Notes - Linear Algebra MT23, Orthogonal, unitary and normal transformations

\[T^\ast T = T^\intercal T = I\]

\[T^\ast T = \overline T^\intercal T = I\]

\[[ \quad e_1 \quad e_2 \quad \cdots \quad e_n \quad ]\]

1. $T$ preserves inner products, i.e. $\langle Tv, Tw \rangle = \langle v, w \rangle$.
2. $T$ preserves lengths, i.e. $ \vert \vert Tv \vert \vert = \vert \vert v \vert \vert $.

\[|\lambda| = 1\]

Let $v \ne 0$ be a $\lambda$-eigenvector. Then

\[\langle v, v\rangle = \langle Tv, Tv\rangle = \langle \lambda v, \lambda v\rangle = \lambda \overline \lambda \langle v, v\rangle\]

So $\lambda \overline \lambda = \vert \lambda \vert = 1$.

Determinant is the product of eigenvalues, all eigenvalues satisfy $ \vert \lambda \vert = 1$.

It is also $T$-invariant.

It is also $T^{-1}$-invariant. This comes in useful when showing that $U^\perp$ is $T$-invariant for orthogonal / unitary maps.

Since $T$ is orthogonal / unitary, we have $T^\star T = I$, so $T^\star = T^{-1}$. Then, if $v \in U^\perp$, for all $u \in U$,

\[\langle u, T(v) \rangle = \langle T^\star(u), v\rangle = \langle T^{-1}(u), v \rangle = \langle u', v\rangle = 0\]

where $u’ \in U$.

So the result follows from showing that $U$ is $T^{-1}$-invariant. This is because

\[m_T(x) = a_0 I + a_1 T + \cdots + a_{m-1}T^{m-1} +T^m\]

Then

\[T \cdot \frac{(a_1 + \cdots + a_{m-1}T^{m-2} + T^{m-1})}{-a_0} = I\]

How can we be sure that $a _ 0$ is non-zero? Since $T$ is orthogonal / unitary, it is in particular invertible. That means that $0$ is not an eigenvalue, so $x$ cannot be a root of $m _ T$, which implies it has a non-zero constant term.

Hence $T^{-1}$ is polynomial is $T$ and since $U$ is $T$-invariant (and so any polynomial combination of $T$ will also be), $U$ is therefore $T^{-1}$ invariant.

There exists an orthonormal basis of $V$ given by eigenvectors of $T$.

There exists an orthonormal basis of $V$ such that

\[T \sim \begin{bmatrix} I & & & \\\\ & -I & & \\\\ & & R_{\theta_1} & & \\\\ & & & \ddots & \\\\ & & & & R_{\theta_\ell} \\\ \end{bmatrix}\]

where each

\[R_{\theta_i} = \begin{bmatrix} \cos \theta_i & -\sin\theta_i \\\\ \sin \theta_i & \cos \theta_i \end{bmatrix} \quad \theta_i \neq 0, \pi.\]

Define

\[\begin{aligned} S &:= T + T^{-1} \\\\ &= T + T^\star \end{aligned}\]

Note then that $S$ is self-adjoint. Hence we know that $S$ is diagonalisable and that there exists a basis of orthonormal eigenvectors such that $V$ decomposes as

\[V = V_1 \oplus \cdots \oplus V_k\]

where each $V _ i$ is the eigenspace of $S$ corresponding to some eigenvalue $\lambda$. Note further that each $V _ i$ is $T$-invariant, since for $v \in V _ i$,

\[\begin{aligned} S(Tv) &= (T + T^\star)(Tv) \\\\ &= (T T + T^\star T)v \\\\ &= T(T + T^\star) v \\\\ &= T(\lambda_i v) \\\\ &= \lambda_i Tv \end{aligned}\]

so then if $v \in V _ i$, we have $Tv \in V _ i$.

For all $v \in V _ i$, we have

\[(T + T^{-1})v = \lambda_i v\]

which implies

\[T^2 - \lambda_i T + I = 0\]

Thus the minimal polynomial of $T \vert _ {V _ i}$ divides $x^2 - \lambda _ i x + 1$ (since this polynomial annihilates) and any eigenvalue of $T \vert _ {V _ i}$ is a root of $x^2 - \lambda _ i x + 1$ (because such an eigenvalue is the root of the minimal polynomial, and since the minimal polynomial divides $x^2 - \lambda _ i x + 1$, it is also a root of this).

If $\lambda _ i = \pm 2$, then $(T + I)^2 = 0$ or $(T - I)^2 = 0$, so the only eigenvalues of $T \vert _ {V _ i}$ are $-1$ or $1$. Since $T \vert _ {V _ i}$ may be diagonalised over $\mathbb C$ (this follows from the fact $T$ is orthogonal, so is unitary, but this theorem is about how it is “almost diagonalisable” over $\mathbb R$), we must then have that $T \vert _ { V _ i } = I$ or $-I$.

If $\lambda _ i \ne \pm 2$, then $T \vert _ {V _ i}$ actually has no real eigenvalues. This is because if it did, then since $T$ is orthogonal, they would have to have absolute value $1$, and since we would require $\lambda _ i \ge 2$ (otherwise the discriminant gives complex roots), they would have to be $\pm 1$, which would force $\lambda _ i = \pm 2$, a contradiction.

Because the eigenvalues are complex, this implies that $\{v, Tv\}$ is a linearly independent set over the reals (otherwise $Tv = \lambda v$ for some real $\lambda$, which we have just ruled out). Then by considering the plane $W = \langle v, Tv \rangle$, we that it is $T$ invariant, since

\[v \mapsto Tv, \quad T(v) \mapsto T^2(v) = \lambda_i T(v) - v\]

Also, since $T$ is normal, $W^\perp$ is also $T$-invariant. So $V _ i = W \oplus W^\perp$ gives a decomposition of $V _ i$ into 2 $T$-invariant subspaces. Then by induction on $\dim V _ i$, $V _ i$ splits into a sequence of $2$-dimensional $T$-invariant subspaces of the form

\[T| _ W = R _ \theta = \begin{bmatrix} \cos \theta_i & -\sin\theta_i \\\\ \sin \theta_i & \cos \theta_i \end{bmatrix} \quad \theta_i \neq 0, \pi\]

(why is $T \vert _ W$ in this form? Because any $2\times 2$ matrix has a basis in this form).

Alternative (though really identical), potentially quicker proof (linear algebra 2020 Q3):

We aim to show the equivalent statement that

\[V = U_1 \oplus \cdots \oplus U_k\]

where $U _ k$ are pairwise orthogonal subspaces and $1 \le \dim U _ i \le 2$, and $T(U _ i) \subseteq U _ i$. (Why is this equivalent? Since the real eigenvalues are $\pm 1$, these correspond to the one-dimensional subspaces, and then two dimensional orthogonal matrices with eigenvalues not $\pm 1$ correspond to rotations).

We induct on $\dim V$. By writing $V = V _ {1} \oplus V _ {-1} \oplus V’$ where we have already taken out all the eigenspaces corresponding to $\pm 1$, we can assume that all eigenvalues of $T$ are complex (since $ \vert \lambda \vert = 1$, and $\lambda \ne \pm 1$).

Consider $S = T + T^\star$. Then $S$ is self-adjoint, and so has real eigenvalues. Take $v \in V \setminus \{0\}$ with $Sv = \lambda v$. Then consider

\[U = \langle v, Tv\rangle\]

We want to show that $T(U) \subseteq U$. If $u = \alpha v + \beta T v$, then

\[Tu = \alpha Tv + \beta T^2 v = \alpha Tv + \beta(\lambda Tv - v) \in U\]

where we use the fact that $\lambda v = Tv + T^{-1}v \implies \lambda Tv = T^2 v + v$, so that $T^2 v = \lambda Tv - v$.

We also need to show that $T(U^\perp) \subseteq U^\perp$. Suppose that $v \in U^\perp$ and $u \in U$. Choose $u’ \in U$ with $T u’ = u$ (this is fine since the assumption that $T$ is orthogonal means it is invertible). Then

\[\langle u, Tv\rangle = \langle Tu', Tv\rangle = \langle u', v\rangle = 0\]

This holds for all $u \in U$, so $Tv \in U^\perp$.

Then $V = U \oplus U^\perp$, and so by induction, we obtain

\[V = U_1 \oplus \cdots \oplus U_k\]

of pairwise orthogonal $T$-invaraint subspaces with $\dim U _ i \le 2$.

You are only guaranteed to have eigenvectors / eigenvalues when working over an algebraically closed field.

• $E _ \lambda$ (the eigenspace) is a non-trivial subspace of $V$.
• $T^\ast \vert _ {E _ \lambda}$ therefore has a non-constant characteristic polynomial, with at least one root. This corresponds to an eigenvector in $E _ \lambda$.
• Hence there exists some shared $v \in E _ \lambda$ such that $v$ is an eigenvector for both $T$ and $T^\ast$.

If $v$ is an eigenvector of $T$ with eigenvalue $\lambda$, then $v$ is an eigenvector of $T^\star$ with eigenvalue $\overline \lambda$.

• Case $\lambda = 0$: We have $Tv = 0$ and want to show that $T^\star v = 0$. Since $\langle \cdot, \cdot \rangle$ is positive-definite, $T^\star v = 0 \iff \langle T^\star v, T^\star v \rangle = 0$, which follows from $T$ being normal: $\langle T^\star v, T^\star v \rangle = \langle T T^\star v, w \rangle = \langle T^\star T v, w \rangle = \langle T v, Tv \rangle = 0$.
• Case $\lambda \ne 0$. We reduce this to the other case. Consider $S = T - \lambda I$, then $S$ is normal. Then $S v = 0$, which by the first case implies $S^\star v = 0$, hence $T^\star v = \overline \lambda v$.

The backwards direction is trivial (take $w = v$). For the forward direction, we want to find some way of writing any inner product $\langle v, w\rangle$ in terms of the norm. Note that

\[\begin{aligned} ||v+w||^2 &= \langle v + w, v + w \rangle = ||v||^2 + \langle v, w\rangle + \overline{\langle v, w \rangle} + ||w||^2 \\\\ ||v+iw||^2 &= \langle v+iw, v + iw\rangle = ||v||^2 + i\langle v, w \rangle + \overline{i\langle v, w\rangle} + ||w||^2 \end{aligned}\]

Then since $\langle v, w \rangle + \overline{\langle v, w \rangle} = 2\text{Re}(\langle v, w\rangle)$ and similarly $i \langle v, w \rangle + i \overline{\langle v, w \rangle} = -2\text{Im}(\langle v, w\rangle)$, we can recover the full value of $\langle v, w \rangle$ in terms of the norm:

\[\begin{aligned} \text{Re}(\langle v, w\rangle) &= \frac{1}{2} (||v+w||^2 - ||v||^2 - ||w||^2) \\\\ \text{Im}(\langle v, w\rangle) &= -\frac{1}{2} (||v+iw||^2 - ||v||^2 - ||w||^2) \end{aligned}\]

Then $\langle v, w \rangle = \text{Re}(\langle v, w\rangle) + i\text{Im}(\langle v, w \rangle)$.

We use the fact that the determinant is the product of the eigenvalues.

Let $p(x)$ be the characteristic polynomial of $A$. Since $V$ is real, then $p(x)$ must be a real polynomial. If $\lambda \in \mathbb C$, then $0 = p(\lambda) = p(\overline{\lambda})$ since $p = \overline p$. Thus $\overline \lambda$ is also a root of $p$. If $\lambda$ is not real, then $\lambda \ne \overline \lambda$ and $\lambda \overline \lambda > 0$.

By pairing off complex roots of $p(x)$, we see that $\det A$ is a product of the real roots of $p(x)$ together with a positive real number (the product of all the pairs). Since $\det A < 0$, it must be the case that $p$ has a real negative root $\mu$. Since $\mu$ is an eigenvalue of an orthogonal linear transformation, $\mu = -1$.

There exists an orthonormal basis $D$ of $V$ such that ${} _ D[R] _ D$ is a diagonal matrix with diagonal entries $-1, 1, \cdots, 1$.

The key idea here is that the decomposition allows you to only consider the case where $\dim V \le 2$.

• If $\dim V = 1$, then $A$ is already the identity or a reflection.
• If $\dim V = 2$, then:
  • If $\det A = -1$, then $A$ has two eigenvalues: $-1$ and $1$ (consider the product of the eigenvalues). Then $A$ must already be a reflection.
  • If $\det A = 1$, then consider $R = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$. Then $A = R (R^{-1}A)$ , and since $\det R^{-1}A = \det(R^{-1})\det(A) = -1$ and $\det R = -1$, both $R$ and $R^{-1}A$ are refelections (by the logic above).

1 implies 2:

\[\begin{aligned} \langle Tv, Tw\rangle &= \langle T^\star T v, w \rangle \\\\ &= \langle v, w \rangle \end{aligned}\]

2 implies 1:

\[\begin{aligned} \langle v, w\rangle &= \langle T v, Tw\rangle \\\\ &= \langle T^\star Tv, w\rangle \\\\ &\implies T^\star T v = v&& \text{by non-degen} \\\\ &\implies T^\star T = I \end{aligned}\]

2 implies 3:

\[\begin{aligned} ||Tv||^2 &= \langle Tv, Tv \rangle \\\\ &= \langle v, v \rangle \\\\ &= ||v||^2 \end{aligned}\]

3 implies 2:

\[\begin{aligned} \langle v + w, v + w \rangle &= \langle v, v \rangle + \langle v, w\rangle + \overline{\langle v, w \rangle} + \langle w, w \rangle \\\\ \langle v+iw, v + iw\rangle &= \langle v, v\rangle + i\langle v, w \rangle + i\overline{\langle v, w\rangle} + \langle w, w\rangle \end{aligned}\]

Or, alternatively

\[\begin{aligned} ||v+w||^2 &= ||v||^2 + \langle v, w\rangle + \overline{\langle v, w \rangle} + ||w||^2 \\\\ ||v+iw||^2 &= ||v||^2 + i\langle v, w \rangle + i\overline{\langle v, w\rangle} + ||w||^2 \end{aligned}\]

Then we see that $\langle v, w \rangle + \overline{\langle v, w \rangle} = 2\text{Re}(\langle v, w\rangle)$ and similarly $i \langle v, w \rangle + i \overline{\langle v, w \rangle} = -2\text{Im}(\langle v, w\rangle)$. Thus

\[\begin{aligned} \text{Re}(\langle v, w\rangle) &= \frac{1}{2} (||v+w||^2 - ||v||^2 - ||w||^2) \\\\ \text{Im}(\langle v, w\rangle) &= -\frac{1}{2} (||v+iw||^2 - ||v||^2 - ||w||^2) \end{aligned}\]

completely describes the inner product in terms fo the norm.

\[A = R_\theta = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\\\ \sin(\theta) & \cos(\theta) \end{pmatrix} \text{ or } S_{\theta/2} = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\\\ \sin(\theta) & -\cos(\theta) \end{pmatrix}\]

$A$ is rotation anticlockwise by $\theta$, $S _ {\theta / 2}$ is a reflection.