I have seen a lot of papers mentioning that a certain matrix is rank-$1$. What properties does a rank-$1$ matrix have?

I know that if a matrix is rank-$1$ then there are no independent columns or rows in that matrix.


Any rank one matrix $A$ can always be written $A=xy^\intercal$ for vectors $x$ and $y$. More precisely...

Proposition: A matrix in $\mathbb{C}^{n\times n}$ has rank one if and only if it can be written as the outer product of two nonzero vectors in $\mathbb{C}^{n}$ (i.e., $A=xy^{\intercal}$).

Proof. This follows from the observation $$ \begin{pmatrix}x_{1}y^{\intercal}\\ x_{2}y^{\intercal}\\ \vdots\\ x_{n}y^{\intercal} \end{pmatrix}=xy^{\intercal}=\begin{pmatrix}y_{1}x & y_{2}x & \cdots & y_{n}x\end{pmatrix}. $$

Corollary: The eigenspace of a rank one matrix in $\mathbb{C}^{n\times n}$ is one dimensional.

Proof. If $A$ is a rank one matrix in $\mathbb{C}^{n\times n}$, the previous result tells us that it can be written in the form $A=xy^{\intercal}$. Next, note that for any vector $w$ in $\mathbb{C}^{n}$, $$ Aw=(xy^{\intercal})w=(y^{\intercal}w)x. $$ Since $(y^{\intercal}w)$ is a scalar, it follows that if $w$ is an eigenvector of $A$, it must be a multiple of $x$. In other words, the eigenspace of $A$ is $$ \operatorname{span}(x)=\left\{ \alpha x\colon\alpha\in\mathbb{C}\right\} . $$

  • $\begingroup$ Is there also a geometric interpretation of the linear transformation associated with $A = xy^T$? $\endgroup$ – philmcole Mar 28 '18 at 17:35
  • $\begingroup$ @philmcole: according to the answer, $Aw$ is always a scalar multiple of the vector $x$. The scalar multiple depends on the length of $y$ and $w$ and the angle between them (see en.wikipedia.org/wiki/Dot_product). $\endgroup$ – parsiad Mar 28 '18 at 18:52

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