# How to find eigenvectors/eigenvalues of a matrix where each diagonal entry is scalar $d$ and all other entries are $1$ [duplicate]

How would you find eigenvalues/eigenvectors of a $n\times n$ matrix where each diagonal entry is scalar $d$ and all other entries are $1$ ? I am looking for a decomposition but cannot find anything for this.
For example:

$\begin{pmatrix}2&1&1&1\\1&2&1&1\\1&1&2&1\\1&1&1&2\end{pmatrix}$

## marked as duplicate by Marc van Leeuwen linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 24 '15 at 12:19

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• You should have clarified your original question; finding exact eigenvectors / eigenvalues of structured matrices as opposed to finding approximate eigenvectors / eigenvalues of arbitrary matrices are very different problems. – Qiaochu Yuan Aug 9 '10 at 7:47
• they are different questions. one requires a general process of finding eigenvectors while this is specific to this matrix. – user957 Aug 9 '10 at 8:02

## 1 Answer

The matrix is $(d-1)I + J$ where $I$ is the identity matrix and $J$ is the all-ones matrix, so once you have the eigenvectors and eigenvalues of $J$ the eigenvectors of $(d-1)I + J$ are the same and the eigenvalues are each $d-1$ greater. (Convince yourself that this works.)

But $J$ has rank $1$, so it has eigenvalue $0$ with multiplicity $n-1$. The last eigenvalue is $n$, and it's quite easy to write down all the eigenvectors.

• It really should be hammered often enough: exploit any structure you find in your problem to the hilt. – J. M. is a poor mathematician Aug 9 '10 at 9:28