Linear Algebra - Eigenvalues/Eigenvectors?

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Homework Statement


a.) If A2 = I, what are the possible eigenvalues of A?
b.) If this A is 2 by 2, and not I or -I, find its trace and determinant.
c.) If the first row is (3,-1), what is the second row?


Homework Equations


None was given, but I think:

1. det(A) = 1 for A2 = I or A-1 = A

We're studying about Matrix diagonalization and the topic is called "Diagonalization of a Matrix". The equation S[tex]\Lambda[/tex]S-1 = A is supposed to be relevant.

The Attempt at a Solution


a.) I got the possible eigenvalues to be: [tex]\lambda[/tex]1 x [tex]\lambda[/tex] 2 x ... x [tex]\lambda[/tex]n = 1
b.) tr(A) = [tex]\lambda[/tex] + [tex]1/\lambda[/tex]
det(A) = 1
c.) This is where I'm stuck... I know I'm supposed to use the equation det(A) = 1 but there are two unknowns and I only know one equation. I was going to use the equation S[tex]\Lambda[/tex]S-1 = A but then I realized that I need the eigenvectors and I can't find the eigenvectors since I don't know the whole matrix.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi KIDRoach! Welcome to PF! :smile:

You're making this very complicated. :redface:

Hint: if det(A) = a, and det(B) = b, what is det(AB)? :wink:

(alternatively, just write A relative to a basis in which A is diagonal)
 
  • #3
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Hi Tim!

Thanks for answering my question. I tried that too I think, but I don't think it's right, or at least either that's wrong or one of the earlier answer is wrong. :(

I know that det(AB) = det(A) x det(B) = ab

When I tried it to solve the equation:
[tex]

\[ \left( \begin{array}{ccc}
3 & -1 \\
0 & x \end{array} \right)\]
[/tex]

I get x = 1/3, which, when I square the A matrix, I don't get the I matrix, so the equation is inconsistent/ something is wrong... But I can't figure out which part I'm wrong at... :(
 
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  • #4
tiny-tim
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I know that det(AB) = det(A) x det(B) = ab
ok, so if A2 = I, and if A ≠ I or -I, then det(A) = … ? :smile:
 
  • #5
hunt_mat
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The eigenvalue equation is [tex]Ax=\lambda x[/tex]. Multiply by A to obtain
[tex]
A^{2}x=\lambda Ax=>x=\lambda^{2}x
[/tex]
So for this to hold [tex]\lambda =\pm 1[/tex]
In order to compute the third part, you have two equations, one coming from the determinant equation and the other one coming from the trace equation.
 
  • #6
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Just wondering, so my answers in part a and b is wrong then? No one has mentioned it yet so I'm not sure... I know that part b is wrong now, that det(A) = -1 instead of 1 ?

Tim,

If A is 2x2 and not I or -I, then det(A) = -1 right? I still don't get how to get from that point to the third though. I still get 1/3 when I use det(A) = -1 to solve the diagonal matrix equation:
Sorry for being so dumb... :frown:

hunt_mat
Wouldn't it be A2x = [tex]\lambda^{2}[/tex]x instead?
I tried using the trace equation. The thing is, the trace equation brings in the two [tex]\lambda[/tex] into the equation and adds significant complexity when I'm computing the eigenvectors.
 
  • #7
hunt_mat
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From the question, we know that A^2 =I. One thing which you may not know is that you can diagonalise a matrix via the eigenvectors. So the determinant is just the product of the eigenvalues, hence det(A)=(+1)(-1)=1. Off the top of my head, I think that the tr(A) is given by the sum of the eigenvalues, so tr(A)=0. Write you matrix as:
[tex]
A=\left(\begin{array}{cc}
-3 & 1 \\
x & y
\end{array}\right)
[/tex]
Use the two equations which I mentioned to obtain your solution.
 
  • #8
vela
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Just wondering, so my answers in part a and b is wrong then? No one has mentioned it yet so I'm not sure... I know that part b is wrong now, that det(A) = -1 instead of 1 ?
Yes, they're wrong. Hunt_mat showed you that the eigenvalues have to be +1 or -1. So think about how A looks in the basis where it's been diagonalized. There are only four possible matrices.

Wouldn't it be A2x = [tex]\lambda^{2}[/tex]x instead?
I tried using the trace equation. The thing is, the trace equation brings in the two [tex]\lambda[/tex] into the equation and adds significant complexity when I'm computing the eigenvectors.
Remember that A2=I, so no more A's on the lefthand side.
 

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