# Homework Help: Query about eigenvectors of a matrix

1. Apr 29, 2013

### FatPhysicsBoy

1. The problem statement, all variables and given/known data

Hi so I have the eigenvalue equation $S\vec x = λ\vec x$ where $S = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ I have correctly calculated the eigenvalues to be $λ=±\frac{\hbar}{2}$ and the corresponding normalised eigenvectors to be: $\hat{e}_{1}= \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\hat{e}_{2}= \frac{1}{\sqrt2}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

2. My questions

So I have confirmed the eigenvalues and they are correct since this was a 'show that' question, however, I have a few questions about the eigenvectors.

1) The second eigenvector $\hat{e}_{2}$ was supposed to be shown to be: $\hat{e}_{2}= \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Now I understand that if I think about the line corresponding to $\hat{e}_{2}$ then both of these eigenvectors refer to the same line, so they are both correct. However, is there one which is 'more correct'? I noticed that it's totally arbitrary where the negative sign is since I can just divide the ratio equation by -1 and get the other one and vice versa.

2) I distinctly rememeber writing eigenvectors with a k multiplied by the eigenvector to show that any multiple of it is also an eigenvector. What happens to this k when you normalise? How come we don't still write the k?

3) Is the way you normalise all eigenvectors the same? No matter what they have in them or what size they are is it always the 'Pythagoras' equivalent? I.e. dividing by the norm? $||\vec e_{n}||=\sqrt{a^{2} + b^{2} + ...}$

Thank you

2. Apr 29, 2013

### DimReg

You noticed that a number times an eigenvector is still an eigenvector. So both forms of e_2 are eigenvectors with the same eigenvalue. This is because the mulitplicative factor, k, is under determined. When you add normalization, you gain a new condition, which may make it not under determined. By fixing a normalization, what do you learn about k? To make that question easier to answer, try assuming the eigenvector is normalized to one, but then multiplied by an arbitrary k.

The normalization doesn't have to be so that the inner product is one, but there is usually no reason to do otherwise.

3. Apr 29, 2013

### FatPhysicsBoy

I think it just clicked to me what you were getting at, so what you're saying is that we assign the (arbitrary) value of k such that the eigenvector is normalised and therefore k = the normalisation factor?

So usually a factor of k would scale up or down the eigenvector right? But in this case it's ensuring that the vector has magnitude = unity?

4. Apr 29, 2013

### Fightfish

The significance of choosing $\hat{e}_{2}= \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ is simply a matter of convention. The same sign ambiguities occur for Sy. The way in which the signs are chosen for the eigenvectors of Sx and Sy actually reflect the right-handed coordinate system that we typically use.

5. Apr 30, 2013

### DimReg

Yes, that's exactly what's going on. But there was one more thing I was hoping you would notice. If you do that, then k^2 = 1. But that has more than one solution for k, so even with the normalization condition, you haven't completely determined k yet. (Think about what the solutions are, then notice how they effect you original question)

6. May 3, 2013

### FatPhysicsBoy

Thank you Fightfish I think I understand this now!

I'm not really following this anymore, so wasn't k = 1/sqrt2 ? - If so, then how is k^2 = 1?

7. May 3, 2013

### DimReg

I'm sorry, I was way off with that. What I meant to show you was that there is a minus sign ambiguity when you normalize a vector, because you square the normalizing value.

What I mean to say, was imagine a situation in which you already have normalized vectors. You can restore the factor of k, and recompute the inner product. What you would have in that situation is:

Given $\hat{e} \cdot \hat{e} = 1$ we know that $k \hat{e} \cdot k \hat{e} = k^2 \hat{e} \cdot \hat{e} = k^2 (1) = k^2$

From that we infer that k = +/- 1. Note that the magnitude of k only follows from my original assumption that e was already normalized. In your specific case, k is 1/sqrt 2 because the vector you are normalizing isn't already normalized, but you can still choose it to be positive or negative freely.

8. May 3, 2013

### FatPhysicsBoy

Thank you very much, I understand this now. I think I've pretty much figured out my university/courses convention now but thank you for clearing this up!