1. ## Eigenvalue

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

How to show that:

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

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

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

2. Hello,
Originally Posted by noob mathematician
Let $\displaystyle \lambda$ be an eigenvalue of a square matrix $\displaystyle A$.

How to show that:

1. $\displaystyle \lambda^n$ is an eigenvalue of $\displaystyle A^n$ where n is a positive integer.
If $\displaystyle \lambda$ is an eigenvalue of A, then there exists X such that :
$\displaystyle AX=\lambda X$

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

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

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

2. If $\displaystyle A$ is invertible, show that $\displaystyle \frac {1}{\lambda}$ is an eigenvalue of $\displaystyle A^{-1}$.
$\displaystyle AX=\lambda X$
Multiply to the left by $\displaystyle A^{-1}$ :

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

And same reasoning as above.

3. Originally Posted by noob mathematician
Let $\displaystyle \lambda$ be an eigenvalue of a square matrix $\displaystyle A$.

How to show that:

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

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

To start off: I was told to let $\displaystyle x$ be an eigenvector such that $\displaystyle Ax= \lambda x$
....
Part 1 is very easy, just follow the hint. $\displaystyle Ax = \lambda x$ so $\displaystyle A^2 x \ = \ A \lambda x \ = \ \lambda A x = \lambda^2 x$. it should be obvious that $\displaystyle \lambda^n$ is an eigenvalue of $\displaystyle A^n$ you may want to write down an inductive argument.

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

Bobak

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

4. ...it should be obvious that
$\displaystyle \lambda^n$ is an eigenvalue of $\displaystyle A^n$...
Do that imply that $\displaystyle A^n$ will always have the same eigenvalues as $\displaystyle A^1$

5. Originally Posted by aidan
Do that imply that $\displaystyle A^n$ will always have the same eigenvalues as $\displaystyle A^1$
no, just the same eigenvectors.

Bobak

6. Originally Posted by bobak
no, just the same eigenvectors.

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

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

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

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?

7. Originally Posted by aidan
I did not phrase that question correctly.
What I meant:

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

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

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 ....?)