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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 $\mathbb{M}$?

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(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 $\mathbb M$.

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  • $\begingroup$ Thanks Jon. How can i find the unit eigen-vectors of the matrix $\mathbb{M}$? $\endgroup$ – Ehsan zarei Apr 12 '16 at 16:16
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    $\begingroup$ $\alpha \alpha^*$ is the projection onto $\operatorname{span} \alpha$ and $\mathbb M = I - \alpha \alpha^*$ is the projection onto $(\operatorname{span} \alpha)^\perp$. You should start with $\frac{\alpha}{\|\alpha\|}$; this is a unit eigenvector of eigenvalue $1 - \sum |\alpha_i|^2$. Then, extend this to an orthonormal basis of $\mathbb C^q$ via Gram-Shmidt; the resulting $q-1$ vectors will give you unit eigenvectors of eigenvalue $1$. $\endgroup$ – Jon Warneke Apr 12 '16 at 16:55
  • $\begingroup$ Very thanks jon. My main question is here(math.stackexchange.com/questions/1736481/…). can you help me? $\endgroup$ – Ehsan zarei Apr 12 '16 at 18:33
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    $\begingroup$ Sure. When i say in my comment "you should start with $\frac{\alpha}{\| \alpha \|}$, this corresponds to the unit eigenvector $\delta_q$ given in your other question. The remaining eigenvectors must be obtained somehow by Gram-Schmidt. (Do you know what this is?) What might be a basis they started with to end up with $\delta_1, \dots, \delta_{q-1}$? $\endgroup$ – Jon Warneke Apr 12 '16 at 18:48
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    $\begingroup$ Certainly it's a unit vector. It's an eigenvector of $\mathbb M$ with that eigenvalue because of the projection stuff I mentioned. Alternatively, we can just compute: $(I - \alpha \alpha^*)\frac{\alpha}{\|\alpha\|} = \frac{\alpha}{\|\alpha\|} - \alpha \frac{\|\alpha\|^2}{\|\alpha\|} = \left( 1 - \| \alpha \|^2 \right) \frac{\alpha}{\|\alpha\|}$, which is what we expected (since $\|\alpha\|^2 = \sum_i |\alpha_i|^2$)! $\endgroup$ – Jon Warneke Apr 12 '16 at 19:26
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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

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  • $\begingroup$ Thank you Dr Truhar for your answer. $\endgroup$ – Ehsan zarei Apr 17 '16 at 16:33
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Test $\mathbb{M}$ with $\alpha$ (What is $\mathbb{M}\alpha$?) and all vectors $\beta$ that are orthogonal to $\alpha$.

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