Linear Algebra - Matrix with given eigenvalues

In summary, the conversation discusses the process of coming up with a 2x2 matrix with eigenvalues of 2 and 1 and all positive entries, and then finding a 3x3 matrix with eigenvalues of 1, 2, and 3. Various attempts and suggestions are made, such as using diagonal matrices and graphing to find suitable values for the diagonal entries. Eventually, a 3x3 matrix is found that satisfies the conditions.
  • #1
roto25
11
0

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.
 
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  • #2
Pick a diagonal matrix.
 
  • #3
Does that count for the entries being positive though?
 
  • #4
roto25 said:
Does that count for the entries being positive though?

Not really, no. Sorry. Better give this more thought than I gave this response.
 
  • #5
thanks though!
 
  • #6
The 2x2-case is not so difficult. Remember (or prove) that the characteristic polynomail of a 2x2-matrix A is

[tex]\lambda^2-tr(A)\lambda+det(A)[/tex]

By the way, I think your characteristic polynomial is wrong.
 
  • #7
? Why do the diagonal matrices
[tex]\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}[/tex]
and
[tex]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]
NOT count as "all entries postive"?
 
  • #8
He probably doesn't consider 0 to be positive.
 
  • #9
But it is much easier to claim that 0 is positive!:tongue:

Thanks.
 
  • #10
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
You still have 0 as an entry, you don't want that.
 
  • #12
Yeah, I didn't realize that at first. :/
 
  • #13
roto25 said:
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?
 
  • #14
Yes. Theoretically, I know what it should do. I just can't actually find the right values to do it.
 
  • #15
roto25 said:
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.
 
  • #16
Well, any value of x between 1 and 2 (like 1.1) work.
 
  • #17
roto25 said:
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.
 
  • #18
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
roto25 said:
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.
 
  • #20
I had just figured out that
1.5 0.5
0.5 1.5
worked out! :)
 
  • #21
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
 
  • #22
The 3x3 case is much harder.
I just spent a significant amount of time verifying that I could indeed find a 3x3 solution! :)
 

1. What is a matrix with given eigenvalues?

A matrix with given eigenvalues is a square matrix that has a predetermined set of numbers called eigenvalues. These eigenvalues represent the scaling factor by which a vector is stretched or compressed when multiplied by the matrix.

2. What is the significance of eigenvalues in linear algebra?

Eigenvalues are crucial in linear algebra as they provide information about the behavior of a matrix. They can determine if a matrix is invertible, the dimension of its null space, and its diagonalizability, among other important properties.

3. How do you find the eigenvectors of a matrix with given eigenvalues?

To find the eigenvectors of a matrix with given eigenvalues, you would need to solve the equation (A - λI)x = 0, where A is the given matrix, λ is the eigenvalue, and x is the eigenvector. This will result in a system of linear equations that can be solved to find the eigenvectors.

4. Can a matrix have multiple sets of eigenvalues and eigenvectors?

Yes, a matrix can have multiple sets of eigenvalues and eigenvectors. This is because different eigenvectors can correspond to the same eigenvalue, and a matrix may have repeated eigenvalues.

5. How is the trace of a matrix related to its eigenvalues?

The trace of a matrix is equal to the sum of its eigenvalues. This relationship is useful in determining the eigenvalues of a matrix since the trace can be easily calculated by adding the elements on the main diagonal of the matrix.

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