find the eigenvalues and eigenvectors of the matrix $\mathbb{M}$ Let $\alpha =\begin{bmatrix}
\alpha _{0}\\ 
\alpha _{1}\\
\vdots\\ 
\alpha _{q-1}\\ 
\end{bmatrix}$ and 
$\mathbb{M}:=I_{q}-\alpha\overline{\alpha }^{T}=$\begin{bmatrix}
 1-\left | \alpha_{0} \right |^{2}&-\alpha_{0}\overline{\alpha_{1}}&\cdots &-\alpha_{0}\overline{\alpha_{q-1}}\\
-\alpha_{1}\overline{\alpha_{0}}&1-\left | \alpha_{1} \right |^2& \cdots& -\alpha_{1}\overline{\alpha_{q-1}}\\
\vdots &\vdots&\ddots &\vdots\\
-\alpha_{q-1}\overline{\alpha_{0}}& -\alpha_{q-1}\overline{\alpha_{1}}&\cdots & 1-\left | \alpha_{q-1} \right |^2
\end{bmatrix}
How can I find the eigenvalues and eigenvectors of the matrix $\mathbb{M}$?
 A: (I'll use the common notation $\alpha^* = \overline\alpha^T$.) Suppose that $v$ is an eigenvector of $\mathbb M$ with eigenvalue $\lambda$. Then
\begin{align*}
\mathbb M v &= \lambda v \\
(I - \alpha \alpha^*) v &= \lambda v \\
(1 - \lambda) v &= \alpha \alpha^* v \\
\mu v &= \alpha \alpha^* v, && \mu := 1 - \lambda
\end{align*}
that is, $(1-\lambda) = \mu$ is an eigenvalue of $\alpha \alpha^*$ with eigenvector $v$. So we've reduced to the problem of finding eigenvalues $\mu$ of $\alpha \alpha^*$.
To solve the (eigen)vector equation $\alpha \alpha^* v = \mu v$, note that $\alpha^* v$ is just a scalar, and hence equating scalars and vectors, we have $\alpha = v$ and $\alpha^* v = \mu$. Substituting $\alpha = v$ into the scalar equation gives $\mu = \alpha^* \alpha = \sum_{i=0}^{q-1} |\alpha_i|^2$. It could also be the case that $\alpha^* v = 0$, in which case $\mu = 0$. These are the only possible ways to satisfy our eigenvector equation, and hence the only eigenvalues. Another way to see this: note that $\alpha \alpha^*$ is an unnormalized projection operator onto the line $\operatorname{span} \alpha$, and hence it can only have two eigenvalues: the factor by which it scales vectors already in $\operatorname{span} \alpha$, and $0$ for the vectors in $(\operatorname{span} \alpha)^\perp$ which it kills. The eigenspaces have dimensions $1$ and $q-1$, repectively, so we're not missing anything.
Tracing back, we see that 
$$
\lambda = 1 - \sum_{i=0}^{q-1} \left| \alpha_i \right|^2, \qquad \lambda = 1 - 0
$$
are the eigenvalues of $\mathbb M$.
A: The simplest way, to find the eigenvalues of 
I-a.a^T
would be to calculate QR of a, that is
[Q,R]=qr(a);
then R is a vector (1D) with R(1)\neq 0. The eigenvalues of 
I-a.a^T are
1-R(1)^2, and 1 (n-1 times), while the corresponding eigenvectors are
the columns of the matrix Q.
Regards
N. Truhar 
A: Test $\mathbb{M}$ with $\alpha$ (What is $\mathbb{M}\alpha$?) and all vectors $\beta$ that are orthogonal to $\alpha$. 
