Find the eigenvalues and eigenvectors of the matrix $A = uu^t$, where $u\in\mathbb{R}^n$ 
Find the eigenvalues and eigenvectors of the matrix $A = uu^t$, where
  $u\in\mathbb{R}^n$

The multiplication will give me a $n\times n$ matrix like this:
\begin{bmatrix}
    u_1^2 & u_1u_2 & \cdots & u_1u_n \\
    u_2u_1 & u_2^2 & \cdots & u_2u_n \\
    \cdot & \\
    u_nu_1 & \cdots & \cdots& u_n^2 
  \end{bmatrix}
I suppose there is some trick using the fact that this matrix is symmetric and square. This should help taking the determinant of
\begin{bmatrix}
    u_1^2 -\lambda & u_1u_2 & \cdots & u_1u_n \\
    u_2u_1 & u_2^2 -\lambda & \cdots & u_2u_n \\
    \cdot & \\
    u_nu_1 & \cdots & \cdots& u_n^2 -\lambda
  \end{bmatrix}
 A: $$A = uu^T$$
Notice that 
$$Au = uu^Tu = u(u^Tu) = (u^Tu) u = \lambda u $$
where 
$$\lambda = u^Tu$$
is an eigenvalue corresponding to eigenvector $u$. 
All other eigenvalues are zero because $A$ is an outer-product of $u$ on itself.
Why zero-eigenvalues?
Consider vectors of the form $$v = (I - \frac{1}{u^Tu}uu^T)\alpha$$
then 
$$Av =uu^Tv = uu^T\alpha - \frac{1}{u^Tu}uu^Tuu^T\alpha =uu^T\alpha - \frac{1}{u^Tu}(u^Tu)uu^T\alpha = uu^T\alpha-uu^T\alpha = 0$$
Turns out there are $n-1$ linearly and independent vectors $v_1 \ldots v_{n-1}$ that satisfy the above equation. So $0$ has $n-1$ multiplicity.
A: To elaborate on what Ahmad didn't mention is that if you take the outer product of a vector with itself you'll end up with a rank one matrix because all the columns are linearly dependent on each other. If the rank of the matrix is only 1 you see the rest of the eigenvalues are zero. 
This is detailed here. 
A: 

Added: Observe that $x^tAx=x^t(uu^t)x=(u^tx)^t(u^tx)=\vert\vert u^tx\vert\vert^2\ge 0$ for any $x\in\mathbb R^n$ .
Thus matrix $A$ is semi-positive definite. Also $Rank(A)=1$ and $A$ is symmetric then $1=p+n$ (where $p$ and $n$ respectively denote the number of positive and negative eigenvalues) gives $p=1$.
Using $trace(A)=$sum of eigenvalues you get $\lambda=u_1^2+u_2^2+....+u_n^2$ as the only non-zero eigenvalue and rest $(n-1)$ eigenvalues are zero.


Another way is to consider $u$ as a $1×n$ row vector which gives you $A=[u_1^2+u_2^2+....+u_n^2]$. Then $Ax=\lambda x$ is trivially satisfied for all $x\in \mathbb R$ thereby giving $\lambda=u_1^2+u_2^2+....+u_n^2$ as eigenvalue and $x$ as eigenvector.
