Eigenvalue

**noob mathematician** · April 11th 2009, 05:59 AM

Let $\lambda$ be an eigenvalue of a square matrix $A$ .

How to show that:

1. $\lambda^n$ is an eigenvalue of $A^n$ where n is a positive integer.

2. If $A$ is invertible, show that $\frac {1}{\lambda}$ is an eigenvalue of $A^{-1}$ .

To start off: I was told to let $x$ be an eigenvector such that $Ax= \lambda x$
....

**Moo** · April 11th 2009, 06:22 AM

Hello,

Originally Posted by noob mathematician

Let $\lambda$ be an eigenvalue of a square matrix $A$ .

How to show that:

1. $\lambda^n$ is an eigenvalue of $A^n$ where n is a positive integer.

If $\lambda$ is an eigenvalue of A, then there exists X such that :
$AX=\lambda X$

Now multiply to the left by A :
$A^2X=A \lambda X$
Since $\lambda$ is a scalar, it commutes with A.

Thus $A^2X=\lambda AX=\lambda (\lambda X)=\lambda^2 X$
Hence $\lambda^2$ is an eigenvalue of $A^2$

You can use this process to prove your question by induction.

2. If $A$ is invertible, show that $\frac {1}{\lambda}$ is an eigenvalue of $A^{-1}$ .

$AX=\lambda X$
Multiply to the left by $A^{-1}$ :

$I_nX=X=\lambda A^{-1}X$

And same reasoning as above.

**bobak** · April 11th 2009, 06:23 AM

Originally Posted by noob mathematician

Let $\lambda$ be an eigenvalue of a square matrix $A$ .

How to show that:

1. $\lambda^n$ is an eigenvalue of $A^n$ where n is a positive integer.

2. If $A$ is invertible, show that $\frac {1}{\lambda}$ is an eigenvalue of $A^{-1}$ .

To start off: I was told to let $x$ be an eigenvector such that $Ax= \lambda x$
....

Part 1 is very easy, just follow the hint. $Ax = \lambda x$ so $A^2 x \ = \ A \lambda x \ = \ \lambda A x = \lambda^2 x$ . it should be obvious that $\lambda^n<br />$ is an eigenvalue of $A^n$ you may want to write down an inductive argument.

part 2 also doesn't present much difficulty. $Ax = \lambda x \rightarrow A^{-1} A x = A^{-1} \lambda x$ so $A^{-1} \lambda x = x$ .

Bobak

Edit: i was beaten where had the delete post button gone ?

**aidan** · April 11th 2009, 11:52 AM

...it should be obvious that
$\lambda^n$ is an eigenvalue of $A^n$ ...

Do that imply that $A^n$ will always have the same eigenvalues as $A^1$

**bobak** · April 11th 2009, 12:14 PM

Originally Posted by aidan

Do that imply that $A^n$ will always have the same eigenvalues as $A^1$

no, just the same eigenvectors.

Bobak

**aidan** · April 12th 2009, 02:45 PM

Originally Posted by bobak

no, just the same eigenvectors.

Bobak

I did not phrase that question correctly.
What I meant:

if $A^1$ has eigenvalues
$\lambda_1 \\ \lambda_2 \lambda_3$

then will $A^n$ have eigenvalues
$\lambda_1^n \lambda_2^n \lambda_3^n$

Based on the answer given above to the original question, it just seems that if it works for one eigenvalue, it will work for all.

When a matrix is raised to a power, will the eigenvalues also be raised to the same power?

**mr fantastic** · April 12th 2009, 04:09 PM

Originally Posted by aidan

I did not phrase that question correctly.
What I meant:

if $A^1$ has eigenvalues
$\lambda_1 \\ \lambda_2 \lambda_3$

then will $A^n$ have eigenvalues
$\lambda_1^n \lambda_2^n \lambda_3^n$

Based on the answer given above to the original question, it just seems that if it works for one eigenvalue, it will work for all.

When a matrix is raised to a power, will the eigenvalues also be raised to the same power?

Yes (was anything special assumed about the eigenvalue in your original post ....?)

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