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Please let me know if this problem is duplicated and I will remove it ASAP.

I see this problem on an interview book.

The vector is defined as $u=(u_1,u_2,...,u_n)^T$.

Then the eigenvalues of $uu^T$ are given as $\Sigma_{i=1}^n u_i^2$ with multiplicity 1 and 0 with multiplicity $n-1$.

I try to start with $det(uu^T-\lambda I)$ and try to show this is exactly

($\lambda-\Sigma_{i=1}^n u_i^2)\lambda^{n-1}=0$

Any help will be appreciated!

2 Answers 2

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Note that $$ (uu^T)u=u(u^Tu)=||u||^2u $$ so $||u||^2$ is an eigenvalue with $u$ itself a corresponding eigenvector. Now, consider the $N-1$ dimensional subspace $W$ of $\mathbb{R}^N$ that contains vectors orthogonal to $u$. Then, $$ w\in W\implies u^Tw=0\implies(uu^T)w=u(u^Tw)=0u $$ so $0$ is an eigenvalue with multiplicity $N-1$.

Edit: the above shows that $0$ has geometric multiplicity $N-1$. But $$ N-1=\text{geometric multiplicity of }0\leq\text{algebraic multiplicity of }0\leq N-1 $$ (the last inequality is because there exists another eigenvalue i.e. $||u||^2$) so you actually have $$ N-1=\text{geometric multiplicity of }0=\text{algebraic multiplicity of }0= N-1. $$

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    Thanks for your great answer! Can I understand in this way: after we have the first eigenvalue $||u||^2$; since matrix $A=uu^T$ apparently has rank 1 and hence all other N-1 eigenvalues are just zeros. – 2017-01-16
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First, prove that $u$ an eigenvector corresponding to eigenvalue $\sum_i u_i^2$.

Then extend $u$ into an orthogonal basis $\{u, v_2,\ldots,v_n\}$ and show that the other basis elements are zero-eigenvectors.