Is there a faster way to do this matrix problem?

  • #1
220
0
Is there a faster way to do this matrix problem??

Homework Statement



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

Homework Equations



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

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

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 [tex]\lambda = 36, 18, -27[/tex]. 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.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2


so say
[tex] B = P^{-1}A P[/tex]

then
[tex] B P^{-1}= P^{-1}A[/tex]

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

now consider what [itex] P^{-1}u [/itex] represents
 
  • #3
220
0


so say
(1) [tex] B = P^{-1}A P[/tex]

then
(2) [tex] B P^{-1}= P^{-1}A[/tex]

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

now consider what [itex] P^{-1}u [/itex] represents

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

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

[tex]\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][/tex]

I don't know what that's really telling me, because that's just a shortened form of the original [tex]\mathbf{P}^{-1}\mathbf{A}\mathbf{P}[/tex] 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.
 
Last edited:
  • #4
lanedance
Homework Helper
3,304
2
Consider what (P^-1u) means to B
 
  • #5
220
0


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.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
964


He is suggesting that if v is an eigenvalue of A, such that [itex]Av= \lambda v[/itex], and you define [itex]u= Pv[/itex] 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.
 

Related Threads on Is there a faster way to do this matrix problem?

Replies
4
Views
1K
Replies
17
Views
2K
Replies
0
Views
808
Replies
2
Views
2K
Replies
25
Views
3K
Replies
6
Views
1K
Replies
11
Views
1K
Replies
6
Views
1K
Replies
2
Views
584
M
Top