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Quick question on eigenvalues

434
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Let's say that I have to construct a 2 X 2 matrix from a second-order differential equation, turning it into a system of first order linear equations, and find its eigenvalues. I'll have two variables that correspond to the two columns in the matrix.

If I swap columns, I end up with two different characteristic polynomials, and thus different sets of eigenvalues (this is the problem). How do I know which variable should correspond to the first column, and which variable should correspond to the second column?
 

Answers and Replies

354
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If you switch the columns, you switch the order of the variables.
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
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You spent a course learning linear arithmetic, so use it. :wink:

Swapping columns is a matrix operation: if you started with the equation

[tex]A x = y[/tex]

and you want to replace A with AC, where C is the column swap elementary matrix... i.e.

[tex]
C = \left(
\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right)
[/tex]

Can you think of any way to modify the equation Ax=y (without changing its validity!) to make a C appear somewhere in it? (Preferably just to the right of the A)
 
434
0
Thanks for the replies :)

You spent a course learning linear arithmetic, so use it. :wink:

Swapping columns is a matrix operation: if you started with the equation

[tex]A x = y[/tex]

and you want to replace A with AC, where C is the column swap elementary matrix... i.e.

[tex]
C = \left(
\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right)
[/tex]

Can you think of any way to modify the equation Ax=y (without changing its validity!) to make a C appear somewhere in it? (Preferably just to the right of the A)
My math professor helped me out this morning, and I found out that if I have a second order differential equation corresponding to an equation y(t), and I let dy/dt = v, then the "rule of thumb" is to let the coefficient matrix A have its first column correspond to the variable y and let the second column correspond to the derivative of y(t), v (this way I know which coefficients in A to subtract [tex]\lambda[/tex] from). But now I want to know what you're talking about! So I'll take a shot at your question :)

I believe your question is, how can I place the column swap matrix C into the equation Ax = y, without the equation losing its validity. To do this, I believe I would multiply both sides of Ax = y by C, such that C appears between A and x, and to the left of y. So I would have ACx= Cy. I worked this out on paper and I see that this does indeed swap the columns in A and the rows in y, which wouldn't change my answer, so it works out. Please let me know if I'm mistaken.
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
I believe your question is, how can I place the column swap matrix C into the equation Ax = y, without the equation losing its validity. To do this, I believe I would multiply both sides of Ax = y by C, such that C appears between A and x, and to the left of y. So I would have ACx= Cy. I worked this out on paper and I see that this does indeed swap the columns in A and the rows in y, which wouldn't change my answer, so it works out. Please let me know if I'm mistaken.
That doesn't quite work -- remember that the operations you have available are "left multiply" and "right multiply"... so you can't just insert it anywhere you want.

The method I was hinting at is this idea: If I want to right-multiply A by C, I can undo that by right-multiplying by C-inverse. In this case, the inverse of C is itself, so I have the following derivation:

Ax = y
A (CC) x = y
(AC) (Cx) = y

The procedure you used shouldn't work in general: if Ax=y, then usually ACx=Cy will be false. In fact, that implies:

ACx = Cy
ACx = C(Ax)
ACx = CAx
(AC - CA) x = 0

so it will only work when the solution for x happens to be a nullvector of (AC - CA).
 
434
0
Ax = y
A (CC) x = y
(AC) (Cx) = y
I see, so this works out because C*C = I, where I would be the 2 X 2 identity matrix, and AI = A.
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
I see, so this works out because C*C = I, where I would be the 2 X 2 identity matrix, and AI = A.
Right. And if you did a more complicated column operation to A, you can hopefully work out what happens to x to neutralize it.
 

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