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Lecture 22

# MATA23H3 Lecture Notes - Lecture 22: Diagonal Matrix, Diagonalizable Matrix, Eigenvalues And EigenvectorsPremium

Department
Mathematics
Course Code
MATA23H3
Professor
Chrysostomou( G)
Lecture
22

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MATA23 - Lecture 22 - Properties of Eigenvalues, Diagonalization, and Multiplicity
Properties of Eigenvalues
If Ahas eigenvalues λ1, λ2, . . . , λn(repeated according to multiplicity), then detA =λ1λ2. . . λn
and trA =λ1+λ2+· · · +λn
det(AλI)is a polynomial of degree n
λ1, λ2, . . . , λnare roots of det(AλI)=0
det(AλI)=(λ1λ)(λ2λ). . . (λnλ)
Set λ= 0, detA =λ1λ2. . . λn
Example 3: Let A= [aij ]be a 3×3matrix and have its characteristic polynomial det(AλI) =
λ3+ 3λ2. Find detA and trA
p(λ) = det(AλI) = λ3+ 3λ2
Set p(λ)=0
p(1) = 1+32 = 0
p(λ) = λ3+λ+ 2λ2
=λ(λ2+ 1) + 2(λ1)
=λ(λ+ 1)(λ+ 1) + 2(λ1)
= (λ+ 1)[λ(λ+ 1) 2]
= (λ+ 1)(λ2+λ2)
= (λ+ 1)(λ1)(λ+ 2),set = 0
λ1= 1, λ2= 1, λ3=2
detA =λ1λ2λ3= (1)(1)(2) = 2
trA =λ1+λ2+λ3= 1 + 1 2=0
Diagonalization
Let Aand Dbe n×nmatrices, and Dbe a diagonal matrix. Ais said to be diagonalizable
if there exists an invertible n×nmatrix Psuch that P1AP =D
Let ~v1, ~v2, . . . , ~vkbe eigenvectors corresponding to the distinct eigenvalues λ1, λ2, . . . , λkof
the square matrix A, then ~v1, ~v2, . . . , ~vkare linearly independent
A~vi=λi~vi, i = 1,2, . . . , k
λi6=λj,if i6=j
A~v =λ1~v
A~v =λ2~v
λ16=λ2
λ1~v λ2~v =~
0
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~v(λ1λ2) = ~
0
(λ1λ2)6=~
0
=~v =~
0,which is impossible
n=1:~v16=~
0, A~v1=λ~v
{~v1}is linearly independant
n=2:A~v1=λ1~v1, A~v2=λ2~v2
Let r1~v1+r2~v2=~
0(#1)
r1, r2R
A(r1~v1+r2~v2) = A~
0 = ~
0
rkA~v1+r2A~v2=~
0
r1λ1~v1+r2λ2~v2=~
0(#2)
By #1: #2 =λ1~
0 + λ2~
0 = ~
0(#3)
#2 - #3: r2(λ2λ1)~v2=~
0
λ2λ16= 0
~v26= 0
=r2= 0
equation #1 becomes r1~v1=~
0
~v16=~
0 =r1= 0
{~v1, ~v2}linearly independent
n=j1, j < k
Assume that {~v1, . . . , ~vj}linearly independent
WTS {~v1, . . . , ~vj1, ~vj}linearly independent
Let r1~v1+r2~v2+· · · +rj1~vj1+rj~vj=~
0(#1’)
riR, i = 1, . . . , j
r1A~v1+r2A~v2+· · · +rj1A~vj1+rjA~vj=A~
0 = ~
0
A~vi=λi~vi, i = 1, . . . , j
r1λ1~v1+· · · +rj1λj1~vj1+rjλj~vj=~
0(#2’)
λj·#1’ =#2’ (#3’)
#2’ - #3’: r1(λ1λj)~v1+· · · +rj1(λj1λj)~vj1=~
0
{~v1, ~v2, . . . , ~vj1}linearly independent
r1(λ1λj)=0, . . . , rj1(λj1λj) = 0
#1’ =rj~vj=~
0
λiλj6= 0 =i6= 0
ri= 0, i = 1, . . . , j 1
~vj6= 0, rj= 0
{~v1, . . . , ~vj1, ~vj}linearly independent
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