Linear Algebra - Matrix with given eigenvalues

[roto25]

Homework Statement

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.

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.

 

[Dick]

Pick a diagonal matrix.

 

[roto25]

Does that count for the entries being positive though?

 

[Dick]

Does that count for the entries being positive though?

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

 

[roto25]

thanks though!

 

[micromass]

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.

 

[HallsofIvy]

? 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 positive"?

 

[micromass]

He probably doesn't consider 0 to be positive.

 

[HallsofIvy]

But it is much easier to claim that 0 is positive!

Thanks.

 

[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?

 

[micromass]

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

 

[roto25]

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

 

[Dick]

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?

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?

 

[roto25]

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

 

[Dick]

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

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.

 

[roto25]

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

 

[Dick]

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

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

 

[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

 

[Dick]

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

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.

 

[roto25]

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

 

[Ray Vickson]

The 3x3 case seems much harder. In fact, is it even obvious that there exists a 3x3 matrix A, with all entries > 0, that has the eigenvalues you specified?

RGV

 

[I like Serena]

The 3x3 case is much harder.
I just spent a significant amount of time verifying that I could indeed find a 3x3 solution! :)