Is there a faster way to do this matrix problem?

[TheFerruccio]

Is there a faster way to do this matrix problem??

Homework Statement

Verify that \mathbf{A} and \mathbf{\hat{A}} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P} have the same spectrum.

Homework Equations

\mathbf{A} =<br /> \left[<br /> \begin{array}{ccc}<br /> -22 &amp; 20 &amp; 10 \\<br /> -4 &amp; 20 &amp; -8 \\<br /> 28 &amp; -14 &amp; 29 \\<br /> \end{array} \right]

\mathbf{P} =<br /> \left[<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 2 \\<br /> 0 &amp; 2 &amp; 4 \\<br /> 2 &amp; 8 &amp; 0 \\<br /> \end{array} \right]

The Attempt at a Solution

The problem is asking whether two similar matrices have the same set of eigenvalues.

Conceptually, I would first find the eigenvalues of A, by finding the characteristic equation, which will be a cubic equation resulting in 3 eigenvalues as the solutions. In this case, I used a computer to find \lambda = 36, 18, -27. I then find the eigenvectors of A to verify that the matrix P represents the eigenvectors of A.

After that, I would compute the inverse of P using Gauss-Jordan elimination, then multiply the matrices out to find the similarity transform of A. Then, I would do the same method previously stated to find the eigenvalues of A.

My question is: Is there ANY nice, fast way to do all of this by hand? This seems like an extremely arduous process. The first time I found the eigenvectors of A, it resulted in filling up a page with text 4 times (then erasing) and finally getting the right answer on the 5th attempt, purely due to my error rate with the arithmetic.

I surely must not be properly understanding the concept if I am doing all this work to achieve the answers.

 

[lanedance]

so say
B = P^{-1}A P

then
B P^{-1}= P^{-1}A

now multiply by an eigenvector of A, u:
B P^{-1}u = P^{-1} A u
B P^{-1}u = P^{-1} \lambda u

now consider what P^{-1}u represents

 

[TheFerruccio]

so say
(1) B = P^{-1}A P

then
(2) B P^{-1}= P^{-1}A

now multiply by an eigenvector of A, u:
(3) B P^{-1}u = P^{-1} A u
(4) B P^{-1}u = P^{-1} \lambda u

now consider what P^{-1}u represents

For the second part, you are making (3) and (4) into another eigenvalue problem?

P^{-1}u, if u is one of the three eigenvectors of A, the solution to P^{-1}u will be either

\left[\begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 0 \\<br /> \end{array}\right],\left[\begin{array}{c}<br /> 0 \\<br /> 1 \\<br /> 0 \\<br /> \end{array}\right],\left[\begin{array}{c}<br /> 0 \\<br /> 0 \\<br /> 1 \\<br /> \end{array}\right]

I don't know what that's really telling me, because that's just a shortened form of the original \mathbf{P}^{-1}\mathbf{A}\mathbf{P} which describes the behavior for three of the eigenvectors. Multiply by the corresponding lambda, and it would result in a diagonal matrix of the eigenvalues.

 

[lanedance]

Consider what (P^-1u) means to B

 

[TheFerruccio]

Consider what (P^-1u) means to B

I'm sorry. I don't see what it means to B. I'm not seeing the connection.

 

[HallsofIvy]

He is suggesting that if v is an eigenvalue of A, such that Av= \lambda v, and you define u= Pv Then
v= P^{-1}u[/itex] so Av= AP^{-1}u= \lambda v= \lambda P^{-1}u= P^{-1}(\lambda u)<br /> <br /> Now Take P of both sides.