[Linear Algebra] #13 Eigenbases, Diagonalization, Quadratic forms

1.  Eigenbases

 

Linear transformation, y= Ax 에서,

eigenvectors 들은 $R^n$ space 를 위한 bases 를 형성할 수도 있고 안할 수도 있으나, 일단 형성한다고 가정하자.

 

Assumption. A : n x n  →   x : n x 1

eigenvectors 들이 n개가 있다고 가정하자. → non-defective matrix / diagonalizable A

 

$x$ = $c_{1} x_{1}$ + $c_{2} x_{2}$ + ··· +  $c_{n} x_{n}$    →    eigenvectors 들은 무조건 linearly independent.

 

대입 → $y$ = $Ax$ = $A$( $c_{1} x_{1}$ + $c_{2} x_{2}$ + · · · +  $c_{n} x_{n}$ )

                                = $c_{1} A x_{1}$ + $c_{2} A x_{2}$ + ··· +  $c_{n} A x_{n}$

                                = $c_{1} \lambda_{1} x_{1}$ + $c_{2} \lambda_{2} x_{2}$ + ··· +  $c_{n} \lambda_{n} x_{n}$

                                = $c'_{1} x_{1}$ + $c'_{2} x_{2}$ + ··· +  $c'_{n} x_{n}$

 

의미 → $Ax$ 라는 linear transformation 은 eigenvectors 들의 간단한 linear combination 으로 표현된다는 것을 보여준다.

 

 

 

 

* $Ax = xD$ 유도

 

Assumption. Linear transformation $y$ = $Ax$ 에서,eigenvectors 들은 $R^n$ space 를 위한 bases 를 형성한다고 가정하자.  → non-defective matrix A

 

 

 

 

Theorem 1」 Basis of eigenvectors

If an n x n matrix A has n distinct eigenvalues, the A has a basis of eigenvectors

$x_{1}$  $x_{2}$  ···  $x_{n}$ for $R^{n}$

 

→ Algebraic multiplicity, $M_{\lambda} = 1$

    Geomatric multiplicity, $m_{\lambda} = 1$

→ $\Delta_{\lambda}$ = 0   ( ∵ $M_{\lambda}$ ≥ $m_{\lambda}$ ≥ 1 )

 

 

 

 

Theorem 2」 Symmetric matrix

A symmetric matrix has an ortonormal basis of eigenvectors for $R^n$

 

Note.

symmetric matrix 는 orthogonal eigenvector 들을 갖는다.

 

 

 

 

 

2. Similarity of matrices. Diagonalizagion

 

목적 : Eigenbasis → A 를 diagonalize 하자.

 

Def」 similarity matrices, similarity transformation

An n x n matrix $\hat A$  is called similar to an n x n matrix $A$  if  $\hat A$ = $P^{-1} A P$  for some (nonsingular) n x n matrix $P$.

This transformation, which gives $\hat A$ from $A$, is called a similarity transformation.

 

 

 

Theorem 3」 Eigenvalues and Eigenvectors of similar matrices

If $\hat A$ is similar to $A$, then $\hat A$ has the same eigenvalues as A.

Furthermore, if $x$ is an eigenvector of $A$,  then $y = P^{-1} x$ is an eigenvector of $\hat A$ corresponding to the same eigenvalue.

 

Proof」

· 주어진 조건 : ① $Ax = \lambda x$      ② $\hat A$ = $P^{-1} A P$                *$P$ : 임의의 nonsingular matrix

· 증명 방향 :  $\lambda x$($A$)  →  $\lambda P^{-1} x$($\hat A$)

 

①  →  $Ax = \lambda x$

 

      →  $P^{-1} Ax$   =  $P^{-1}$ $\lambda x$

 

      →  $P^{-1} A P$$P^{-1} x$    =    $\lambda P^{-1} x$

 

    →  $\hat A$$P^{-1} x$   =   $\lambda P^{-1} x$   by  ②

 

∴ $\hat A$ 의 eigenvalue 는 $A$ 와 같은 $\lambda$ 가 되고, 그 $\lambda$에 대응하는 eigenvector는 $P^{-1} x$가  된다.

 

 

 

 

Theorem 4」 Diagonalization of Matrix

If an n x n matrix A has a basis of eigenvectors, then $D = X^{-1}AX$ is diagonal.

Here, X is the matrix with these eigenvectors as column vectors.

Also, $D^m = X^{-1}A^{m}X$    ($m = 2, 3,$ ···)

 

Proof」