Math 202A: Lecture 6

Last lecture, we gave a definition of what it means for a Hilbert space to be finite-dimensional. Let $\mathbf{FVec}$ be the category whose objects are finite-dimensional Hilbert spaces $V$ and whose morphisms are linear transformations. Note that we do not require maps in $\mathrm{Hom}(V,W)$ to interact with the scalar products on $V$ and $W$ in any particular way. This means that even though the objects in $\mathbf{FVec}$ are both algebraic and geometric in nature, as described in Lecture 4, the morphisms of $\mathbf{FVec}$ are only required to respect their algebraic structure.

Of course, we are free to consider special linear transformations in $\mathrm{Hom}(V,W)$ which do interact with the geometry of the source and target spaces. In particular, the following definition is quite important.

Definition 6.1. A transformation $A \in \mathrm{Hom}(V,W)$ is said to be an isometry if $\|Av\|=\|v\|$ for all $v \in V.$

Thus an isometry is a linear transformation which preserves norms and hence distances.

Theorem 6.2. A transformation $A \in \mathrm{Hom}(V,W)$ is an isometry if and only if $\langle Av_1,Av_2\rangle = \langle v_1,v_2\rangle$ for all $v_1,v_2 \in V.$

Proof: It is clear that if $A$ preserves scalar products it preserves norms. That the converse also holds follows from the fact that norms uniquely determine scalar products via polarization. $\square$

Proposition 6.3. If $A \in \mathrm{Hom}(V,W)$ is an isometry then it is injective.

Proof: Suppose $v_1,v_2 \in V$ are such that $Av_1=Av_2$ . Then, $A(v_1-v_2) = 0_w$ and consequently $\|A(v_1-v_2)\|=\|v_1-v_2\|=0.$ $\square$

For any $A \in \mathrm{Hom}(V,W)$ we define the kernel of $A$ by

$\mathrm{Ker}(A) = \{v \in V \colon Av = 0_W\}.$

Certainly you already know that $\mathrm{Ker}(A)$ is a subspace of $V$ . We will discuss subspaces in a systematic way soon. For now, let us simply note that in the course of the preceding argument we observed the following.

Proposition 6.4. A transformation $A \in \mathrm{Hom}(V,W)$ is injective if and only if $\mathrm{Ker}(A) = \{0_W\}.$

Now let us discuss isomorphism in the category $\mathbf{FVec}.$ The general categorical definition is that two Hilbert spaces $V$ and $W$ are isomorphic if and only if there exists transformations $A \in \mathrm{Hom}(V,W)$ and $B \in \mathrm{Hom}(W,V)$ such that

$B \circ A = I_V \quad\text{and}\quad A \circ B=I_W,$

where $I_V \in \mathrm{End}(V)$ and $I_W \in \mathrm{End}(W)$ are the identity operators on $V$ and $W$ , respectively. We showed already in a previous homework problem that $A \in \mathrm{Hom}(V,W)$ is an isomorphism if and only if it is bijective. Here is a restatement of this as an extremal property of linear transformations.

Proposition 6.5. A transformation $A \in \mathrm{Hom}(V,W)$ is an isomorphism if and only if its kernel is minimal and its image is maximal.

Now we come to the fact that dimension completely characterizes isomorphism in $\mathbf{FVec}.$

Theorem 6.6. Two Hilbert spaces $V$ and $W$ are isomorphic if and only if they have the same dimension.

Proof: Suppose first that $V$ and $W$ are two Hilbert spaces of the same dimension, and let $X \subset V$ and $Y \subset W$ be two orthonormal bases. Then $|X|=|Y|.$ Let $f \colon X \to Y$ be any bijective function (i.e. set isomorphism) and define a corresponding linear transformation

$A_f \colon V \longrightarrow W$

$A_f(v) = \sum\limits_{x \in X} \langle x,v\rangle f(x).$

Since $f$ is a bijection from $X$ to $Y$ , we also have

$A_f(v) = \sum\limits_{y \in Y} \langle f^{-1}(y),v\rangle y,$

so that $A_f(v_1) = A_f(v_2)$ if and only if $v_1,v_2 \in V$ have the same coordinates relative to $X$ , in which case they are the same vector. Next, let $w \in W$ be an arbitrary vector and let

$w = \sum\limits_{y \in Y} \langle y,v \rangle y$

be its coordinate expansion relative to the orthonormal basis $Y$ . Then, the vector $v \in V$ defined by

$v = \sum\limits_{x \in X} \langle y,v\rangle f^{-1}(x)$

satisfies $A_f(v)=w$ , whence $A_f$ is surjective as well as injective.

Conversely, suppose that $A \colon V \to W$ and $B \colon W \to V$ are linear transformations such that $AB=I_W$ and $BA=I_V,$ where $I_V$ is the identity transformation of $V$ and $I_W$ is the identity transformation of $W.$ Let $X \subset V$ be a basis of $V$ and consider the $|X|=\dim V$ vectors in $W$ defined by

$Ax, \quad x \in X$

These vectors form the image of $X$ under $A$ in $V,$ and we claim they are linearly independent. Indeed, suppose that

$\sum\limits_{x \in X} \alpha_x Ax = 0_W.$

Then, applying $B$ to each side of this equation we get

$\sum\limits_{x \in X} \alpha_x x=0_V,$

whence $\alpha_x=0$ for each $x \in X.$ Thus $Y=\{Ax \colon x \in X\}$ is a linearly independent set in $W$ of cardinality $|X|$ , so that $\dim W \geq |X|=\dim V$ . Applying exactly the same argument with $B$ in place of $A$ we get that $\dim V \geq \dim W.$ We thus conclude $\dim V=\dim W.$

-QED

Having discussed isomorphisms in $\mathbf{FVec},$ let us now consider isomorphisms which preserve geometric structure.

Definition 6.7. A transformation $U \in \mathrm{Hom}(V,W)$ is called an isometric isomorphism if it is both an isometry and an isomorphism.

Isometric isomorphisms in $\mathbf{FVec}$ are also referred to as unitary transformations. Since every isometry is injective we have the following.

Proposition 6.8. An isometry $A \colon V \to W$ is an isomorphism if and only if it is surjective.

Having discussed isomorphisms of Hilbert spaces, we move on to automorphisms. As we have already seen, the automorphisms of any object in a locally small category form a group under composition. The automorphisms group $\mathrm{Aut}(V)$ of a Hilbert space is usually called the general linear group of $V$ and denoted $\mathrm{GL}(V).$ Thus, automorphisms $A \in \mathrm{GL}(V)$ are linear bijections $A \colon V \to V.$

Proposition 6.9. A transformation $A \in \mathrm{End}(V)$ is bijective if and only if it is injective.

Proof: One direction is obvious: bijections are injective. Conversely, suppose that $A \colon V \to V$ is an injective linear operator and let $X \subset V$ be a basis. Then, $A(X) = \{Ax \colon x \in X\}$ is also a basis of $V.$ Indeed, injectivity insures that $|A(X)|=|X|$ , and if $\alpha_x,$ $x \in X$ are scalars such that

$\sum\limits_{x \in X} \alpha_x Ax = 0_V$

then by linearity we have

$A \sum\limits_{x \in X} \alpha_x x=0_V.$

Since $A$ has minimal kernel this forces

$\sum\limits_{x \in X} \alpha_x x = 0_V$

which in turn implies $\alpha_x=0$ for all $x \in X$ because $X$ is a linearly independent set. $\square$

Automorphisms of $V$ are not required to preserve scalar products, but nevertheless there is a characterization of automorphisms which does involve the scalar product.

Theorem 6.10. An operator $A \in \mathrm{End}(V)$ is an automorphism if and only if the function

$\langle \cdot,\cdot \rangle_A \colon V \times V \longrightarrow \mathbb{C}$

defined by $\langle v,w \rangle_A = \langle Av,Aw\rangle$ is a scalar product on $V.$

Proof: Suppose first that $\langle \cdot,\cdot\rangle_A$ is a scalar product. Then $\langle v,v \rangle_A = 0$ if and only if $v=0_V.$ Consequently, $\|Av\|=0$ if and only if $v = 0_V$ , hence $A$ is injective and thus also bijective by Proposition 6.9.

Now suppose $A \in \mathrm{GL}(V)$ . We have to check that $\langle v,w\rangle_A$ is consistent with the scalar product axioms. First, observe that the set of vectors $v \in V$ such that $\langle v,v\rangle_A = \langle Av,Av\rangle=0$ is exactly the kernel of $A,$ which is minimal by hypothesis. Hermitian symmetry is clear,

$\langle w,v\rangle_A = \langle Aw,Av\rangle = \overline{\langle v,w\rangle_A}.$

Linearity in the second slot follows from the fact that $A$ is linear and $\langle \cdot,\cdot \rangle$ is a scalar product. $\square$

Having discussed automorphisms we move on to isometric automorphisms, i.e. unitary operators. Let $\mathrm{U}(V)$ be set of operators $Y \in \mathrm{End}(V)$ such that $\langle Uv,Uw \rangle = \langle v,w\rangle$ for all $v,w \in V.$

Proposition 6.11. The set $\mathrm{U}(V)$ is a subgroup of $\mathrm{GL}(V).$

The proof of Proposition 6.11 is straightforward and left as an exercise. You can also check for yourself that $\mathrm{U}(V)$ is not a normal subgroup of $\mathrm{GL}(V)$ .

Math 202A: Lecture 6

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