Lecture 31 - Finishing Complexification

From last time:

complexification of $V$ , $V_{C}$

Suppose $V$ is a real vector space.

The complexification of $V$ , denoted $V_{C}$ , equals $V \times V$ . An element of $V_{C}$ is an ordered pair $(u, v)$ where $u, v \in V$ , but we'll write this as $u + i v$ .
Addition on $V_{C}$ is defined by:
$(u_{1} + i v_{1}) + (u_{2} + i v_{2}) = (u_{1} + u_{2}) + i (v_{1} + v_{2})$
for $u_{1}, v_{1}, u_{2}, v_{2} \in V$ .
Complex scalar multiplication on $V_{C}$ is defined by:
$(a + b i) (u + i v) = (a u - b v) + i (a v + b u)$
for $a, b \in R$ and $u, v \in V$ .

... (see the Chapter 9 notes or last lecture for more details).

9.B: Normal Operators and Isometries on Real Inner Product Spaces

Let $V$ be a real inner product space with $T \in L (V)$ . Then:

$T$ is normal iff $\exists$ an orthonormal basis $β$ for which $M (T, β)$ is block diagonal, where each block is $1 \times 1$ or $2 \times 2$ .
1. If a block is $2 \times 2$ then it looks like $[\begin{matrix} a & - b \\ b & a \end{matrix}]$ with $b > 0$ . (this is because the matrix has to equal its conjugate transpose to be normal, which in the real vector spaces is just the transpose)
$T$ is an isometry iff $\exists$ an orthonormal basis $β$ for which $M (T, β)$ is block diagonal, where each block is $1 \times 1$ or $2 \times 2$ , EXCEPT:
1. The $1 \times 1$ blocks are $\pm 1$
2. The $2 \times 2$ blocks are of the form $[\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}]$

Looking at:

Every operator has an invariant subspace of dimension 1 or 2.

Every operator on a nonzero finite-dimensional vector space has an invariant subspace of dimension 1 or 2.

This is where the blocks of these sizes come from. Namely for $T \in L (V)$ that's normal:

If we have $2 \times 2$ blocks that represents a Self-Adjoint transformation we have $[\begin{matrix} a & 0 \\ 0 & b \end{matrix}]$
If it's not self-adjoint then consider normal. We require that $b = \pm c, \neq 0$ for the matrix $[\begin{matrix} a & c \\ b & d \end{matrix}]$ (since we can equate terms in the transpose). Calculating $M (T^{*} T) = M (T T^{*})$ shows that the matrix has to be $[\begin{matrix} a & - b \\ b & a \end{matrix}]$

Let's prove (1) above:
Proof
Let $T \in L (V)$ be normal and $V$ is real. Then $V$ has a dimension 1 or 2 subspace that's $T$ -invariant. Then:

V = U \oplus U^{⊥}

It turns out that $T |_{U}$ is normal and $T |_{U^{⊥}}$ is also normal. Thus, we can reapply the argument all the way down until we have only $\dim = 1, 2$ subspaces, which we know are of the form above:

V = ⨁_{j = 1}^{m} U_{j}

where $T |_{U_{j}}$ is normal and $\dim U_{j} = 1, 2$ .
☐

For the isometry, we can do something similar. If $T$ is an isometry where $\dim (V) = 2$ then we require that $T^{*} T = T T^{*} = I$ so then we require that:

M (T^{*} T) = M (I) \Leftrightarrow a^{2} + b^{2} = 1, c^{2} + d^{2} = 1

But since $T$ is not self-adjoint then $b \neq 0$ , so it seems clear that we should define $θ = \arccos (a)$ then have:

b = \sin (θ)

as a result, giving the matrix we desired. Notice that since $θ \in [0, π]$ then that makes $b > 0$ as required.