Important Example Solved Problems - Engineering Maths or Mathematics - Eigenvalues and Eigenvectors: Theorem 2 - Example Solved Problems
EIGENVALUES
AND EIGENVECTORS – THEOREM 2
EXAMPLE PROBLEMS
Example 2
Let T be a linear operator
on V, and let λ be an eigenvalue of T. Prove that a vector v ∈ V is an eigenvector of T
corresponding to λ if and only if v ≠ 0 & v ∈
N(T−λI).
Solution:
Let
T be a linear operator on vector space V.
Let
λ be an eigenvalue of T.
Let
v ∈ V be an eigenvector of
T corresponding to the eigenvalue λ.
To
prove v ∈
N (T − λI)
Since
v is an eigenvector of T corresponding to λ.
We
have Tv=λ v
⇒ Tv − λv = (T− λI)v = 0
v ∈
N(T−λI)
Conversely
if v ∈ N(T−λI) then (T−λI)v=0.
⇒ (T−λI)v=0
⇒ Tv=λv
⇒ T=λ
v is an
eigenvector of T corresponding to eigenvalue 2.
Example 3
Prove that similar
matrices have the same characteristic polynomial.
Solution:
Let
A and B be two similar matrices and p(k) denote the kth degree polynomial.
Now
we have to prove that p(A) and p(B) are similar matrices.
Since
A and B are similar matrices, B = P‒1AP for some matrix P.
Now
let
|B| = |P‒1AP| = |P‒1| |A|
|P|
=
|A| |P‒1| | P |
=
|A| |P‒1P|
=
|A| |I|
=
|A|
|B|
= |A|
Now
P(B)
= | B−λI |
(B ‒
λI) = P‒1 AP − λI = P‒1 AP − λ(P‒1IP)
=
P‒1(A−λI)P
Since
|B−λI| is same as A−λI
⇒ P(A) and P (B) are
similar matrices.
The similar matrices have the same
characteristic polynomial.
Example 4
Prove that similar
matrices have the same trace.
Solution:
Trace
of [A]n×n Sum of its diagonal elements.
Let
A & B be two matrices of same order n.
Trace
(A) =
Aii
Let
A and B be two similar matrices.
Trace
(AB) = Trace (BA)
Then
there exists an invertible matrix P such that B = PAP−1.
Trace
(B) = Trace (P A P‒1) = Trace (P‒1AP)
=
Trace (P−1 PA)
=
Trace (IA)
=
Trace (A)
Trace
(B) = Trace (A)
Hence,
similar matrices have the same trace.
Linear Algebra: UNIT II: Linear Transformations and Diagonalization : Tag: maths, mathematics : - Eigenvalues and Eigenvectors: Theorem 2 - Example Solved Problems
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