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I find in Wikipedia that a matrix $A\in\mathbb{M}^{N\times n}$ has rank one if it can be written as the following tensor product

$$A=a\otimes b,$$ with $b\in \mathbb{R}^N$ and $a\in\mathbb{R}^n$.

I know that a matrix has rank one if one is the maximum number of linearly independent colums of the matrix.

Why a rank one matrix can be written in that way?

Thank you!

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    $\begingroup$ Could you give the Wikipedia reference you are mentionning ? There is a little caution to be taken about the $\otimes$ notation in the matrix domain which can designate the (so important) Kronecker product. Here, you could use $U.V^T$ where $U$ and $V$ are column vectors (as @ekkilop does). $\endgroup$
    – Jean Marie
    Commented Sep 6, 2016 at 14:34
  • $\begingroup$ @JeanMarie I looked at that en.wikipedia.org/wiki/Outer_product $\endgroup$
    – GGG
    Commented Sep 6, 2016 at 15:06

4 Answers 4

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Let a matrix $A$ have columns $\vec{a}_i$, $i = 1, \dots, N$. A rank one matrix is characterised by the fact that all columns $\vec{a}_i$ are linearly dependent, that is $$ \vec{a}_i = b_i \vec{a} $$ for some vector $\vec{a}$ and scalars $b_i$. Thus, the matrix $A$ takes the form $$ A = (b_1 \vec{a}, \dots, b_N \vec{a}) = \vec{a} \vec{b}^T = \vec{a} \otimes \vec{b}, $$ where $\vec{b}^T = (b_1, \dots, b_N)$.

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The matrix $A$ is matrix of some linear map $\mathcal{A} \in \{\mathbb R^n \to \mathbb R^N\}$. Having rank $1$ means that in some bases $\{a_1,a_2,...,a_n\}$ and $\{b_1,b_2,...,b_N\}$ of respective spaces its matrix is $$\pmatrix{ 1 & 0 & ... & 0 \\ 0 & 0 &... & 0 \\ ... & ... & ... & ... \\ 0 & 0 & ... & 0 }$$ and thus $\mathcal A = a_1 \otimes b_1$. This equality holds for any numeric representation of $\mathcal A, a_1, b_1$ including initial basis you wrote matrix $A$ in.

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For example $$\begin{pmatrix}1&1&1\\2&2&2\\3&3&3\end{pmatrix}$$ has rank one and is $(1,1,1)\otimes (1,2,3)$

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    $\begingroup$ I believe OP asked about a general proof, not a specific example. $\endgroup$ Commented Sep 6, 2016 at 13:40
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Look at the first non-zero column of $A$. That is $b$. Then every other column of $A$ is a multiple $b$, since the rank of $A$ is $1$. $a$ consists of these proportionality constants, in order.

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