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

$$\mathbf{A} = \left[ \begin{array}{ccc} -22 & 20 & 10 \\ -4 & 20 & -8 \\ 28 & -14 & 29 \\ \end{array} \right]$$

$$\mathbf{P} = \left[ \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 4 \\ 2 & 8 & 0 \\ \end{array} \right]$$

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

$$\left[\begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array}\right],\left[\begin{array}{c} 0 \\ 1 \\ 0 \\ \end{array}\right],\left[\begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array}\right]$$

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
[tex]v= P^{-1}u[/itex] so [itex]Av= AP^{-1}u= \lambda v= \lambda P^{-1}u= P^{-1}(\lambda u)[/tex]

Now Take P of both sides.