Eigenvalues and Eigenvectors

• Eigenvalues
• The characteristic equation
• Similarity
• Diagonalization
• Eigenvectors and Linear Transformations
• Linear Transformations on $R^n$
• Similarity of Matrix Representattions
• Iterative Estimates For Eigenvalues
• The inverse Power Method For estimating An Eigenvalue $\lambda$ Of $A$

Eigenvalues

The Eigenvalues of a triangular matrix are the entries on its main diagonal.
If $v_1,\cdots,v_r$v1​,⋯,vr​ are eigenvectors that correspond to distinct eigenvalues $\lambda_1,\cdots,\lambda_r$λ1​,⋯,λr​ of an $n\times n$n×n matrix $A$A, then the set $\{v_1,\cdots,v_r\}${v1​,⋯,vr​} is linearly independent.

The characteristic equation

Let A be an $n \times n$n×n matrix. Then $A$A is invertible if and only if:
s.the number 0 is not a eigenvalues of$A$A
t.the determinant of $A$A is not zero
A scalar $\lambda$λ is an eigenvalue of an $n \times n$n×n matrix $A$A if and only if $\lambda$λ satisfies the characteristic equation $det(A-\lambda I)=0$det(A−λI)=0

Similarity

$A$A is similar to $B$B if there is an invertible matrix $P$P that $P^{-1}AP=B$P−1AP=B
If $n \times n$n×n matrix $A$A and $B$B are similar, then they have same characteristic polynomial and hence the same eigenvalues(with same multiplicities).

Diagonalization

An $n \times n$n×n matrix $A$A is diagoalizable if and only if $A$A has $n$n linearly independent eigenvectors.
In fact, $A= PDP^{-1}$A=PDP−1, with $D$D a diagonal matrix, if and only if the columns of $P$P are linearly independent eigenvectors of $A$A. In this case, the diagonal entries of $D$D are eigenvalues of $A$A that correspond, respectively, to the eigenvectors in $P$P.
An $n \times n$n×n matrix with $n$n distinct eigenvalues is diagonalizable
Let $A$A ba an $n \times n$n×n matrix whose distinct eigenvalues are $\lambda_1 , \cdots, \lambda_p$λ1​,⋯,λp​.
a. For $1 \leq k \leq p$1≤k≤p, the dimension of the eigenspace for $\lambda_k$λk​ is less than or equal to the multiplicity of the eigenvalue $\lambda_k$λk​.
b. The matrix $A$A is diagonalizable if and only if the sum of the dimensions of the eigenspaces equals $n$n, and this happens if and only if $(i)$(i) the characteristic polynomial factors completely into linear factors and $(ii)$(ii) the dimension of the eigenspace for each $\lambda_k$λk​ equals the multiplicity of $\lambda_k$λk​
c. If $A$A is diagonalizable and $\beta_k$βk​ is a basis for the eigenspace corresponding to $\lambda_k$λk​ for each $k$k, then the total collection of vactors in the sets $\beta_1,\cdots,\beta_p$β1​,⋯,βp​ forms an eigenvectors basis for $R^n$Rn.

Eigenvectors and Linear Transformations

there are two vector space $V$V $R^n$Rn and $W$W $R^m$Rm, coordinate vector $[x]_{\beta}$[x]β​ is of $R^n$Rn and $[T(x)]_{C}$[T(x)]C​ is in $R^m$Rm. ${b_1,\cdots,b_n}$b1​,⋯,bn​ is the basis $\beta$β for $V$V. Then the matrix $M=[[T(b_1)]_c,\cdots,[T(b_n)]_c]$M=[[T(b1​)]c​,⋯,[T(bn​)]c​] The action
of T on $X$X may be viewed as left-multiplication by $M$M.

Linear Transformations on $R^n$Rn

Suppose $A=PDP^{-1}$A=PDP−1, where $D$D is a diagonal $n\times n$n×n matrix. If $B$B is the basis for $R^n$Rn formed from the columns of $P$P, then $D$D is the $\beta$βmatrix for the transformation $x\rightarrow Ax$x→Ax.

Similarity of Matrix Representattions

$A=PCP^{-1}$A=PCP−1

Iterative Estimates For Eigenvalues

If there is an power method applies to an $n \times n$n×n matrix $A$A with a strictly domaint eigenvalue $\lambda_1$λ1​ meaning that $\lambda_1$λ1​ must be larger in aboluate value than all the other eigenvalues. That is $|\lambda_1|>|\lambda_2|\geq |lambda_3| \cdots \geq |\lambda_n|$∣λ1​∣>∣λ2​∣≥∣lambda3​∣⋯≥∣λn​∣ Then through the below method it is easy for us to get a eigenvector:
1. Select an initial vector $x_0$x0​ whose largest entry is 1.
2. For $k=0,1,\cdots,$k=0,1,⋯,
a.compute $Ax_k$Axk​
b.Let $\mu_k$μk​ be an entry in$Ax_k$Axk​ whose abolute value is as large as possible
c.Compute $x_{k+1}=(1/\mu_k)Ax_k$xk+1​=(1/μk​)Axk​
3. For almost all choices of $x_0$x0​, the sequence $\{\mu_k\}${μk​} approaches the domaint eigenvalue, and the sequence $\{x_k\}${xk​} approaches a corresponding eigenvector.

The inverse Power Method For estimating An Eigenvalue $\lambda$λ Of $A$A

1. Select an initial estimate $\alpha$α sufficiently close to $\lambda$λ
2. select an initial vector $x_0$x0​ whose largest entry is 1
3. For $k = 0,1,\cdots,$k=0,1,⋯,
   a.slove $(A-\alpha I)y_k=x_k$(A−αI)yk​=xk​ for $y_k$yk​
   b.Let $\mu_k$μk​ be an entry in $y_k$yk​ whose absolute value is as large as possible.
   c.Compute $v_k=\alpha+(1/\mu_k)$vk​=α+(1/μk​)
   d.Compute $x_{k+1}=(1/\mu_k)y_k$xk+1​=(1/μk​)yk​
4. For almost all choices of $x_0$x0​, the sequence $\{v_k\}${vk​} approaches the eigenvalue $\lambda$λ of $A$A, and the sequence $\{x_k\}${xk​} approaches a corresponding eigenvector.