# Linear Algebra - Matrix with given eigenvalues

1. ### roto25

11
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. ### Dick

25,822
Pick a diagonal matrix.

3. ### roto25

11
Does that count for the entries being positive though?

4. ### Dick

25,822
Not really, no. Sorry. Better give this more thought than I gave this response.

5. ### roto25

11
thanks though!

6. ### micromass

18,695
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. ### HallsofIvy

40,386
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. ### micromass

18,695
Staff Emeritus
He probably doesn't consider 0 to be positive.

9. ### HallsofIvy

40,386
Staff Emeritus
But it is much easier to claim that 0 is positive!:tongue:

Thanks.

10. ### roto25

11
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. ### micromass

18,695
Staff Emeritus
You still have 0 as an entry, you don't want that.

12. ### roto25

11
Yeah, I didn't realize that at first. :/

13. ### Dick

25,822
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. ### roto25

11
Yes. Theoretically, I know what it should do. I just can't actually find the right values to do it.

15. ### Dick

25,822
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. ### roto25

11
Well, any value of x between 1 and 2 (like 1.1) work.

17. ### Dick

25,822
Ok, so you just need to fill in the rest of the matrix.

18. ### roto25

11
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. ### Dick

25,822
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. ### roto25

11
I had just figured out that
1.5 0.5
0.5 1.5
worked out! :)