Eigenvalues / eigenvectors for squared matrix

A_Fool's_Paradise

Hi, I'm new here, hello. :smile:

Sorry to resurrect an old thread, but would it be correct to say that

B^2{\bf v} = k^2{\bf v}

? I can't see how to prove this, my matrix algebra is really lacking. Could somebody show me?

Thanks.

If you can find the eigenvalue of a matrix or operator for a given eigenvector, you can just replace wherever the operator/matrix acting on the eigenvector with the eigenvalue times the eigenvector, so if:

\hat{H} \psi = E \psi \to \hat{H^2} \psi = \hat{H} E \psi = E^2 \psi

The hat on the H denotes that it is an operator/matrix, and

\hat{H}

and E commute (there order of operation doesn't matter) because E is just a scalar.

Does this help any?

Reply 23

0 div curl F

If you can find the eigenvalue of a matrix or operator for a given eigenvector, you can just replace wherever the operator/matrix acting on the eigenvector with the eigenvalue times the eigenvector, so if:

\hat{H} \psi = E \psi \to \hat{H^2} \psi = \hat{H} E \psi = E^2 \psi

The hat on the H denotes that it is an operator/matrix, and

\hat{H}

and E commute (there order of operation doesn't matter) because E is just a scalar.

Does this help any?

Thanks everyone for your replies, helped a lot, however...

I'm having trouble seeing how you got from

\hat{H} E \psi

to

E^2 \psi

.

Reply 24

A_Fool's_Paradise

Thanks everyone for your replies, helped a lot, however...

I'm having trouble seeing how you got from

\hat{H} E \psi

to

E^2 \psi

.

Well you can switch the order of the

\hat{H}

and the E since they commute. so:

\hat{H} E \psi = E \hat{H} \psi

Then we just use that

\hat{H} \psi = E \psi

from before, so

E \hat{H} \psi = E E \psi = E^2 \psi

Is that any better?

Reply 25

Ergh, I feel like an idiot. Thanks for your help.

Reply 26

No problem, glad I could help. This type of mathematics is a very powerful tool, I'd encourage anybody to read up on this linear algebra because you'd be surprised at the types of problems you can solve with this type of formalism

Reply 27

It's definitely useful, but a total bastard to get my head around! Another question from this damn book that has me stumped:

So I have these iterative formulae:

x_{n+1} = x_n + 3y_n

y_{n+1} = 2x_n + \frac{1}{2}y_n

Which Ive expressed in the matrix form Ax = b, I've then taken eigenvalues and eigenvectors of the matrix A... these are...

Eigenvector of eigenvalue -1 = [-3; 2]
Eigenvector of eigenvalue 4: = [1; 1]

The question then asks "Using the initial values

x_0 = -3

and

y_0 = 2

, and that

A^n x = \lambda^n x

, find an expression for

y_n

in terms of n."

I am utterly utterly lost.

I'd like to work through this myself, but could anybody give me a nudge in the right direction??!

Reply 28

DFranklin

18

A_Fool's_Paradise

It's definitely useful, but a total bastard to get my head around! Another question from this damn book that has me stumped:

So I have these iterative formulae:

x_{n+1} = x_n + 3y_n

y_{n+1} = 2x_n + \frac{1}{2}y_n

Which Ive expressed in the matrix form Ax = b, I've then taken eigenvalues and eigenvectors of the matrix A... these are...

Eigenvector of eigenvalue -1 = [-3; 2]
Eigenvector of eigenvalue 4: = [1; 1]

The question then asks "Using the initial values

x_0 = -3

and

y_0 = 2

, and that

A^n x = \lambda^n x

, find an expression for

y_n

in terms of n."

I am utterly utterly lost.

I'd like to work through this myself, but could anybody give me a nudge in the right direction??!

I don't get the eigenvalues you do (and your eigenvectors don't seem to be valid eigenvectors for the matrix I have in mind) - on the other hand, my eigenvalues are "not nice" (involving sqrt(97)), so I suspect either I'm wrong or I'm misunderstanding which matrix you've formed, or you've written the question incorrectly.

I suggest you post more detailed working.

Reply 29

So did you end up with a matrix equation which looked like this:

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n+1}}\\\noalign{\medskip}y_{{n+1}} \end {array} \right] = \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&1/2\end {array} \right] \cdot \left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right]

If so I don't get the same eigenvalues as you did?

Reply 30

DFranklin

18

0 div curl F

So did you end up with a matrix equation which looked like this:

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n+1}}\\\noalign{\medskip}y_{{n+1}} \end {array} \right] = \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&1/2\end {array} \right] \cdot \left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right]

If so I don't get the same eigenvalues as you did?

That's what I did, and I ended up with 2t^2-3t-11 for the characteristic poly. If the matrix is right, I'm sure I made a mistake, but I'm not seeing it.

Edit: just did a numerical calc of the largest eigenvalue, and it tallies with my answer of

(3+\sqrt{97})/4

. My guess is there's a typo somewhere.

Reply 31

Apologies, my mistake. That 1/2 should just be a 2.

x_{n+1} = x_n + 3y_n

y_{n+1} = 2x_n + 2y_n

Giving...

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n+1}}\\\noalign{\medskip}y_{{n+1}} \end {array} \right] = \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&2\end {array} \right] \left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right]

Reply 32

So for that matrix I do indeed agree with your eigenvalues.

As for the question, the initial conditions are an eigenvector of the matrix, and a eigenvector is a special direction that is mapped on to itself with its matrix, i.e. its direction does not change, just its length changes. so

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right] = { \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&2\end {array} \right] }^{n} \cdot \left[ \begin {array}{c} x_{{0}}\\\noalign{\medskip}y_{{0}}\end {array} \right]

from before, and since this vector is an eigenvector with eigenvalue -1, we can replace the matrix with its eigenvalue, just as I did in the example I wrote up last night, so we get (-1)^n

so I make this to be true:

y_n = (-1)^n y_0

Does this help any? I think its correct

Reply 33

0 div curl F

So for that matrix I do indeed agree with your eigenvalues.

As for the question, the initial conditions are an eigenvector of the matrix, and a eigenvector is a special direction that is mapped on to itself with its matrix, i.e. its direction does not change, just its length changes. so

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right] = { \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&2\end {array} \right] }^{n} \cdot \left[ \begin {array}{c} x_{{0}}\\\noalign{\medskip}y_{{0}}\end {array} \right]

from before, and since this vector is an eigenvector with eigenvalue -1, we can replace the matrix with its eigenvalue, just as I did in the example I wrote up last night, so we get (-1)^n

so I make this to be true:

y_n = (-1)^n y_0

Does this help any? I think its correct

Thanks, that is the right answer. I understand that the initial conditions are an eigenvector, but I'm having problems following why

Unparseable latex formula:

\left[ \begin {array}{c} x_{{n}}\\\noalign{\medskip}y_{{n}} \end {array} \right] = { \left[ \begin {array}{cc} 1&3\\\noalign{\medskip}2&2\end {array} \right] }^{n} \cdot \left[ \begin {array}{c} x_{{0}}\\\noalign{\medskip}y_{{0}}\end {array} \right]

is true. Could you explain?

Thanks.

Reply 34

DFranklin

18

That would be true whatever the matrix and vector. You've set things up so that

v_{n+1} = Av_n

(A is the matrix, V is the vector), and it's immediate that

v_n = A^n v_0

(if you really want to, you can prove it by induction on n).

Reply 35