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Let $b$ be an $M \times 1$ complex vector and matrix $R = {{\alpha}^2}{b}{b^t}$.

Prove that $R$ has $(M - 1)$ zero eigenvalues and that the distinct eigenvector of $R$ with non-zero eigenvalue is $b$.

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    This assumes $\alpha \neq 0$ and $b \neq \vec{0}$. It is best for your to show your thoughts. For this problem, you just need to show there are $M-1$ linearly independent vectors $v_k$, $k \in \{1, ..., M-1\}$, that satisfy $Rv_k=\vec{0}$. – 2017-02-21
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    What is $\;\alpha\;$ ? is $\;b^H\;$ to mean $\;b^T=$ the transpose of $\;b\;$ ? – 2017-02-21
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    @DonAntonio $H$ is often used for the "Hermitian transpose" (conjugate transpose). – 2017-02-21

3 Answers 3

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Let

$$b=\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\implies b^H=( \overline x\; \overline y\; \overline z\; \overline w)$$

and thus

$$bb^H=\begin{pmatrix} |x|^2&x \overline y&x \overline z&x \overline w\\ \overline xy&|y|^2&y \overline z&y \overline w\\ \overline xz& \overline yz&|z|^2&z \overline w\\ \overline xw& \overline yw& \overline zw&|z|^2\end{pmatrix}$$

Now, call the above matrix's rows $\;c_1,...c_4\;$ resp. , and then observe that

$$c_2=\frac yxc_1\;,\;\;c_3=\frac zxc_1\;,\;\;c_4=\frac wxc_1$$

(what happens if $\;x=0\;$ ?) , so that we get that rank$\,bb^H=1\;$ , and thus there are three linearly independent vectors $\;u_i\;,\;\;i=1,2,3\;$ such that $\;bb^Hu_i=0\;$.

Finally, check directly that

$$Rb=\alpha^2 bb^Hb=\alpha^2(|x|^2+|y^2+|z|^2+|w|^2)b$$

and assuming $\;\alpha\neq0\;$ there we have our eigenvector for our non-zero eigenvalue.

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Hint: Calculate $R^2$ and see how it relates to $R$. What does this tell you about the minimal polynomial of $R$. Then show that $R$ has rank $1$ by inspecting the rowspace of $R$. What do you know about projections of rank $1$?

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Hint: Any (non-zero) vector orthogonal to $b$ is an eigenvector of $R$. $b$ is itself also an eigenvector of $R$. With this, we have accounted for $M$ eigenvalues.