# How can I quickly calculate the eigenvalue and eigenvector for this matrix?

I wonder if there is any trick to calculate the eigenvalue and eigenvectors for the all-1 matrix, namely $A=% \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1% \end{bmatrix}%$

More over, suppose I have a matrix which has a form $A=U\cdot V^{T}$, that U and V are low rank, but not necessarily orthogonal basis, is there any trick that I can quickly get the eigenvalue?

Thanks a lot.

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There is no trick, but there is method... –  Lierre Nov 5 '12 at 19:29
A method is just a fancy trick :-). –  copper.hat Nov 5 '12 at 19:30
Or, a trick is just a special method ;) –  Arkamis Nov 5 '12 at 19:32
Fancy is the inverse of special... –  copper.hat Nov 5 '12 at 19:33
I'm pretty sure I've seen a very similar question in the last couple months... i.e., how to calculate eigenvalue and eigenvector of some all 1s square matrix. –  Graphth Nov 5 '12 at 19:36

You can write $A=e e^T$, where $e = (1,1,1,1,1)^T$. Hence $Ae = 5 e$, and any vector $v$ that is orthogonal to $e$ satisfies $Av = 0$. This gives you all the eigenvalues ($0,5$), and the eigenvectors (or eigenspaces, really).

An eigenvector of the eigenvalue 5 is $e$, the eigenspace for the eigenvalues 0 is $\{e\}^\bot$.

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Thanks a lot dude. –  SoManyProb_for_a_broken_heart. Nov 6 '12 at 21:04
You are very welcome. –  copper.hat Nov 7 '12 at 1:58

The rank of the matrix is 1, so 0 is definitely an eigenvalue. Use whatever method you like, it is clear that $e_1-e_2$, $e_1-e_3$, $e_1-e_4$, $e_1 - e_5$ are all eigenvectors for 0.

To find the remaining eigenvalue, note that sum of eigenvalues is the trace which is 5. So the remaining eigenvalue is 5, and by symmetry it is clear that $[1,1,1,1,1]$ is an eigenvector.

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Thanks a lot dude. –  SoManyProb_for_a_broken_heart. Nov 6 '12 at 21:00

You have four eigenvectors corresponding to the eigenvalue $0$ by choosing a $+$ and a $-$ sign. For example $(1,0,0,0,-1)$. You also have an eigenvector $(1,1,1,1,1)$ with eigenvalue $5$.

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Thanks a lot dude. –  SoManyProb_for_a_broken_heart. Nov 6 '12 at 21:02