# What makes this an Eigen value problem and why?

1. May 24, 2012

### svishal03

I'm undergoing a self study course on Structural Dynamics and have arrived at the following equation:

KR = w^2 M R

where; M and K are positive definite matrices.

1) Why is this classified as a Eigen value problem? I will be grateful if you give a mathematical explanation here.

2) Why is this classified as a Eigen value problem? I will be grateful if you give a physical interpretation/ explanation here.

2. May 25, 2012

Mathematically, assuming you're looking for vectors R that solve this equation, as every positive definite matrix is invertible, any vector that solves that equation will also solve $M^{-1} K R = w^2 R$. Hence R must an eigenvector of $M^{-1}K$ with eigenvalue w. So then solving it is merely a task of finding the eigenvectors and respective eigenvalues. On the other hand, assuming you only have M and K, you can find multiple R and w that will satisfy this equation, assuming the matrix has a complete set of eigenvectors (as I think it should).

I'm not sure this answers your question, but it should provide some definite insight, I suppose.

3. May 25, 2012

### AlephZero

Padfoot said everything except one important word: You are looking for non-zero vectors R that solve the equation.

Non-zero solutions only exist for special values of w, which are the eigenvalues. "Eigen" in German means "separate", "strange", "peculiar", "particular", etc.

Physically, the values of w are the natural frequencies of vibration of a mechanical system, and the corresponding values of R are the vibrating shapes. A system like a stretched string, a cantilever beam, etc, with no applied forces can only vibrate at certain "special" frequencies, sometimes called "harmonics".

4. May 26, 2012

### svishal03

Thank you padfoot yes it was useful.

Alephzero,

Thank you as well for the physcial interpretation. But, I have a question:

Why do you say:

Because:

See attached extract from the book on Dynamics of Structures by Prof.Anil K Chopra from University of Berkeley. The 2 storey frame can vibrate at any frequency which is not necessarily a simple harmonic motion but only for certain distribution of displacement can result in a simple harmonic motion. That is: wouldn't it be better to put as:

A system like a stretched string, a cantilever beam, etc, with no applied forces can only vibrate in simple harmonic motion at certain "special" frequencies, sometimes called "harmonics".

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5. May 26, 2012

### svishal03

If R_s and R_r are the eigen vectors of the matrix we were just discussing i.e. the matrix M^-1K

and if we prove that;

R_s^T M R_r =0

R_s^T K R_r =0

Then,

Is,

R_s^T M R_s =1

and R_S^T K R_s = w_s^2

Why?

The author says that this is the normality relation, I do not follow how?

6. May 27, 2012

### svishal03

I shall be grateful if anyone can help with my queries in response to your answers...

Much appreciated,

Vishal

7. May 27, 2012

I'm not sure if this is the answer you're looking for, but it might provide some insight into what's going on. So, with this normalization condition, we require that the eigenvectors $R_s,R_t$ are normalized. Hence we want the inner product of the vectors with themselves to be 1, or$<R_s|R_s> = 1$ and $<R_t|R_t> = 1$.

Note then, since we have $M^{-1}K R_t = w^2 R_t$, it then follows (using this normality condition) that (using expectation value notation, with respect to R_t, but the same would be true with R_s), $<M^{-1} K > = w^2$. However, if one were to move $M$ over to the other side, one would find that it must also be true that $\frac{<K>}{<M>} = <M^{-1} K > = w^2$. Hence, if we choose $<K> = w^2$ and $<M> = 1$, then we satisfy the normality condition (inner product of the vector with itself is 1).

Note that there is a little bit of ambiguity in this situation, as one could choose the vectors such that the individual expectation values come out differently (e.g. 2 $w^2$ and 2, respectively) and still get the same normality condition. However, it's likely that $w^2$ and 1 are chosen as the convention to keep things simple as possible.

Also, I don't see where this relies on the fact that if you have two eigenvectors with eigenvalues w^2, they have to be orthogonal, however that's probably chosen to keep the math simple, as after doing this you would then have a orthonormal space with eigenvalues of w^2.

I don't know that this helped answer all of your questions, but perchance it helped.

8. May 29, 2012

### svishal03

Thank you very much for the reply.

I shall be grateful if you help me here

You said:

1) $M^{-1}K R_t = w^2 R_t$

How did you arrive at this?

2)You then said:

$M^{-1} K = w^2$

Is this because:

From 1;

$M^{-1}K =w^2 R_t R_t{-1}$

Hence:

$M^{-1} K = w^2$

Vishal

9. May 29, 2012

We get (1) from assuming that R_t is an eigenvector of of the matrix (M^-1 * K) with eigenvalue w^2. 1 is the definition of this. The matrix acting on R_t returns R_t multiplied by w^2.

As you get further into Linear Algebra, and understanding notation, (2) will eventually make your stomach hurt. The equation you have for (2) says a matrix is equal to a scalar. This is a big no-no. So we have to look back a little more closely at my notation.

The < > brackets mean an "expectation value" (which may or may not be purely a physics notation). Expectation values are taken with respect to certain vectors (one of the R's that has eigenvalue w^2, in this case), and are applied on matrices. So It means, in similar notation to what you used, $R^T( M^{-1} K) R$ where $R^t$ is the transpose of the R vector. Hence, following the matrix multiplication through, we see that this expression yields a scalar, as we'd demand for obvious reasons. After understanding this notation the rest is simply algebra. If the notation is still confusing for you, I'd recommend googling expectation values and look at the linear algebra involved (most of the links you pull up will be quantum mechanics, but just ignore the quantum speak and look at the math of what they're doing).

Let me know if this helps.

10. May 30, 2012

### svishal03

Thank you again!

Did you mean:

$M^{-1} K R_t= w_t^2 R_t$ i.e. did you miss the _t for w?

Yes, if it is so I get the above.

Following this, I shall be grateful if you can expand the math steps to get :

$R_s^T M R_s =1$ and

$R_S^T K R_s = w_s^2$

Yes-i do not intend going into details of expectation values at the moment- the math steps to arrive at $R_s^T M R_s =1$ and $R_S^T K R_s = w_s^2$ will be sufficient.

11. May 30, 2012

Yes, I just dropped the _t. Since I basically explained the answer to this in my second reponse, I think that rather than give you the solution, it will be universally more beneficial to you for you to work it out yourself.

However, here's an outline of the solution. Start with $K R = w^2 M R$ and hit both sides of the equation with the transpose of R from the left side. You'll get things that you're looking for on both sides. You can then solve this for a ratio and it's here that you'll see that choosing the vector such that the conditions your author suggests are satisfied is the easiest.

For a further treat, if you suppose that the norm of the vector is 1, starting with $M^-1 K R = w^2 R$, you'll see why this is the "normality condition". Using this and my second post, you shouldn't have any trouble seeing why you should choose the vectors to satisfy the conditions your author suggests.

12. May 30, 2012

### svishal03

Thanks a lot for the reply.

Did you mean this:

1) $K R = w^2 M R$

2) $R^T K R = R^ T w^2 M R$

3) $R^T K R / R^ T M R = w^2$

4) If we put $R^ T M R = 1$

then;

5) $R^T K R = w^2$

Is this what you would derive?

13. May 30, 2012

Yes. As you can see, there's some built in ambiguity in satisfying equation 3, but by choosing equation 4, you've chosen the simplest way of satisfying 3. So it works out really nicely. This ratio (equation 3) being equal to w^2 is what ensures that you have normality of the vector, as well. Hence why it's the normality condition. Hope this all helped.