Eigenvalues of the power of a matrix

IniquiTrance

Homework Statement

If $\lambda_i$ are the eigenvalues of a matrix $A^2$, and $A$ is symmetric, then what can you say about the eigenvalues of A?

The Attempt at a Solution

I know how to prove that if $\sqrt(\lambda_i)$ is an eigenvalue of A, then $\lambda_i$ is an eigenvalue of $A^2$.

I don't know how to prove the converse though.

Thanks!

Staff Emeritus
Homework Helper
Let $\lambda$ be an eigenvalue of $A^2$. What can you say about $A^2-\lambda I$?

Now, use that

$$A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)$$

IniquiTrance
Let $\lambda$ be an eigenvalue of $A^2$. What can you say about $A^2-\lambda I$?

Now, use that

$$A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)$$

Then:

$$det(A-\sqrt{\lambda}I)*det(A+\sqrt{\lambda} I) = 0$$

Not sure how to proceed...

Staff Emeritus
Homework Helper
If xy=0, then what can you say about x or y?

IniquiTrance
If xy=0, then what can you say about x or y?

Oh so all we can say is that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of A, but not both?

Staff Emeritus
Homework Helper
Oh so all we can say is that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of A, but not both?

Right.

IniquiTrance
Thank you.

IniquiTrance
One more question, how do we know that all the eigenvalues of $A^2$ are positive? Since A is symmetric, if $\lambda < 0$ were an eigenevalue of $A^2$ we'd run into a problem...

Staff Emeritus
Homework Helper
We can do the same with $\sqrt{\lambda}$ complex numbers, no?

IniquiTrance
We can do the same with $\sqrt{\lambda}$ complex numbers, no?

But since A is symmetric, it must have real eigenvalues, no?

Staff Emeritus
Homework Helper
But since A is symmetric, it must have real eigenvalues, no?

Yes, sure. So what we do is we argue the same argument where $\lambda$ might be negative and where $\sqrt{\lambda}$ might be complex. Then we can deduce that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ are eigenvalues of A. Since A is symmetric, we have that $\sqrt{\lambda}$ must be real, so $\lambda$ was positive to begin with.

IniquiTrance
Yes, sure. So what we do is we argue the same argument where $\lambda$ might be negative and where $\sqrt{\lambda}$ might be complex. Then we can deduce that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ are eigenvalues of A. Since A is symmetric, we have that $\sqrt{\lambda}$ must be real, so $\lambda$ was positive to begin with.

Ah ok, thanks again!

$$A = \pmatrix{2 & 0 \\0 & -2}$$ has
$$B = A^2 = \pmatrix{4 & 0 \\0 & 4},$$
and so both $\sqrt{4}$ and $-\sqrt{4}$ are eigenvalues of $A.$