T1: Eigenvalues and Eigenvectors

July 16, 2024

• Eigenvalues and Eigenvectors
• Symmetric Function $s_k$
• Reference

This post presents basics about eigenvalues and eigenvectors.

Eigenvalues and Eigenvectors

For a matrix $A \in \mathbb{C}^{n\times n}$, scalars $\lambda$ and vectors $\mathsf{x}_{n\times 1}$ satisfying $A\mathsf{x} = \lambda \mathsf{x}$ are eigenvalues and eigenvectors of $A$. The set of distinct eigenvalues, $\delta(A)$ is the spectrum of $A$.

The characteristic polynomial of $A_{n\times n}$ is $p(\lambda) = \text{det}(A-\lambda \textup{I})$.

Let $\{ x \neq 0 \mid x \in N(A -\lambda \textup{I})\}$ be the set of all eigenvectors associated with eigenvalues $\lambda$. We say that $N(A - \lambda \text{I})$ is the eigenspace for $A$.

$\textbf{Definition}$[Spanning Sets]. For a set of vectors $\mathcal{S} = \{v_1,\ldots, v_t\}$, the subspace

$$span(\mathcal{S}) = \{\alpha_1v_1+\cdots+\alpha_t v_t\}$$

that forms all linear combinations of vectors of $\mathcal{S}$ is called the space spanned by $\mathcal{S}$.

Symmetric Function $s_k$

We define the $k$-th symmetric function of $\lambda_1,\lambda_2,\ldots,\lambda_n$ to be the sum of the product of eigenvalues multiplication for $k$ times, i.e.,

$$s_k = \sum_{1\leq i_1<\cdots<i_k \leq n}\lambda_{i_1}\ldots \lambda_{i_k}$$

Example: $n = 3$

$$s_1 = \lambda_1 + \lambda_2 + \lambda_3$$ $$s_2 = \lambda_1\lambda_2+\lambda_1\lambda_3 + \lambda_2\lambda_3$$ $$s_3 = \lambda_1\lambda_2\lambda_3$$

We expand $p(\lambda) = \text{det}(A-\lambda\text{I}) =\left \lvert\begin{matrix} a_{11}-\lambda & a_{12} &\cdots & a_{1n}\\ a_{21} & a_{22}-\lambda &\cdots &a_{2n}\\ \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2}& \cdots & a_{nn}-\lambda \end{matrix} \right \rvert$

as follows.

$$p(\lambda) = (-1)^n[\lambda^n+c_1\lambda^{n-1}+\cdots+c_{n-1}\lambda+c_n]$$

By setting $\lambda = 0$, we have:

Proposition 1: $c_n = (-1)^n \text{det}(A)$.

Now, let us see the value of $c_1$.

$$p(\lambda) = (a_{11}-\lambda)\cdots(a_{nn}-\lambda)+\sum_{i=2}^{n-2}c'_i\lambda^{n-i}+c'_n$$

It is easy to see that we only need to care about $(a_{11}-\lambda)\cdots(a_{nn}-\lambda)$ to find $c_1$.

We can see that there are $n$ ways of obtaining $(-1)^{n-1}\lambda^{n-1}$.

Then

$$c_1 = (-1)^{n-1}/(-1)^n \cdot \sum_{i=1}^n a_{ii} = -\text{trace}(A)$$

Next,

$\textbf{Theorem}$[The fundamental Theorem of Algebra]. If $P(x)$ is a polynomial of degree, then the equation $P(x) = 0$ has exactly $n$ roots.

By the above theorem,

$$\begin{align} \begin{aligned} p(\lambda) = (-1)^n(\lambda-\lambda_1)\cdots(\lambda-\lambda_n) \\ =(-1)^n\left[ \lambda^n - \lambda^{n-1}\sum_{i=1}^n \lambda_i+\cdots + \lambda^{n-k}\sum_{(i_1,\cdots,i_k) \in \delta_k}\prod_{j=1}^k \lambda_{i_j} \right] \end{aligned} \end{align}$$

So, $c_1 = -\sum_{i=1}^n \lambda_i$.

Since $c_1 = -\text{trace}(A)$, we obtain that

$$\text{trace}(A) = \sum_{i=1}^n \lambda_i$$

Also, we find that $s_k =(-1)^k c_k$.

$\prod_{i=1}^n \lambda_i = (-1)^n c_n$.

By setting $\lambda = 0$, we obtain

$$\text{det}(A) = (-1)^n c_n =\prod_{i=1}^n \lambda_i$$

Reference

[1]. Meyer, Carl D. Matrix analysis and applied linear algebra. Society for Industrial and Applied Mathematics, 2023.