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· by Xianbin
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.