Find eigenvalues, eigenvectors and rank of matrix $A$.

$$\textbf{a}=\begin{bmatrix}a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{bmatrix}, \quad \textbf{b} \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{bmatrix}$$

$$A = \textbf{a} \cdot \textbf{b}^T$$

I tried finding the rank through $\dim V=\operatorname{rank} A+\dim(\ker A)$ and

$v \in \ker A $


but haven't produced result.

I was also thinking about $a(b^Tx)=\lambda x$

Any ideas?

  • $\begingroup$ Obviously, if $v$ is perpendicular to $b$, it is a $0$-eigenvector.. and $a$ itself is the remaining eigenvector. $\endgroup$ – Peter Franek Aug 15 '14 at 16:48
  • $\begingroup$ @PeterFranek But vector $b$ is transposed can we claim that $b^T v=0$? What about other eigenvalues? $\endgroup$ – Žan Žurič Aug 15 '14 at 17:02

Hint: (Rank part) Consider the formula about rank of product of matrices $$\text{rank}(AB)\leq\min\{\text{rank}(A),\text{rank}(B)\}$$ (Eigenvalue part) Note that $(ab^T)x=a(b^Tx)=(b^Tx)a=\lambda x$ implies $a\parallel x$.

  • $\begingroup$ So $rankA=1$. Haven't come across that formula, thank you. Ok, $a$ is parallel to $x$, but I don't see the connection to eigenvalues $\endgroup$ – Žan Žurič Aug 15 '14 at 18:16
  • $\begingroup$ @ŽanŽurič If $a\parallel p x$, then $x$ has to be multiple of $a$, say $x=ka$. Then what are eigenvalues? $\endgroup$ – Shuchang Aug 16 '14 at 1:35
  • $\begingroup$ I get the x=ka, ofcourse, however $Ax=\lambda x$, matrix $A$ acts as if it's multiplying vector $x$ with the scalar $λ$. But if we take $(b^Tx)a=\lambda x$ there is no matrix $A$ to begin with. $\endgroup$ – Žan Žurič Aug 17 '14 at 13:57
  • $\begingroup$ So $k$ is one of the eigenvalues, but what are the others? $\endgroup$ – Matt G Jan 19 '18 at 20:45

The set of $0$-eigenvectors is the null-space of the $1\times n$ matrix $b^T$

  • $\begingroup$ Doesn't need to be. $\endgroup$ – Peter Franek Aug 15 '14 at 17:14
  • $\begingroup$ Unless $a=0$, I suppose $\endgroup$ – Empy2 Aug 15 '14 at 17:17
  • $\begingroup$ Ah, sorry, right :) $\endgroup$ – Peter Franek Aug 15 '14 at 17:18
  • $\begingroup$ Why is that the case? $\endgroup$ – Matt G Jan 19 '18 at 20:46

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