in W$. Consider the linear form on $V$ that maps $v\in V$ to $\langle f(v),w\rangle$. Let $f^{*}w$ be the unique vector in V such that this linear form is given by taking inner products with $f^{*}w$. So, the $f^{*}w$ is the unique vector such that $\langle f(v),w\rangle= \langle v, f^{*} w\rangle \; \forall v\in V.$

I think the existence of the adjoint (as described above) depends a lot on the fact that

$f^{*}w$ is the unique vector in V such that this linear form is given by taking inner products with $f^{*}w$

which you seem to also use in your fist try. Does this help?

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  • $\begingroup$ It doesn't answer my question: I want to prove that such a $f^*$ exists. Indeed, it depends on equality of the two linear forms, but how to show that such a $f^*$ exists? $\endgroup$ – Mussé Redi Dec 2 '13 at 17:21
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This essentially follows from Riesz Representation theorem. That is, each linear functional $f(x)$ may be written as $<x,w_f>$ for some vector $w_f$. To prove Riesz Representation theorem, fix an orthonormal basis and write $x$ as linear combination of those...

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