高等代数学 Chapter 1,2,6
Theorem 1.7.1
Laplace Theorem
\[ \det (A) = \sum_{\text{fix i or j}} A\begin{pmatrix} i_1 & i_2 & \cdots & i_k \\ j_i & j_2 & \cdots & j_k \end{pmatrix} \hat{A} \begin{pmatrix} i_1 & i_2 & \cdots & i_k \\ j_i & j_2 & \cdots & j_k \end{pmatrix} \]
where
\[ A\begin{pmatrix} i_1 & i_2 & \cdots & i_k \\ j_i & j_2 & \cdots & j_k \end{pmatrix} = \text{keep row i and column j}\]
\[ A\begin{pmatrix} i_1 & i_2 & \cdots & i_k \\ j_i & j_2 & \cdots & j_k \end{pmatrix} = (-1)^{\sum i+j}\text{remove row i and column j}\]
helpful to calculate det for matrix that have alot of zeros
helpful to deduce an algebric expression of something
Theorem 2.7.1
Cauchy-Binet Theorem
\[\det(AB) = \sum_{j}A\begin{pmatrix} 1 & 2 & \cdots & m \\ j_i & j_2 & \cdots & j_m \end{pmatrix} B \begin{pmatrix} j_i & j_2 & \cdots & j_m \\ 1 & 2 & \cdots & m \end{pmatrix} \]
This means find the common length of matrix A and B. And Iterate all possible choices of the other dimension.
Definition 4.3.1
if
\[ B = P^{-1}AP\]
then \[B \sim A\]
Finding a good similar matrix
let P be an array of eigenvectors \[ P = (v_1,v_2,\cdots,v_n) \]
then
\[P^{-1}AP = \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)\]
\[AP = P\text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)\]
\[AP = \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)P\]
so all matrix in this equation have a good shape
Step by Step method:
Find the eigenvalues
Find the eigenvectors
Match the eigenvalues with eigenvectors
\(P\) is eigenvectors, \(P^{-1}AP\) is eigenvalues
Exercise
Find \(A^{10}\)
\[ A =\begin{pmatrix} 1 & 0 \\ 1 & -2 \end{pmatrix} \]
\[ |A-\lambda I| = \begin{pmatrix} 1-\lambda & 0 \\ 1 & -2-\lambda \end{pmatrix} = 0\]
\[ (1-\lambda)(-2-\lambda)=0\]
\[ \lambda_1 = 1, \lambda_2 = -2\]
Step 1.
\[P^{-1}AP = \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix}\]
\[\begin{pmatrix} 0 & 0 \\ 1 & -3 \end{pmatrix}v_1=0\]
\[v_1 = (3,-1)^{T}\]
\[\begin{pmatrix} 3 & 0 \\ 1 & 0 \end{pmatrix}v_2=0\]
\[v_1 = (0,114514)^{T}\]
Step 2. \[ P = \begin{pmatrix} 3 & 0 \\ -1 & 1 \end{pmatrix}\]
\[A^{10} = P(P^{-1}AP)^{10}P^{-1} = \begin{pmatrix} 1 & 0 \\ 1-2^{10} & 2^{10} \end{pmatrix} \]
Theorem 6.3.1
Cayley-Hamilton Theorem
let the eigenvalues of matrix A form
\[f(x) = \prod_{i=1}^n (\lambda_i-x)\]
then
\[f(A) = O\]
Proofing this consider an upper triangular Matrix B that is similar to A