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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:


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marked as duplicate by Marc van Leeuwen linear-algebra 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
up vote 6 down vote accepted

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.

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

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