TITLE: What are the eigenvalues of matrix $uu^T$ where $u$ is a column vector?
QUESTION [5 upvotes]: 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!

REPLY [5 votes]: 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
$$
N-1=\text{geometric multiplicity of }0=\text{algebraic multiplicity of }0= N-1.
$$