Linear Algebra - Matrix with given eigenvalues

1. Apr 3, 2012

roto25

1. The problem statement, all variables and given/known data
Come up with a 2 x 2 matrix with 2 and 1 as the eigenvalues. All the entries must be positive.
Then, find a 3 x 3 matrix with 1, 2, 3 as eigenvalues.

3. The attempt at a solution
I found the characteristic equation for the 2x2 would be λ2 - 3λ + 2 = 0. But then I couldn't get a matrix with positive entries to work for that.

Last edited: Apr 4, 2012
2. Apr 3, 2012

Dick

Pick a diagonal matrix.

3. Apr 3, 2012

roto25

Does that count for the entries being positive though?

4. Apr 3, 2012

Dick

Not really, no. Sorry. Better give this more thought than I gave this response.

5. Apr 3, 2012

roto25

thanks though!

6. Apr 4, 2012

micromass

Staff Emeritus
The 2x2-case is not so difficult. Remember (or prove) that the characteristic polynomail of a 2x2-matrix A is

$$\lambda^2-tr(A)\lambda+det(A)$$

By the way, I think your characteristic polynomial is wrong.

7. Apr 4, 2012

HallsofIvy

Staff Emeritus
??? Why do the diagonal matrices
$$\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$$
and
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}$$
NOT count as "all entries postive"?

8. Apr 4, 2012

micromass

Staff Emeritus
He probably doesn't consider 0 to be positive.

9. Apr 4, 2012

HallsofIvy

Staff Emeritus
But it is much easier to claim that 0 is positive!:tongue:

Thanks.

10. Apr 4, 2012

roto25

Oh, I had typed 3 instead of 2 for the characteristic polynomial. I ended up looking at this from a Hermitian matrix point of view.
And then I got the matrix:
0 i +1
i-1 3
And I did get the right eigenvalues from that. Does that work?

11. Apr 4, 2012

micromass

Staff Emeritus
You still have 0 as an entry, you don't want that.

12. Apr 4, 2012

roto25

Yeah, I didn't realize that at first. :/

13. Apr 4, 2012

Dick

I don't think i+1 would be considered a positive number either. Stick to real entries. Your diagonal entries need to sum to 3, and their product should be greater than 2. Do you see why?

14. Apr 4, 2012

roto25

Yes. Theoretically, I know what it should do. I just can't actually find the right values to do it.

15. Apr 4, 2012

Dick

Call one diagonal entry x. Then the other one must be 3-x. Can you find a positive value of x that makes x*(3-x)>2? Graph it.

16. Apr 4, 2012

roto25

Well, any value of x between 1 and 2 (like 1.1) work.

17. Apr 4, 2012

Dick

Ok, so you just need to fill in the rest of the matrix.

18. Apr 4, 2012

roto25

but if I set x to be 1.1, my matrix would be
1.1 __
__ 1.9

And those two spaces have to be equivalent to 1.1*1.9 - 2, right?
because no matter what values I try, when the eigenvalues are getting closer to 1 and two, the matrix is just getting closer to the matrix of:
1 0
0 2

19. Apr 4, 2012

Dick

The two spaces multiplied together have to give you 1.1*1.9 - 2. How about putting one blank to be 1 and the other to be 1.1*1.9 - 2? The eigenvalues should work out to EXACTLY 1 and 2. Try it with x=3/2.

20. Apr 4, 2012

roto25

I had just figured out that
1.5 0.5
0.5 1.5
worked out! :)