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Find eigenvalues and eigenvectors of a matrix $A_{n\times n}$ where elements $a_{ij} $ of $A_{n\times n}$ are given as

\begin{cases} \alpha, & \text{if }i=j \\[2ex] 1, & \text{if }|i-j|=1\\[2ex] 0 & \text{otherwise} \end{cases}

where $\alpha$ is a constant.

I tried finding out the eigenvalues by finding the polynomial equation of this equation and the result which I was getting was of the form:-

$|A_{n\times n}-\lambda I_{n\times n}|=(\alpha-\lambda)(|A_{(n-1)\times (n-1)}-\lambda I_{(n-1)\times (n-1)}|-|A_{(n-2)\times (n-2)}-\lambda I_{(n-2)\times (n-2)}|)$

But I was not able to go further.

Rathi_Ji
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    As has been posted on your previous questions, please keep in mind that askers are expected to provide some context to their questions. As the messages on your closed questions say: please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. – Ben Grossmann May 25 '20 at 05:50
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    That said: in all seriousness, a reasonable amount context is important and is generally appreciated. For this problem, even saying something like "this is a problem I found while self-studying from textbook X" or "I computed the eigenvalues for the $2 \times 2$ case with $\alpha = 0$ but couldn't get much further" can be very helpful since it gives us an indication of the type and degree of guidance you might need. – Ben Grossmann May 25 '20 at 06:28
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    I will surely take care of the comments from the next time. Will also edit this question a bit to set the progress that I made in this question. – Rathi_Ji May 25 '20 at 06:31

1 Answers1

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A formula for the eigenvalues and eigenvectors of such a matrix is given here.

We can deduce these eigenvalues and eigenvectors nicely, however, if we correctly "guess" that we can find a complete set of eigenvectors in which each eigenvector is of the form $$ v = (\sin(\theta), \sin(2 \theta), \dots , \sin (n \theta)). $$ See this post for details on this approach.

Note that it suffices to consider the case of $\alpha = 0$, since for any matrix $M$, the matrices $M$ and $M + \alpha I$ have the same eigenvalues (where $I$ denotes the identity matrix).

Ben Grossmann
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