Linear Algebra: UNIT II: Linear Transformations and Diagonalization
Characteristic Polynomial: Example Solved Problems
Important Example Solved Problems - Engineering Maths or Mathematics - Linear Transformations and Diagonalization: Characteristic Polynomial
CHARACTERISTIC POLYNOMIAL
EXAMPLE PROBLEMS
Example 5
Let A be a square
matrix. Prove that A and AT have the same eigenvalues.
Solution:
To
prove A and AT have the same eigenvalues let λ be the eigenvalue of
A.
Then
its characteristic equation is |A−λI|.
| A − λI | = | (A − λI)T | = | (AT
– λIT) |
|A− λI|
= |AT − λI |
Since
the characteristic equations of A and AT are identical in any field,
the eigenvalues are same.
Thus
A and AT have the same characteristic equations and have the same
eigenvalues.
Example 6
Find the eigenvalues of the matrix A = 
∈
M2×2(R).
Solution:
Let
A = 
then the characteristic equation is
f(t) = |A − λI| = 0
= (1 – λ)2 ‒ 4 = 0
1+ λ2−2λ−4 = 0
λ2−2λ−3=0
(λ
− 3) (λ + 1) = 0
The
eigenvalues are 3,− 1.
Example 7
Let T be the linear
operator on P2(R) defined by T (f(x))
= f(x)+(x+1)f '(x). Let β be the standard ordered basis for P2(R).
Compute characteristic polynomial and eigenvalue of T.
Solution:
Let
us consider β = { 1, x, x2 }
T(1)=1+0
(1+1)=1=1+0x+0x2
T(x)=x+1(1+x)=1+2x=1+2x+0x2
T(x2) = x2 + (x + 1)2x =
3x2+2x=0+2x+3x2
f(t) = |A−λ| = 
=
− λ3 + 6λ2 − 11λ + 6
=
−(λ−1) (λ−2) (λ−3)
⇒ λ=1, 2, 3
The
eigenvalues are 1, 2, 3.
Linear Algebra: UNIT II: Linear Transformations and Diagonalization : Tag: maths, mathematics : - Characteristic Polynomial: Example Solved Problems
Linear Algebra: UNIT II: Linear Transformations and Diagonalization
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Linear Algebra
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