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Eigenvectors form an ONB

Module by: Richard Baraniuk

Summary: Shows the concepts behind eigenvectors forming an ONB.

Let us focus on normal matrices. A matrix A A is normal if it commutes with AH A That is,

AAH=AHA A A A A (1)

Example 1

A=010001100 A 0 1 0 0 0 1 1 0 0

AAH ≟ AHA A A ≟ A A (2)

010001100001100010 ≟ 001100010010001100 0 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 1 0 ≟ 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0 (3)

100010001=100010001 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 (4)

note:

We will use normal matrices a lot in ELEC 301

A special case of a normal matrix is a symmetric matrix where

A=AH A A (5)

ie: row i = i column i i.

Facts on Eiegenanalysis of Normal Matrices

  1. When normalized to ∥vi∥2=1 2 x v i 1 , their eigenvectors form an ONB for ℂN N .
  2. They can be diagonalized
    A=VΛVH A V Λ V (6)
    where Λ Λ is a diaglonal matrix of eigenvalues λ 1 0000 λ 2 0000... ...00... λ N - 1 λ 1 0 0 0 0 λ 2 0 0 0 0 ... ... 0 0 ... λ N - 1

Example 2

Show that 2 is true using 1.

Avi=λivi A v i λ i v i (7)
for i=0 i 0 , 1 1, 2 2, ..., N-1 N 1

∥vi∥2=1 2 v i 1 (8)
V=v0v1v2...v N - 1 V v 0 v 1 v 2 ... v N - 1

ONB meatix implies

VHV=I V V I (9)
VVH=I V V I (10)
Σ= λ 1 0000 λ 2 0000... ...00... λ N - 1 Σ λ 1 0 0 0 0 λ 2 0 0 0 0 ... ... 0 0 ... λ N - 1

Example 3

Show that

A=VΛVH A V Λ V (11)
AV=VΛVHV A V V Λ V V (12)
AV=VΛ A V V Λ (13)
look in detail at the LHS
AV=Av0v1v2...v N - 1 A V A v 0 v 1 v 2 ... v N - 1 (14)
AV=Av0Av1Av2...Av N - 1 A V A v 0 A v 1 A v 2 ... A v N - 1 (15)
why?
AV=λ0v0λ1v1λ2v2... λ N - 1 v N - 1 A V λ 0 v 0 λ 1 v 1 λ 2 v 2 ... λ N - 1 v N - 1 (16)
AV= λ 1 0000 λ 2 0000... ...00... λ N - 1 v0v1v2...v N - 1 A V λ 1 0 0 0 0 λ 2 0 0 0 0 ... ... 0 0 ... λ N - 1 v 0 v 1 v 2 ... v N - 1 (17)
AV=ΛV A V Λ V (18)
= RHS.

Example 4

A=3-1-13 A 3 -1 -1 3 since symmetric, A A is normal.

From earlier we have V=v0v1=12111-1 V v 0 v 1 1 2 1 1 1 -1

Λ=2004 Λ 2 0 0 4

compute VΛVH V Λ V

=12111-12004111-112 1 2 1 1 1 -1 2 0 0 4 1 1 1 -1 1 2 (19)
=12111-1224-4 1 2 1 1 1 -1 2 2 4 -4 (20)
=126-2-26=3-1-13=A 1 2 6 -2 -2 6 3 -1 -1 3 A (21)

Bottom Lines

  1. Every normal matrix A A has a natural basis (ONB) formed by its eigenvectors.
  2. We can diagonalize any normal matrix A A by
    A=VΛVH A V Λ V (22)
    Sometimes it is much more efficient to apply VH V then Λ Λ then V V than A A.
  3. A Ais completely characterized by its eigenvectors vi v i and eigenvalues λi λ i

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