Notes - Linear Algebra MT23, Inner product spaces

Created: November 23, 2023

Flashcards

What does it mean for a bilinear form $B : V \to V$ to be nondegenerate?
\[B(v, w) = 0 \text{ } \forall w \in V \implies v = 0\]

Let $V$ be a vector space over $\mathbb R$. What properties must an inner product $\langle \cdot, \cdot \rangle$ have?
  • Symmetric
  • Bilinear
  • Postive definite

What does non-degeneracy of a bilinear form imply about the associated Gram matrix?

It is invertible.

Let $V$ be a vector space over $\mathbb C$. What properties must an inner product $\langle \cdot, \cdot \rangle$ have?
  • Sesquilinear (bilinear, but scalar $\lambda$ in first argument becomes $\bar \lambda$).
  • Conjugate symmetric ($\langle v, w \rangle = \overline{\langle w, v \rangle}$).
  • Positive definite (note $\langle v, v \rangle$ is necessarily real).

A complex inner product $\langle \cdot, \cdot \rangle$ is sesquilinear form. Is it the first or second argument that conjugates scalars?

The first, i.e.

\[\langle \lambda v, \mu u \rangle = \bar \lambda \mu \langle v, u\rangle\]

Can you give an example of a map that is $\mathbb R$-linear but not $\mathbb C$-linear?
\[z \mapsto \overline z\]

Quickly prove that having an inner product $\langle \cdot, \cdot \rangle$ induces a natural injective $\mathbb R$-linear map $\phi : V \to V’$

\[\phi(v) = \langle v, \cdot \rangle\]

which is an isomorphism between real vector spaces when $V$ is finite-dimensional.

Note that this map is $\mathbb R$-linear as

\[\phi(\lambda v + w) = \langle \lambda v+w, \cdot\rangle = \lambda \langle v, \cdot \rangle + \langle w, \cdot \rangle\]

(note that this is not $\mathbb C$-linear, since if $\lambda \in \mathbb C$, then $\langle \lambda v, \cdot \rangle = \overline \lambda \langle v, \cdot \rangle$). It is also injective as if $\langle v, w \rangle = 0$ for all $w$, then $v = 0$ by non-degenerancy.

If $V$ is finite dimensional, then $\dim _ {\mathbb R} V = \dim _ {\mathbb R} V’$, so $\text{Im } \phi = V’$.

(This argument shows that if $V$ and $V’$ are real vector spaces, i.e. the field of scalars is $\mathbb R$, then $V$ and $V’$ are isomorphic. It also shows the existence of a bijective $\mathbb R$-linear map between complex vector spaces, but this is not necessarily an isomorphism since an isomorphism between complex vector spaces needs to be $\mathbb C$-linear).



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