Lecture 10

[[lecture-data]]

EXAM 1

Exam 1 will cover chapter 1, 2, 3 and will likely be early-to-mid october. Finishing Chapter 2 today

2024-09-18

Readings

• a

2. Chapter 2

Recall the definition of normal matrices: $A{A}^{\ast }=A^{\ast }A$ and recall triangular matrices are normal iff diagonal.

Lemma

Suppose $A,B\in M_n$ are unitarily similar. Then $A$ is normal if and only if $B$ is normal.

$A=UB{U}^{\ast }$ for $U\in M_n$ unitary. $A$ normal means $A{A}^{\ast }=A^{\ast }A$. ie,

Say

UBU^*(UBU^*)^* &= (UBU^*)^*UBU^* \\ UBU^*UB^*U^* &= UB^*U^*UBU^* \\ UB B^*U^* &= UB^*BU^* \\ BB^* &= B^*B \end{aligned}$$ ie, $B$ is normal

(see unitarily similar matrices are simultaneously normal)

Unitarily Diagonalizable

$A\in M_n$ is unitarily diagonalizable precisely when there exists $U\in M_n$ unitary and $D\in M_n$ diagonal such that $A=UD{U}^{\ast }$

(see unitarily diagonalizable)

This means

1. there exists not just linearly independent eigenvectors, but orthonormal eigenvectors!
2. There is some rigid transformation from the standard basis to the “basis of the matrix”

Spectral Theorem for Normal Matrices

Consider any $A\in M_n$. Say the eigenvalues are ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$. Then the following are equivalent:

1. $A$ is normal
2. $A$ is unitarily diagonalizable
3. $||A|{|}_F^2={\sum }_{i=1}^n|{\lambda }_i{|}^2$

(see spectral theorem for normal matrices)

Proof

(2) $⟹$ (1),(3) Suppose $A$ is unitarily diagonalizable, say $A=VD{V}^{\ast }$ where $V\in M_n$ is unitary and $D\in M_n$ is diagonal. Then $D$ is normal and $⟹A$ is normal by the lemma above. Also, $||A|{|}_F^2=||VD{V}^{\ast }|{|}_F^2=||D|{|}_F^2$ (see the lemma from the beginning of the chapter in Lecture 07) and note that this is precisely ${\sum }_{i=1}^n|{\lambda }_i{|}^2$

(1) $⟹$ (2) Suppose $A$ is normal. Let $A=UT{U}^{\ast }$ be its Schur decomposition. If $A$ is normal, then $T$ is normal by the lemma above. And since triangular matrices are normal iff diagonal, this implies that $A=UT{U}^{\ast }$ is a unitary diagonalization!

(3) $⟹$ (3) Suppose $||A|{|}_F^2={\sum }_{i=1}^n|{\lambda }_i{|}^2$. Let $A=U{T}^{\ast }U$ be the Schur decomposition. Then

\sum_{i=1}^n \lvert \lambda_{i} \rvert^2 &= \lvert \lvert A \rvert \rvert _{F}^2 \\ &= \lvert \lvert UTU^* \rvert \rvert _{F}^2 \\ &= \lvert \lvert T \rvert \rvert _{F}^2 \\ &= \sum_{i=1}^n \lvert t_{ii} \rvert^2 + \sum_{j > i}\lvert t_{ij} \rvert^2 \\ \implies \forall_{i\neq j}\;\;t_{ij} = 0 \end{aligned}$$ $\implies T$ is diagonal! So $A=UTU^*$ is a [[Concept Wiki/unitarily diagonalizable\|unitary diagonalization]]

$A$ is normal, then its Schur decomposition is automatically a diagonalization. Thus, if $A$ and $B$ are normal and unitarily similar, then they are simultaneously diagonalizable and thus they commute!

If

(see also normal, unitarily similar matrices are simultaneously diagonalizable)

(may help homework)

We call $||A|{|}_F^2-{\sum }_{i=1}^n|{\lambda }_i{|}^2$ the “defect from normality”. And if this defect is $0$, then $A$ is normal.

(see defect from normality)

Spectral Theorem for Hermitian Matrices

Let $A\in M_n$. $A$ is hermitian if and only if

1. $A$ is unitarily diagonalizable and
2. the spectrum of $A$, $\sigma (A)$ is real.

(see spectral theorem for hermitian matrices)

$⟹$) Suppose that $A$ is hermitian. Then $A$ is normal and hence $A$ is unitarily diagonalizable (by the spectral theorem for normal matrices).

(

Say $A=UD{U}^{\ast }$ where $U\in M_n$ unitary and $D\in M_n$ diagonal. Then

A &= A^* \\ UDU^* &= (UDU^*)^* \\ &= U^*D^*U \\ \implies D &= D^* \\ \implies d_{ii} &\in \mathbb{R} \;\;\forall\;\;i \end{aligned}$$ ($\impliedby$) Suppose $A$ is [[Concept Wiki/unitarily diagonalizable]] and the [[Concept Wiki/eigenvalue\|eigenvalues]] of $A$ are real. Then $A = UDU^*$. Eigenvalues of $A$ are real means that $D^*=D$. Thus $$A^* = (UDU^*)^* = UD^*U^* = UDU^* = A$$ ie, $A$ is [[Concept Wiki/hermitian]]! $\blacksquare$

Theorem

Suppose $A\in M_n$. $A$ is symmetric if and only if $A$ is real-orthogonally diagonalizable. That is, if we can write $A=QD{Q}^T$ with $Q\in M_n(\mathbb{R})$ orthogonal and $D\in M_n(\mathbb{R})$ diagonal.

$(⟹)$ Suppose $A$ is real-orthogonally diagonalizable. Say that $A=QD{Q}^T$ for $Q\in M_n(\mathbb{R})$ orthogonal and $D\in M_n(\mathbb{R})$ diagonal. Then $A^T=(QD{Q}^T{)}^T=QD{Q}^T=A$

$(⟸)$ Suppose $A$ is symmetric. Then $A=A^T$ and $A$ is hermitian $⟹$ the eigenvalues of $A$ are real by the spectral theorem for hermitian matrices. Since $A$ is real and has real eigenvalues, this implies that $A$ has a real Schur decomposition. Say then that $A=QT{Q}^T$ with $Q\in M_n(\mathbb{R})$ orthogonal and $T\in M_n(\mathbb{R})$ upper triangular.

Since $A$ is hermitian, $A$ is normal by the spectral theorem for normal matrices. Note then that $A$ is unitarily similar to $T$ and unitarily similar matrices are simultaneously normal, we can say that $T$ is normal. Then, since triangular matrices are normal iff diagonal, this means that $T$ is diagonal!

That is, when we write $A=QT{Q}^T$ this is a real orthogonal diagonalization! $■$

(see matrices are symmetric if and only if they are real orthogonally diagonalizable)

$A^{\ast }=-A$ are called skew hermitian. (see skew hermitian)

Exercise

Skew hermitian matrices have pure imaginary eigenvalues