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I need to find the eigenvalues and vector for the operator in hilbert space $H$:

$$Tx=\langle x,v\rangle w+\langle x,w\rangle v,$$

where $\langle v,w \rangle =0$, but not necessarily with norm=1.

I already checked that it is compact and self adjoint so there must be eigenvalues

I tried the form of $\alpha w+\beta v$ which works only if $\alpha =\beta$ and if both $w$ and $v$ are normalize, and they are not

Theo Bendit
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Benuci
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  • Hint: the eigenvectors must either be in the kernel or the range of $T$; the former is the eigenspace corresponding to $0$. To find the other eigenvectors, try considering vectors of the form $\alpha w + \beta v$. – Theo Bendit Jul 09 '21 at 18:48
  • this is the first thing I tried, if alpha=betta it would have work only if both v and w are normalize, which they aren't – Benuci Jul 09 '21 at 18:53
  • Ah, then this is exactly the kind of context you should be writing into your questions! If you tried something sensible, and it went nowhere, it's good to mention it in the question. More efforts shown will make your questions better received, and also stop people like me delivering useless comments. :-) – Theo Bendit Jul 09 '21 at 18:55
  • edited, do you have any lead? or at least a way to describe this operator so I can find this problem in the site – Benuci Jul 09 '21 at 19:01
  • You already have my +1. Now you have my answer too! – Theo Bendit Jul 09 '21 at 19:17

1 Answers1

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I will assume $v, w \neq 0$. The $v = 0$ and/or $w = 0$ cases are easier, and can be decided separately. Note then that $\langle v, w \rangle = 0$ implies $v$ and $w$ are linearly independent.

Using my hint in the comments, I do believe we can indeed find the other eigenvectors (other than the ones in the kernel, i.e. the vectors perpendicular to both $v$ and $w$). Let $u = \alpha v + \beta w$, and assume it is an eigenvector with respect to $\lambda$. As $v, w$ are perpendicular, we have $$\langle u, v\rangle = \langle \alpha v + \beta w, v\rangle = \alpha \|v\|^2,$$ and similarly $\langle u, w\rangle = \beta\|w\|^2$. We then have, by the linear independence of $v$ and $w$, \begin{align*} T(u) - \lambda u = 0 &\iff \langle u, v\rangle w + \langle u, w \rangle v - \lambda(\alpha v + \beta w) = 0 \\ &\iff (\alpha\|v\|^2 - \lambda \beta)w + (\beta\|w\|^2 - \lambda\alpha)v = 0 \\ &\iff \begin{cases} \alpha\|v\|^2 - \lambda \beta = 0 \\\beta\|w\|^2 - \lambda \alpha = 0 \end{cases} \\ &\iff \begin{cases} \alpha\|v\|^2 \|w\|^2 - \lambda \beta \|w\|^2 = 0 \\\beta\|w\|^2 = \lambda \alpha \end{cases} \\ &\iff \begin{cases} \alpha\|v\|^2 \|w\|^2 - \lambda^2 \alpha = 0 \\\beta\|w\|^2 = \lambda \alpha \end{cases}. \end{align*} From here, we can deduce that $\alpha = 0$ or $\lambda = \pm \|v\|\|w\|$. If $\alpha = 0$ were true, then $\beta\|w\|^2 - \lambda \alpha$ would imply $\beta = 0$ too, in which case we don't have an eigenvector. Thus, the only possible eigenvalues are $\lambda = \pm \|v\|\|w\|$.

We now need to confirm that these are not just possible eigenvalues, but actual eigenvalues, by finding actual eigenvectors. If $\lambda = \|v\|\|w\|$, then $\beta\|w\|^2 - \lambda \alpha = 0$ implies $$\beta\|w\|^2 = \|v\|\|w\|\alpha \implies \alpha = \beta\frac{\|w\|}{\|v\|}.$$ We are not going to get better than this, as $\alpha$ and $\beta$ can be scaled by any non-zero scalar, and we will still get an eigenvector. So, I will choose $\alpha = \|w\|$ and $\beta = \|v\|$. This gives us the eigenvector $u = \|w\|v + \|v\|w$. We can check this: \begin{align*} T(\|w\|v + \|v\|w) &= \langle \|w\|v + \|v\|w, v\rangle w + \langle \|w\|v + \|v\|w, w \rangle v \\ &= \|w\|\langle v, v\rangle w + \|v\|\langle w, w\rangle v \\ &= \|v\|\|w\|(\|v\|w + \|w\|v), \end{align*} as required. Thus, we have an eigenvector, and so $\|v\|\|w\|$ is definitely an eigenvalue of $T$.

I'll leave the $-\|v\|\|w\|$ case to you.

Theo Bendit
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