Lecture 07

[[lecture-data]]

2024-09-11

Readings

• a

Recall: complex numbers.

Complex Conjugate

Let $k=\alpha +\beta i$ be a complex number. Then the complex conjugate is $\bar{k}=\alpha -\beta i$

(see complex conjugate)

Note that $\overline{k_1+k_2}=\overline{k_1}+\overline{k_2}$ in this case

Now, suppose we have a complex number in polar coordinates $z=\rho e^{i\theta }$. Then the complex conjugate is $\overline{z}=\rho e^{-i\theta }$

Note that $\overline{z_1{z}_2}=\overline{z_1}\cdot \overline{z_2}$ in this case

2. Chapter 2

Conjugate Transpose, Hermitian

Let $A\in M_{m\times n}(\mathbb{C})$. Define the conjugate transpose of $A$: $A^{\ast }=\overline{A^T}$

If $A^{\ast }=A$, then we call $A$ hermitian

(see conjugate transpose, hermitian)

Note

If $A\in M_{m,n}$ and $B\in M_{n,p}$ then $(AB{)}^T=B^T{A}^T$ and $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$

Length

The euclidean length of a vector $z$ is $||z|{|}_{(2)}=\sqrt{z^{\ast }z}$

(see euclidean length of a vector)

Orthogonal, Orthonormal Vectors

If $x,y\in \mathbb{C}$, then $x\perp y$ or ”$x$ is orthogonal to $y$” means $y^{\ast }x=0$.

$\{x_1,\dots ,x_n\}\subset {\mathbb{C}}^n$ “are orthogonal” means the $x_i$ are pairwise orthogonal. We say they are orthonormal if for all $i$, we have that $x_i^{\ast }{x}_i=1$

(see orthogonal vectors)

Unitary

A matrix $U\in M_n$ is called unitary precisely when $U^{\ast }U=I$.

• ie, $U$ is invertible
• And its inverse is its conjugate transpose

If $Q\in M_n(\mathbb{R})$, then $Q$ is (real) orthogonal precisely when $Q^{T}Q=I$

(see unitary matrices)

Note

A matrix $U\in M_n$ is unitary if and only if the columns of $U$ are orthonormal.

$U$ unitary means that $U^{\ast }U=I$ if and only if for all columns $u_k$ of $U$:

1 ;;\text{ if } i=j \ 0 ;;\text{ otherwise} \end{cases}$$

which means that the columns are 1) orthogonal and 2) normal

(see unitary matrices have orthonormal columns)

Note

If $U,V\in M_n$ are unitary,then $UV$ is unitary.

$(UV{)}^{\ast }(UV)=V^{\ast }{U}^{\ast }U{V}^{\ast }=I$

(see the product of unitary matrices is unitary)

Fact

Unitary matrices form a group

Note

If $U\in M_n$ is unitary, then $U:{\mathbb{C}}^n\to {\mathbb{C}}^n$ is an isometry. This means $\forall x\in {\mathbb{C}}^n$, the length $||Ux|{|}_2=\sqrt{(Ux{)}^{\ast }Ux}=\sqrt{x^{\ast }{U}^{\ast }Ux}=||x|{|}_2$That is, the transformation preserves the length of any input vector.

Additionally, for any pair of vectors $x,y\in {\mathbb{C}}^n$ we have $||Ux-Uy|{|}_2=||U(x-y)|{|}_2=||x-y|{|}_2$ by the result immediately above, so the transformation also preserves pairwise distances between input vectors!

(see unitary matrices define isometries)

Unitarily Similar

$A,B\in M_n$ are unitarily similar (or unitarily equivalent)means that there exists $U\in M_n$ unitary such that $A=UB{U}^{\ast }$

This is a similarity where the change of basis is a distance-preserving one.

(see unitarily similar)

Frobenius Norm

“treat the matrix like a long euclidean vector” $||A|{|}_F=\sqrt{{\sum }_{i,j}|a_{ij}{|}^2}$

(see frobenius norm)

Note

Suppose $A,B$ are unitarily similar. Then $||A|{|}_F=||B|{|}_F$

$A=UB{U}^{\ast }$ for $U$ unitary

We have that

\lvert \lvert A \rvert \rvert _{F}^2 &= \mathrm{Tr}(A^A) \ &= \mathrm{Tr}((UBU^)^UBU^) \ &= \mathrm{Tr}(UB^*U^UBU^) \ &= \mathrm{Tr}(UB^BU^) \ & = \mathrm{Tr}(U^*UB^*B) \ &= \mathrm{Tr}(B^*B) &= \lvert \lvert B \rvert \rvert _{F}^2 \end{aligned}$$

(see unitarily similar matrices have the same frobenius norm)

Householder Transformation

Suppose we have $w\in {\mathbb{C}}^n$ with $||w||=1$. The Householder transformation is given by $H_w:=I-2w{w}^{\ast }$

(see householder transformation)

$H_w$ is hermitian and unitary!

(I-2ww^*)(I-2ww^*) &= I-4ww^*+4ww^*ww^* \\ &= I - 4ww^* + 4ww^* \\ &= I \end{aligned}$$ >[!theorem] Note > >Given any unit vector $x \in \mathbb{C}^n$, there exists a unitary matrix $U \in M_{n}$ where the first column is $x$. > >>[!proof]+ >>Gram-Schmidt: start with the first column $x$ and any orthonormal basis for the rest of the matrix. > >(see [[Concept Wiki/we can find a unitary matrix for any unit vector]]) >[!theorem] Note > >Let $x,y \in \mathbb{R}^n$ such that $x \neq y$ but $\lvert \lvert x \rvert \rvert = \lvert \lvert y \rvert \rvert$. If we take $w = \frac{x-y}{\lvert \lvert x-y \rvert \rvert}$, then $H_{w}x = y$ and $H_{w}y = x$ if $x$ is real, then $$Hx = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$ use the unitary matrix $U=I$. Otherwise, take $U=H_{w}$ such that $H_{w}[1 \;0\;0 \dots{0}]^T = x$ and use $$w = \frac{1}{\lvert \lvert \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} - x \rvert \rvert_{2} } \left(\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} - x\right)$$ central result for chapter 2: Schur's Theorem (next lecture)