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

25,887
Pick a diagonal matrix.

3. roto25

11
Does that count for the entries being positive though?

4. Dick

25,887
Not really, no. Sorry. Better give this more thought than I gave this response.

5. roto25

11
thanks though!

6. micromass

19,347
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

40,769
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

19,347
Staff Emeritus
He probably doesn't consider 0 to be positive.

9. HallsofIvy

40,769
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

19,347
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

25,887
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

25,887
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

25,887
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

25,887
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! :)