# Random math: 19

Source of most content: Gunhee Cho

# Diagonalization

Recall that the orthonormal basis(ONB) consisting of eigenvectors of $T$ exists iff $T$ is normal as shown in RM 18.

## Corollary

$A \in Mat_{n \times n} (\mathbb{C}) \text{ is normal iff } \exists \text{ unitary } Q \text{ s.t. } Q^*AQ \text{ is a diagonal matrix.} \nonumber$

### Proof

Since $A$ is normal, lin. op. $L_A : \mathbb{C}^n \rightarrow \mathbb{C}^n$ is also normal. Thus, there exists ONB $\beta = \{ v_1, \cdots, v_n \}$ of $\mathbb{C}^n$ consisting of eigenvectors of $L _A$. $\Leftrightarrow$ $[ L_A ] _{\beta}$ is a diagonal matrix s.t. $[ L_A ] _\beta = [ Id ]^\beta _{\epsilon _n} [ L_A ]^{\epsilon _n} _{\epsilon _n} [ Id ]^{\epsilon _n} _\beta = Q^{-1} A Q = Q^* A Q$. Thus, there exists a unitary matrix $Q$ s.t. $Q^* A Q$ is a diagonal matrix.

NB1; $A \in Mat_{n \times n} (\mathbb{R})$. Then, $A$ is self-adjoint iff $A=A^T$, i.e., $A$ is symmetric.
NB2; If $A\in Mat_{n \times n} (\mathbb{R})$ is symmetric, then $A$ is normal. But note that converse is not true.
NB3; If for $A\in Mat_{n \times n} (\mathbb{C})$, $A=A^H$ holds, then $A$ is normal. But symmetric $A$ does not mean $A$ is normal.

## Lemma

$\text{Assume } T: V \rightarrow V \text{ is self-adjoint (S-A). Then, } \nonumber$
1. Every eigenvalue of $T$ is real.
2. Let $V$ be f.d.i.p.s over $\mathbb{R}$. Then, $\varphi_T (t)$ splits. (thus we can use Schur’s thm)

### Proof

1. Let $T(v) = \lambda v$ since $T$ is normal ($\because T$ is self-adjoint). Then, $T(v)= T^* (v) = \bar{\lambda} v = \lambda v = T(v)$ $\Leftrightarrow (\lambda - \bar{\lambda})v = 0 \Leftrightarrow \lambda = 0 \in \mathbb{R}$.

### Proof of lemma

Since $T$ is S-A, there exists ONB $\beta = \{ v_1, \cdots, v_n \}$ consisting of eigenvectors. By Asm, $\forall i, < v_i, T(v_i) > = 0$ $= < v_i, \lambda_i v_i > = \bar{\lambda}_i < v_i, v_i > = \bar{\lambda}_i$. Thus, $\forall i, \lambda_i = 0 \implies [ T ] _\beta = 0$.

## Corollary

$\text{TFAE} \nonumber$
1. $\forall x \in \mathbb{R} \text{ or } \mathbb{C},\ \Vert Ax \Vert = \Vert x \Vert$.
2. $A$ is unitary/orthogonal.
3. $\forall x,y , \ < Ax, Ay > = < x, y >$.

# Projection

Def. Let $(V, \langle , \rangle)$ be f.d.i.p.s over $F$. Then, $T: V\rightarrow V$ is called projection if

1. $V = \text{Im}(T) \oplus \ker (T)$.
2. For $x \in \text{Im} (T), \ T(x) x$, i.e., $T \mid_{\text{Im}T}=Id$.

Def. Let $(V, \langle , \rangle)$ be f.d.i.p.s over $F=\mathbb{R}$ or $\mathbb{C}$. Then, a projection $T: V\rightarrow V$ is called orthogonal if $\ker (T) = \text{Im}(T)^\perp$.

## Proposition

$T: V\rightarrow V \text{ is a projection iff } T\circ T = T^2 = T. \nonumber$

### Proof

1. $(\Rightarrow)$ Since $T$ is a projection, $V= \text{Im}(T) \oplus \ker(T)$. Thus, $\forall (x\in \text{Im}(T),y\in\ker(T)) \in V$, $T^2((x,y))= T \circ T(x,y) \stackrel{y\in\ker(T)}{=} T((x,0))\stackrel{x\in\text{Im}(T)}{=}(x,0)=T(x,y)$.
2. $(\Leftarrow)$ It suffices to show $\text{Im}(T) \cap \ker(T) = \{ 0 \}$ to prove $V = \text{Im}(T) \oplus \ker(T)$ by the dimension formula, $dim V = dim \text{Im}(T) + dim \ker (T)$. If $x \in \text{Im}(T) \cap \ker(T)$, then $x = T(y)$ for some $y \in V$. Since $x \in \ker T$, $T(x) = 0 = T(T(y)) \stackrel{T^2 = T}{=} T(y) = x$.

$T: V\rightarrow V \text{ is an orthogonal projection iff } T\circ T = T^2 = T^*. \nonumber$

### Proof

1. $(\Leftarrow)$ $\ker(T) \stackrel{NB1}{=} (\text{Im}(T^* ))^\perp \stackrel{ T=T^* }{=} (\text{Im}(T))^\perp \leadsto \ker(T) = (\text{Im}(T ))^\perp$.
2. $(\Rightarrow)$ For any $x, y \in V$, let $x= x_1 + x_2$ and $y= y_1 + y_2$ for $x_1, y_1 \in \text{Im}(T)$ and $x_2, y_2 \in \ker (T)$. Then, $< T(x), y > = < x_1, y_1+y_2 > = < x_1, y_1 >$ and $< T^* (x), y > = < x, T(y) > = < x_1 + x_2, y_1 > = < x_1 , y_1 >$. Thus, $T = T^*$.

NB1; For f.d.i.p.s $V$, $\ker (T) = (\text{Im}(T^* ))^\perp$: $x \in \ker(T) \Leftrightarrow T(x) = 0 \Leftrightarrow \forall y \in V, < T(x), y > = 0 \Leftrightarrow \forall y \in V, < x, T^* (y) > = 0 \Leftrightarrow x \perp \text{Im}(T^* ) \Leftrightarrow x \in (\text{Im}(T^* ))^\perp$

# Spectral Theorem

$\text{Given a f.d.i.p.s } (V, \langle , \rangle) \text{ over } F,\ \text{let linear operator } T: V\rightarrow V. \text{Let } \{ \lambda_i \}_{i=1}^k \text{be mutually distinct eigenvalues} \nonumber$ $\text{and } \{ W_i \}_{i=1}^k \text{ be eigenspaces of } T \text{ corresponding to } \lambda, \text{ i.e., } W_i = \ker (T- \lambda_i Id). \nonumber$ $\text{Assume } T \text{ is normal/self-adjoint based on the field. Then, following holds: } \nonumber$
1. $V = W_1 \oplus \ldots \oplus W_k$.
2. $W_1 \perp W_j$ if $i \ne j$.
3. $\forall i, \ W_i^\perp = \ldots \oplus W_{i-1} \oplus W_{i+1} \oplus \ldots \oplus W_k$.
4. Define $T_i : V \rightarrow V \left( (x_1, \ldots, x_k) \mapsto (0, \ldots, 0, x_i, 0, \ldots, 0) \right)$. Then, $T_i$ is orthogonal projection satisfying

a. $T_i \circ T_j = \delta_{ij}T_i$,
b. $\sum_{i=1}^k T_i= Id$,
c. $\sum_{i=1}^k \lambda_i T_i = T$.

NB2; This indeed implies that there exists orthonormal basis consisting of eigenvectors and its proof is given in RM 18.
NB3; Let $\forall i, \beta_i$ be an orthonormal basis of $W_i$, i.e., $\beta=\cup_{i=1}^k \beta_i$. Then, $[ T ] _{\beta} = diag( \lambda _1 I _{n _1}, \ldots, \lambda _k I _{n_k} )$, where $n_i = dim W_i$.