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}
Best Answer
$$A = uu^T$$ 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.