# Eigenvalues of outer product matrix of two N-dimensional vectors

I have a vector $\textbf{a}=(a_1, a_2, ....)$, and the outer product $M_{ij}=a_i a_j$. What are the eigenvalues of this matrix? and what can you say about the co-ordinate system in which $M$ is diagonal? I have proved that the only eigenvalue of the matrix is the norm of the vector squared, and that one of the eigenvectors is $\textbf{a}$ itself.

$$Mu=a a^{T}u=\lambda u\\\ \implies a^{T}(aa^{T})u=a^{T}\lambda u \implies a^{T}a (a^{T}u)=\lambda a^{T}u \implies ||a||^2=\lambda$$

Also it is obvious that $a$ is the eigenvector of $aa^{T}$, which implies $M= \begin{pmatrix} ||a||^2 &0 &0 &..... \\0 & ||a||^2\\.\\.\\\\. \end{pmatrix}$

Is this correct? What are all the eigenvectors of this the outer product? and what intuitively happens in the transformed basis when M is diagonal?

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All vectors orthogonal to $a$ will be eigenvectors with eigenvalue zero. – Raskolnikov May 27 '13 at 8:08

Say $b \perp a$ then $a^{\text{T}}b=0$ and therefore $Mb=aa^{\text{T}}b=0$. So, just find $n-1$ independent vectors that are orthogonal to $a$ and you have $n-1$ new eigenvectors of $M$, all with eigenvalue $0$.
So the spectrum of your matrix $M$ is $(\|a\|,0,\ldots,0)$. The transformed basis is just what happens when you realign your axes to correspond with the eigenvectors. That's when you see that $M$ projects on one direction, the direction of $a$.