Roots over $~\mathbb{C}$
As you might have seen, for a quadratic polynomial $ax^2+bx+c$, we have a formula for the roots \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) However, we cannot make sense of this as a real number if \(b^2 - 4ac < 0\)
As mentioned briefly before, $\mathbb{R}$ is a subfield of a larger field, the complex numbers $\mathbb{C}$.
In general, to build new fields from old ones, we pretend we have a root to a polynomial which actually has no root. This process is called adjoining a root.
The complex numbers come from adjoining a root to the equation \(x^2 + 1 = 0\) We introduce a new symbol $\imath$ which satisfies the algebraic identity \(\imath^2 = -1\)
With this symbol in hand, we can now make sense of any negative root in $\mathbb{C}$. For example, \(\sqrt{-5} = \imath \sqrt{5}\)
The elements of $\mathbb{C}$ are sums \(a + \imath b\) with $a,b \in \mathbb{R}$. We add two complex numbers as if we were adding a two-dimensional vector: \(a + b \imath + c + d \imath = (a+c) + (b+d)\imath\) We mulitply using distributivity and remembering $\imath^2 = -1$: \((a + b \imath)(c + d\imath) = ac + ad\imath + bc \imath + bd\imath^2 = (ac-bd) + (ad+bc)\imath\)
This is indeed a field though we won’t provide all the details of the proof. The additive inverse is \(-(a + b \imath) = -a - b\imath\) which is, again, the same as if we treated $a + b\imath$ as the vector $(a,b)$. The zero element $0$ of $\mathbb{C}$ is $a=b=0$. The multiplicative identity $1$ is $a=1$ and $b=0$.
Lemma. Let $a+b\imath$ be a nonzero complex number. Then, \((a+b\imath) \left(\frac{a-b\imath}{a^2+b^2} \right) = 1\)
Therefore, \((a+b\imath)^{-1} = \frac{a-b\imath}{a^2+b^2}\)
Proof. (Expand to view)
Let’s first note that if $a +b \imath \neq 0$, then either $a \neq 0$ or $b \neq 0$. In either case $a^2 + b^2 \neq 0$ so the division in the formula is well-defined. We have \((a+b\imath)(a-b\imath) = a^2 + b^2\) so \((a+b\imath) \left(\frac{a-b\imath}{a^2+b^2} \right) = 1\) ■
Definition. In general, the complex number $a-b\imath$ is called the complex conjugate of $a+b\imath$. The complex conjugate of $z$ is denoted by $\overline{z}$.
Thus, \(z^{-1} = \frac{\overline{z}}{z\overline{z}}\)
Example. So \((1-\imath)^{-1} = \frac{1}{2}(1 + \imath)\)
It turns out that adjoining a root of $x^2+1$ to $\mathbb{R}$ actually gives all roots to any polynomial. The following result is the strongest argument for the utility of complex numbers (with their use in quantum mechanics a close second).
As convention, instead of using $x$ to denote a variable, we use $z$ to denote a complex variable. If we write $z = a+b\imath$, then we say $a$ is the real part of $z$ and $b$ is the imaginary part of $z$. We denote the real part by $Re(z)$ and the imaginary part by $Im(z)$.
Theorem (The Fundamental Theorem of Algebra). Let $p(z)$ be a polynomial with complex coefficients. Then there is a $z_0 \in \mathbb{C}$ with \(p(z_0) = 0\) In other words, any polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$.
Proof. (Expand to view)
There are multiple interesting proofs of the Fundamental Theorem of Algebra, some algebraic, some analytic. ■
Corollary. Let $p(z)$ be a nonconstant polynomial one complex variable. Then, $p(z)$ factors completely: \(p(z) = a \prod_{i=1}^d (z-z_i)\) for $a,z_1,\ldots,z_d \in \mathbb{C}$.
Proof. (Expand to view)
We proceed by induction on the degree $d$ of $p(z)$. For $d=1$, we already have $p(z) = az+b$ and we take $z_1 = -b/a$.
Now assume we know the result is true for degree $d-1$ and let $p(z)$ have degree $d$. From the Fundamental Theorem of Algebra, we have a root $z_d$ of $p(z)$. Thus, \(p(z) = (z-z_d)q(z)\) for a polynomial of degree $d-1$. Applying the induction hypothesis, we get \(p(z) = a \prod_{i=d} (z-z_i)\) ■
For us, we are interested in characteristic polynomials $\chi_A$ of $n \times n$ matrices $A$.
We have already seen that we need not have an eigenvector in general for a matrix over $\mathbb{R}$. This is not the case for $\mathbb{C}$.
Proposition. Any $n \times n$ matrix $A$ with entries in $\mathbb{C}$ has an eigenvector.
Proof. (Expand to view)
From the Fundamental Theorem of Algebra, we know that $\chi_A(z)$ has root $\lambda$. Thus, $A$ has an eigenvector with eigenvalue $\lambda$. ■
Given a complex matrix $A$, its complex conjugate $\overline{A}$ is given by taking the complex conjugates of its entries \(\overline{A}_{ij} = \overline{A_{ij}}\)
The next lemma shows that honest complex eigenvalues come in pairs.
Lemma. Assume we have a real $n \times n$ matrix. If $\lambda$ is a complex eigenvalue of $A$, then so is $\overline{\lambda}$. Moreover, \(E_{\overline{\lambda}}(A) = \overline{E_{\lambda}(A)}\)
Proof. (Expand to view)
Taking the complex conjugate of \(A v = \lambda v\) gives \(A \overline{v} = \overline{Av} = \overline{\lambda v} = \overline{\lambda} \overline{v}\) which shows that $\overline{v}$ is also an $A$-eigenvector with eigenvalue $\overline{\lambda}$.
This also shows that \(\overline{E_{\lambda}(A)} \subseteq E_{\overline{\lambda}}(A)\) But, we can apply complex conjugation here to get \(E_{\lambda}(A) = \overline{\overline{E_{\lambda}(A)}} \subseteq \overline{E_{\overline{\lambda}}(A)}\) Applying this, but replacing $\overline{\lambda}$ with $\lambda$ gives \(\overline{E_{\lambda}(A)} \superseteq E_{\overline{\lambda}}(A)\) ■
We also have a simple criteria to determine whether there is basis of eigenvectors.
Proposition. Let $A$ be a $n \times n$ matrix over a field $k$. Assume that $\chi_A(x)$ has $n$ distinct roots. Then, there is a basis of $k^n$ consisting of $A$-eigenvectors.
In particular, $A$ is diagonalizable.
Proof. (Expand to view)
We have to have at least on eigenvector for each distinct eigenvalue, ie each distinct root of $\chi_A(x)$. As we saw, if we have a set of eigenvectors with distinct eigenvalues, then that set is linearly independent. If we have $n$ distinct roots, then we have a linearly independent set of size $n$ in an $n$-dimensional vector space. This must span, so it is a basis.
The matrix representation of $A$ in its basis of eigenvectors is a diagonal matrix, where the entries on the diagonal are the eigenvalues. ■
We work through an example.
Example. Let’s compute the eigenspaces of \(A = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}\) The characteristic polynomial is \(\chi_A(z) = x^2 + 2x + 2\) Using the quadratic formula, we have roots \(\frac{-2 \pm \sqrt{-4}}{2} = -1 \pm \imath\) Note that since we have two distinct eigenvalues we already know we have a basis of $\mathbb{C}^2$ consisting of $A$-eigenvectors.
Let’s compute the eigenspaces. We want the null space of \(A - (-1 + \imath)I_2 = \begin{pmatrix} 1 - \imath & -1 \\ 2 & -1 - \imath \end{pmatrix}\) For simplicity let’s scale the first row so that the $(1,1)$ entry is $1$. We have \((1-\imath)^{-1} = \frac{1}{2}(1+\imath)\) Scaling we get \(\begin{pmatrix} 1 & -\frac{1}{2}(1+\imath) \\ 2 & -1-\imath \end{pmatrix}\) Eliminating the second row using the first gives \(\begin{pmatrix} 1 & -\frac{1}{2}(1+\imath) \\ 0 & 0 \end{pmatrix}\) Thus, the $(-1+\imath)$-eigenspace is \(E_{-1+\imath}(A) = \left\lbrace \begin{pmatrix} \frac{1}{2}(1+\imath)w \\ w \end{pmatrix} \mid w \in \mathbb{C} \right\rbrace\)
As we saw above, we can get the eigenspace for the complex conjugate eigenvalue via complex conjugation: \(E_{-1 - \imath}(A) = \overline{E_{-1+\imath}(A)} = \left\lbrace \begin{pmatrix} \frac{1}{2}(1-\imath)w \\ w \end{pmatrix} \mid w \in \mathbb{C} \right\rbrace\)