Ch.14 Complex Vector Spaces

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Polynomial Division

A polynomial has the form $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0$ with $c_n\ne0$. It has degree $n$. If $n=0$, it is a constant polynomial $p(x)=c_0$. If $c_0\ne0$ it has degree $0$, if $c_0=0$ we define it as having degree $-\infty$.

Let $p(x)$ be a polynomial. If $d(x)$ is a nonzero polynomial, then there are quotient and remainder polynomials $q(x)$ and $r(x)$ such that
$p(x)=d(x)\cdot q(x)+r(x)$
where the degree of $r(x)$ is strictly less than $d(x)$. This is polynomial division.
If we choose $d(x)=x-\lambda$, then $r(x)$ is a degree zero polynomial, i.e. a constant, and we can see $r(x)=p(\lambda)$.

If divisor $d(x)$ goes into $p(x)$ evenly, so that $r(x)=0$, then $d(x)$ is a factor of $p(x)$. Any root of the factor $\lambda\in\mathbb{R}$ such that $d(\lambda)=0$ is also a root of $p(x)$ because $p(\lambda)=d(\lambda)\cdot q(\lambda)=0$
If $\lambda$ is a root of $p(x)$, then $x-\lambda$ divides $p(x)$ because
$p(x)=(x-\lambda)q(x)+p(\lambda)=(x-\lambda)q(x)+0=(x-\lambda)q(x)$
because $\lambda$ is a root.

An irriducible polynomial over a field (e.g. $\mathbb{Z},\mathbb{R}$) is one that cannot be factored into lesser-degree polynomials over that field.
A degree $0$ or $1$ polynomial is always irriducible over the reals. A degree $2$ polynomial is irriducible over the reals iff the discriminant is negative. No degree $n$ polynomial $n>3$ is irriducible over the reals.

Example 14.1

$x^4+1$ can be factored into $(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}+1)$.

This implies that every polynomial can be factored into linear and irriducible quadratic polynomials with real coefficients, uniquely. Complex numbers are required to factor irriducible quadratics.

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Complex Numbers

To solve the problem that there is no real solution to $x^2+1=0$, we declare by fiat that $i=\sqrt{-1}$ exists.
This creates numbers of the form $a+bi$, with $a,b\in\mathbb{R}$.
A complex number isa number of the form $a+bi$ for $a,b\in\mathbb{R}$. The set of complex numbers is denoted $\mathbb{C}$.
Note $\mathbb{R}\subset\mathbb{C}$: theyre all in the form $a+0i$

We can identity $\mathbb{C}$ with $\mathbb{R}^2$ by $a+bi\leftrightarrow\begin{pmatrix}a\\b\end{pmatrix}$ and draw $\mathbb{C}$ as a plane.

Operations on Complex Numbers

Addition

$(a+bi)+(c+di)=(a+c)+(b+d)i$

Multiplication

$(a+bi)(c+di)=ac+adi+bci+bd(-1)=(ac-bd)(ad+bc)i$

Complex Conjugation

$\overline{a+bi}=a-bi$
Note that this implies for $z,w\in\mathbb{C}$
$\overline{z+w}=\overline{z}+\overline{w}\text{ and }\overline{zw}=\overline{z}\cdot\overline{w}$
Proof of above:
Let $z=a+bi$ and $w=c+di$
$\overline{z+w}=\overline{(a+bi)+(c+di)}=\overline{(a+c)+(b+d)i}=(a+c)-(b+d)i=a-bi+c-di=\overline{z}+\overline{w}$
$\overline{zw}=\overline{(a+bi)(c+di)}=\overline{(ac-bd)+(ad+bc)i}=(ac-bd)-(ad+bc)i=\overline{z}\cdot\overline{w}$

Absolute Value

$|a+bi|=\sqrt{a^2+b^2}$
Note this is a real number.
Also, $|z|=\sqrt{z\overline{z}}$ since
$z\overline{z}=(a+bi)(a-bi)=a^2-b^2(-1)=a^2+b^2$
and $|zw|=|z||w|$ since
$|zw|=|(ac-bd)+(ad+bc)i|=\sqrt{(ac-bd)^2+(ad+bc)^2}\\=\sqrt{a^2c^2+b^2d^2-2abcd+a^2d^2+b^2c^2+2abcd}\\=\sqrt{(a^2+b^2)(c^2+d^2)}=\sqrt{a^2+b^2}\cdot\sqrt{c^2+d^2}=|z||w|$

Division by a nonzero real number

$\frac{a+bi}{c}=\frac{a}{c}+\frac{b}{c}i$

Division by a nonzero complex number

$\frac{z}{w}=\frac{z\overline{w}}{w\overline{w}}=\frac{z\overline{w}}{|w|^2}$
Note that $|w|^2$ is a real number

Real and Imaginary Part

$\mathscr{R}(a+bi)=a$
$\mathscr{I}(a+bi)=b$

Polar Coordinates for Complex Numbers

Any complex number $z$ also has polar coordinates
$z=|z|(\cos\theta+i\sin\theta)$
where $\theta$ is the argument of $z$, denoted $\theta=\arg{(z)}$.
Note $\arg{(\overline{z})}=-\arg{(z)}$

When you multiply, complex numbers, you multiply the absolute values and add the arguments.
$|zw|=|z||w|\space\space\arg{(zw)}=\arg{(z)}+\arg{(w)}$

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Fundamental Theorem of Algebra

Fundamental Theorem of Algebra: Every degree $n$ polynomial has exactly $n$ complex roots, counting multiplicity.
In other words, for any degree $n$ polynomial $f(x)=x^n+a_{n_1}x^{n-1}+\cdots+a_1x+a_0$, there exists (not necessarily distinct) $\lambda_1,.,,,\lambda_n$ such that $f(x)=(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n)$

If $f(x)$ has real coefficients an $\lambda$ is a root, then $\overline{\lambda}$ is also a root since if $f(\lambda)=0$,
$0=\overline{f(\lambda)}=\overline{\lambda^n+a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda+a_0}=\overline{\lambda}^n+a_{n-1}\overline{\lambda}^{n-1}+\cdots+a_1\overline{\lambda}+a_0=f(\overline{\lambda})$
So for polynomials with real roots, solutions come in conjugate pairs.

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Complex Vector Spaces

We define a complex vector space in the same way we defined vector spaces for $\mathbb{R}$ with the same operations, including vector addition, scalar multiplication, matrix operations, etc.

Like $\mathbb{R}^n$, we denote $\mathbb{C}^n$ the set of ordered $n$-tuples of complex numbers
$\mathbb{C}^n=\{(z_1,...,z_n)\space|\space z_1,...,z_n\in\mathbb{C}\}$
They can also be written as a column vector of $n$ complex entries.
$C^n=\{\begin{pmatrix}z_1\\\vdots\\z_n\end{pmatrix}\space|\space z_1,...,z_n\in\mathbb{C}\}$
Note that since $\mathbb{R}\subset\mathbb{C}$, $\mathbb{R}^n\subset\mathbb{C}^n$

If $\vec{v}=\begin{pmatrix}z_1\\\vdots\\z_n\end{pmatrix}\in\mathbb{C}$, we can write
$z_1=a_1+ib_i,..., z_n=a_n+ib_n\text{ with } a_1,...,a_n,b_1,...,b_n\in\mathbb{R}$
Thus,
$\vec{v}=\begin{pmatrix}z_1\\\vdots\\z_n\end{pmatrix}=\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}+i\begin{pmatrix}b_1\\\vdots\\b_n\end{pmatrix},\text{ with }\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix},\begin{pmatrix}b_1\\\vdots\\b_n\end{pmatrix}\in\mathbb{R}^n$

So, every vector $\vec{v}\in\mathbb{C}^n$ can be uniquely written as a linear combination $\vec{v}=\vec{a}+i\vec{b}$ with $\vec{a},\vec{b}\in\mathbb{R}^n$

Proof

We have shown we can write it, so it remains to show it is unique.
Recall that in $\vec{v}$, each $z_k=a_k+ib_k$ for $k\in\{1,...,n\}$, where $a_k,b_k\in\mathbb{R}$ are the $k$-th elements of $\vec{a}$ and $\vec{b}$ respectively. Therefore, we must have $\mathscr{R}(z_k)=a_k$ and $\mathscr{I}(z_k)=b_k$, which is a well-defined function, making $\vec{a}$ and $\vec{b}$ unique.

From that, we define $\mathscr{R}(\vec{v})=\vec{a}$ and $\mathscr{I}(\vec{v})=\vec{b}$, so that $\vec{v}=\mathscr{R}(\vec{v})+i\mathscr{I}(\vec{v})$

Some Facts

$\mathscr{R}(\vec{0})=\mathscr{I}(\vec{0})=\vec{0}\\ \begin{array}{cc}\mathscr{R}(\vec{v}+\vec{w})=\mathscr{R}(\vec{v})+\mathscr{R}(\vec{w}),&\mathscr{I}(\vec{v}+\vec{w})=\mathscr{I}(\vec{v})+\mathscr{I}(\vec{w})\end{array}\\ \begin{array}{cc}\mathscr{R}(r\vec{v})=r\mathscr{R}(\vec{v}),&\mathscr{I}(r\vec{v})=r\mathscr{I}(\vec{v}),&\text{for all }r\in\mathbb{R}\end{array}\\ \begin{array}{cc}\mathscr{R}(i\vec{v})=-\mathscr{I}(\vec{v}),&\mathscr{I}(i\vec{v})=\mathscr{R}(\vec{v})\end{array}$

The fact that $\vec{v}=\mathscr{R}(\vec{v})+i\mathscr{I}(\vec{v})$ and $\mathscr{R}(\vec{v}),\mathscr{I}(\vec{v})\in\mathbb{R}^n$ implies $\mathbb{R}^n$ spans $\mathbb{C}^n$ over the scalars $\mathbb{C}$. So, if a set $S\subseteq\mathbb{R}^n$ spans $\mathbb{R}^n$ over real scalars, then it also spans $\mathbb{C}^n$ over complex scalars.

If $\vec{a}_1,...,\vec{a}_k\in\mathbb{R}$ is a linearly independent subset of the real vector space $\mathbb{R}^n$, then $\vec{a}_1,...,\vec{a}_k$ is a linearly independent subset of the complex vector space $\mathbb{C}^n$.

Proof

Suppose $z_1\vec{a}_1+\cdots+z_k\vec{a}_k=\vec{0}$ with $z_1,...,z_k\in\mathbb{C}$. Then using $z_i=\mathscr{R}(z_i)+i\mathscr{I}(z_i)$ for $i\in\{1,...,k\}$, we have
$[\mathscr{R}(z_1)\vec{a}_1+\cdots+\mathscr{R}(z_k)\vec{a}_k]+i[\mathscr{I}(z_1)\vec{a}_1+\cdots+\mathscr{I}(z_k)\vec{a}_k]=\vec{0}$
Since both $\mathscr{R}(z_1)\vec{a}_1+\cdots+\mathscr{R}(z_k)\vec{a}_k$ and $\mathscr{I}(z_1)\vec{a}_1+\cdots+\mathscr{I}(z_k)\vec{a}_k$ are in $\mathbb{R}^n$, we must have
$\mathscr{R}(z_1)\vec{a}_1+\cdots+\mathscr{R}(z_k)\vec{a}_k=\mathscr{R}(z_1\vec{a}_1+\cdots+z_k\vec{a}_k)=\mathscr{R}(0)=\vec{0}$
$\mathscr{I}(z_1)\vec{a}_1+\cdots+\mathscr{I}(z_k)\vec{a}_k=\mathscr{I}(z_1\vec{a}_1+\cdots+z_k\vec{a}_k)=\mathscr{I}(\vec{0})=\vec{0}$
Since $a_1,...,a_k$ is linearly independent and $\mathscr{R}(z_1),...,\mathscr{R}(z_k),\mathscr{I}(z_1),...,\mathscr{I}(z_k)\in\mathbb{R}$, we must have $\mathscr{R}(z_1)=\cdots=\mathscr{R}(z_k)=\mathscr{I}(z_1)=\cdots=\mathscr{I}(z_k)=0$

This implies any basis $B=\langle\vec{b}_1,...,\vec{b}_n\rangle\subset\mathbb{R}^n$ is also a basis of $\mathbb{C}^n$. In particular, the standard basis $\mathcal{E}_n$ of $\mathbb{R}^n$ is also the standard basis of $\mathbb{C}^n$

Complex Linear Maps

If $t:\mathbb{R}^n\to\mathbb{R}^m$ is a linear map, then we can define a map $t^c:\mathbb{C}^n\to\mathbb{C}^m$ by
$t^c(\vec{v})=t(\mathscr{R}\vec{v})+it(\mathscr{I}\vec{v})$
In particular, $t^c(\vec{v})=t(\vec{v})$ for $\vec{v}\in\mathbb{R}^n$.
This map $t^c$ is linear, and has the same matrix representation in the complex standard bases as the matrix representation of $t$ in the real standard bases.
The map $t^c$ is called the complexification of $t$ or $t$ complexified
Proof is relatively simple: prove linearity, prove the matrices are the same.

So, all the results in real vector spaces carry over to complex vector spaces.

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Similarity

Recall two matrices $H$ and $\hat{H}$ are matrix equivalent if there are non singular matrices $P$ and $Q$ such that $\hat{H}=PHQ$ (e.g. $H$ and $\hat{H}$ represent the same map in different bases)
We now consider the special case of transformations, where the codomain equals the domain, and add the requirement that the basis is also the same.
i.e. representations with respect to $B,B$ and $D,D$

The matrices $T$ and $\hat{T}$ are similar if there is a nonsingular $P$ such that $\hat{T}=PTP^{-1}$

Note the zero matrix $Z$ and the identity matrix $I$ is similar to only itself since
$PZP^{-1}=PZ=Z\\ PIP^{-1}=PP^{-1}=I$

Similarity is an equivalence relation, since it is a specific case of matrix equivalence. However, the converse is not true.