Let $\alpha=(m_{0}(p\xi+pu(0)),m_{0}(p\xi+pu(1)),...,m_{0}(p\xi+pu(q-1)) )^T$ and

$\mathbb{M}(\xi)=I_{q}-\alpha\overline{\alpha }^{T}=(\beta_{1},\beta_{2},...,\beta_{q})$

enter image description here

How these eigenvalues of the matrix are obtained?

If we further assume that:

$$\sum_{k=0}^{q-1}\left | m_{0}(p\xi+pu(k)) \right |^2\leq 1$$

Then the unit eigen-vectors of the matrix $\mathbb{M}(\xi)$ can be represented by enter image description here enter image description here

How these unit eigen-vectors are obtained?

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    $\begingroup$ What are your Full attempts ? What I can say is the following: if we where with real components, expression $I-\alpha\alpha^T$ is a projection on the hyperplane defined normal to vector $\alpha$, thus an eigenvalue 1 (associated with eignevector $\alpha$) and (q-1) eigenvalues 0 (asociated with a basis of vectors orthogonal to $\alpha$). This can be slightly different with complex coordinates. $\endgroup$ – Jean Marie Apr 10 '16 at 19:54
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    $\begingroup$ @JeanMarie: $\alpha$ is not necessarily a unit vector ... $\endgroup$ – Catalin Zara Apr 12 '16 at 14:46
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    $\begingroup$ @JeanMarie, why is it slightly different with complex coordinates? We're just doing orthogonal projections in a Hilbert space. $\endgroup$ – Jon Warneke Apr 13 '16 at 13:41
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    $\begingroup$ @Catalin Zara and Jon Warneke : I agree $\endgroup$ – Jean Marie Apr 13 '16 at 15:27

We know $\alpha$ is an eigenvector of $\mathbb{M}$ by $$ \mathbb{M}\alpha=\alpha - \lvert\alpha\rvert^2 \alpha=(1-\lvert\alpha \rvert^2) \alpha, $$ as long as $\alpha \neq 0$. We note that $\delta_q=\frac{\alpha}{\lvert \alpha\rvert}$.

Further we know, if we have a vector $\eta$ with $\bar{\alpha}^\top\eta=0$ we get $$ \mathbb{M}\eta=\eta - \alpha\cdot 0=1\cdot\eta$$ thus, all vectors $\eta$ that are orthogonal to $\alpha$ are eigenvectors of $\mathbb{M}$ to the eigenvalue $1$.

Additionally, $\mathbb{M}$ is hermitian, i.e. $\bar{\mathbb{M}}^\top=I-\alpha\bar{\alpha}^\top=\mathbb{M}$. Hence, we know that $\mathbb{M}$ is diagonalizable with orthonormal eigenvectors.

Now we investigate, how to define an orthogonal basis for the eigenspace of the eigenvalue $1$, that is also orthogonal to the eigenvector $\alpha$ of the eigenvalue $1-\lvert\alpha \rvert^2$.

For simplicity, we say $$\alpha=\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}.$$ Then the vector $$ \beta = \begin{pmatrix} -\bar{b} \\ \bar{a} \\ 0 \\ 0\end{pmatrix} $$ is orthogonal to $\alpha$. Now we need a third vector that is orthogonal to $\alpha$ and $\beta$. For $\beta$ it is easy, we just take the first two entries of $\alpha$ again $$\hat{\gamma}=\begin{pmatrix} a \\ b \\ \gamma_3 \\ 0\end{pmatrix} $$ but of course $\hat{\gamma}$ is not longer orthogonal to $\alpha$ and we get $$ \bar{\alpha}^\top\hat{\gamma}=\lvert a\rvert^2 +\lvert b\rvert^2 +\bar{c}\gamma_3.$$ Now we have an idea for $\gamma$ $$ \gamma=\begin{pmatrix} a\bar{c} \\ b\bar{c} \\ -(\lvert a\rvert^2 +\lvert b\rvert^2) \\ 0\end{pmatrix} $$ The next (and last vector in this simplified case) has to be orthogonal to the first three, thus we use $$ \delta=\begin{pmatrix} a\bar{d} \\ b\bar{d} \\ c\bar{d}\\ -(\lvert a\rvert^2 +\lvert b\rvert^2+\lvert c\rvert^2)\end{pmatrix}.$$

Now we can divide them by their norm to get an orthonormal basis (see $\Omega_k$).

I hope this clarified one idea, to get an orthogonal basis from a vector.

On the other hand, the vectors $\delta_1$ and $\delta_q$ in your question are not orthogonal, maybe just a typo.

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    $\begingroup$ Edit: added something like 'why are vectors orthogonal to $\alpha$ eigenvectors of $\mathbb{M}$' $\endgroup$ – Hajo Apr 13 '16 at 14:24
  • $\begingroup$ what do you mean by "we know that $\mathbb{M}$ is diagonalizable with orthonormal eigenvectors." ? $\endgroup$ – Ehsan zarei Apr 14 '16 at 5:59
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    $\begingroup$ @ehsan a hermitian matrix is a normal matrix and normal matrices have this property $\endgroup$ – Hajo Apr 14 '16 at 7:26

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