Linear Algebra MT23, Inner product spaces
Flashcards
What does it mean for a bilinear form $B : V \to V$ to be nondegenerate?
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?
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 $\mathrm{Im}\, \phi = V'$.
(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).
Bite-sized
Real inner product axioms: symmetric, bilinear, positive definite. Complex inner product axioms: conjugate-symmetric, sesquilinear, positive definite. The conjugation convention is in the first argument: $\langle \lambda v, w \rangle = \overline{\lambda} \langle v, w \rangle$.
What is the “natural injective map” $\phi : V \to V'$ induced by an inner product?
$\phi(v) = \langle v, \cdot \rangle$ — sends each vector to “take the inner product with me”. Over $\mathbb R$ this is $\mathbb R$-linear and injective; in finite dimension it’s an isomorphism by a dimension count. Over $\mathbb C$ it’s only $\mathbb R$-linear (not $\mathbb C$-linear, because the first argument conjugates scalars), so it’s a real-linear bijection rather than a complex-linear iso. This is the foundation of Riesz representation in finite dimension.