Riesz Representation Theorem

For any finite-dimensional inner product space $\mathbf{V}$ and any $\phi _v \in V'$, there exist a unique $v \in V$ such that

$$ \phi_v (u) = \lang u \space | \space v\rang \hspace{15pt} \forall u \in V $$

Moreover, since $v$ is unique, this result says that the map $\mu : V \rightarrow V'$, for which $v \mapsto \phi_ v$, is an anti-linear isomorphism ($\phi_{\lambda v} = \bar{\lambda} \lang u \space | \space v\rang$).

The significance of this theorem comes from the fact that because this map is an isomorphism, we can equivalently represent any linear functional $\phi_v$ as a vector in $\mathbf{V}$, namely the vector $v$ for which taking the inner product with any other vector $u$ yields the same result as applying $\phi _v$ to $u$.

• Proof

Adjoints

The adjoint of a linear map $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$ is the unique linear map $T^* \in \mathscr{L}(\mathbf{W}, \mathbf{V})$ with the property that for all $v \in V$ and $w \in W$,

$$ \lang Tv \space | \space w\rang = \lang v \space | \space T^* w\rang $$

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Characteristics of Adjoints.

1. Anti-linearity. $(aS+bT)^* = \bar{a}S^* + \bar{b}T^*$
2. Idempotence. $(T^)^ = T$
3. Order Reversal. $(ST)^* = T^* S^*$
4. Invertibility. If $T$ is invertible, then $T^$ is invertible and $(T^)^{-1} = (T^{-1})^*$

• Proofs </aside>

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Lemma. The null spaces and ranges of $T$ and $T^*$ are related by orthogonal complements.

$$ \text{null } T^* = (\text{range }T)^\perp \\

\text{range } T^* = (\text{null }T)^\perp $$

• Proof </aside>

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Lemma. For orthonormal bases $\beta$ and $\gamma$ of $\mathbf{V}$ and $\mathbf{W}$ respectively,

$$ [T^*]{\gamma} ^\beta = (\overline{[T]{\beta}^\gamma})^T $$

• Proof </aside>

Self-Adjoint and Normal Operators

An operator is said to self-adjoint if it is its own adjoint and normal if it commutes with its adjoint.

$$ \text{$T$ is self-adjoint } \iff T= T^* \\ \hspace{113pt} \text{$T$ is normal } \iff TT^* = T^T \iff [T, T^] =0 $$

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Lemma. If $T$ is self-adjoint, all of its eigenvalues are real.

$$ T=T^* \implies \sigma (T) \sub \R $$

• Proof </aside>

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Lemma. A self-adjoint operator $T$ is $0$ iff any preimage is orthogonal to its image.

$$ T=0\iff \forall v \in V, \lang Tv \space | \space v \rang = 0 $$

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