# Linear Algebra, eigenvectors problem

• Apr 15th 2010, 06:26 AM
ksheehan62
Linear Algebra, eigenvectors problem
If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

(1) v is also an eigenvector of 2A
(2) v is also an eigenvector of A^2
(3) v is also an eigenvector of A^-1

A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 1,2 and 3

I am pretty sure 2 is true because if we look at Av = (lamba)v.
A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

Can someone help show 3? If Av = lambda v
then (A^-1)Av =A^-1(lambda)v
I am not sure how to show it
• Apr 15th 2010, 06:52 AM
HallsofIvy
Quote:

Originally Posted by ksheehan62
If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

(1) v is also an eigenvector of 2A
(2) v is also an eigenvector of A^2
(3) v is also an eigenvector of A^-1

A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 1,2 and 3

I am pretty sure 2 is true because if we look at Av = (lamba)v.
A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

Very good! (Clapping)

Quote:

2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
I presume you meant (1) here.

Quote:

Can someone help show 3? If Av = lambda v
then (A^-1)Av =A^-1(lambda)v
I am not sure how to show it
Well, what is (A^-1)Av? And, of course, because A and $\displaystyle A^{-1}$ are linear operators, $\displaystyle A^{-1}(\lambda v)= \lambda A^{-1}v$.
• Apr 15th 2010, 06:54 AM
HallsofIvy
Quote:

Originally Posted by ksheehan62
If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

(1) v is also an eigenvector of 2A
(2) v is also an eigenvector of A^2
(3) v is also an eigenvector of A^-1

A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 1,2 and 3

I am pretty sure 2 is true because if we look at Av = (lamba)v.
A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

Very good! (Clapping)

Quote:

2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
I presume you meant (1) here.

Quote:

Can someone help show 3? If Av = lambda v
then (A^-1)Av =A^-1(lambda)v
I am not sure how to show it
Well, what is (A^-1)Av? And, of course, because A and $\displaystyle A^{-1}$ are linear operators, $\displaystyle A^{-1}(\lambda v)= \lambda A^{-1}v$.

As long as $\displaystyle \lambda\ne 0$, you can divide both sides by $\displaystyle \lambda$. Of course, if $\displaystyle \lambda= 0$, then A does not have an inverse.
• Apr 15th 2010, 06:55 AM
tonio
Quote:

Originally Posted by ksheehan62
If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

(1) v is also an eigenvector of 2A
(2) v is also an eigenvector of A^2
(3) v is also an eigenvector of A^-1

A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 1,2 and 3

I am pretty sure 2 is true because if we look at Av = (lamba)v.
A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

Can someone help show 3? If Av = lambda v
then (A^-1)Av =A^-1(lambda)v
I am not sure how to show it

$\displaystyle Av=\lambda v\Longrightarrow v=Iv=A^{-1}Av=A^{-1}(Av)=A^{-1}(\lambda v)=\lambda A^{-1}v\Longrightarrow$ multiplying now by $\displaystyle \lambda^{-1}$ in both extremes of the above equality (why

can we do such a thing?) you get that $\displaystyle v$ is also an eigenvector of $\displaystyle A^{-1}$ .

Tonio