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I believe the following fact is true:

Given a $n\times n$ matrix $M$ over a field $F$ with rank less or equal to 1, then there exists two vectors $\alpha,\beta$ such that $\alpha\beta^T=M$.

Is there any theorem saying this?

If we change the field $F$ to a ring $R$, it may not be true, right?

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Hm, don't know about the ring $R$. Frankly, I don't even know the definition of rank in this case. Could someone maybe give a link to that? –  Dan Shved Oct 31 '12 at 7:54
    
@DanShved, maybe as martini said, the image of $M$ has dimension 1. –  hxhxhx88 Oct 31 '12 at 8:28
    
Next challange for you: If rank of $A$ is $r$, then there exist two matrices $B$ (of size $n\times r$ and rank $r$) and $C$ (of size $r\times n$ and rank $r$) such that $A=BC$. (Btw, $A$ is not necessary to be a square matrix.) –  user26857 Oct 31 '12 at 9:08
    
@navigetor23..oh..your enlightened me! This general case is easy.. we can write $A=PJQ$ where $P$ and $Q$ are invertible $n\times n$ and $m\times m$ matrix and $J$ is $n\times m$ diagonal matrix with first $r$ entries being $1$ and others being $0$... Then split $J$ into $J_1J_2$ which is $n\times r$ and $r\times m$ matrix..and the rest is clear.. –  hxhxhx88 Oct 31 '12 at 14:49

3 Answers 3

up vote 1 down vote accepted

$\def\rk{\operatorname{rank}}\def\im{\operatorname{im}}$If $\rk M = 0$, then $M = 0$, hence $\alpha = \beta = 0$ will work. If $\rk M = 1$, choose $x \in K^n$ with $Mx \ne 0$ and let $\alpha = Mx$. As $\dim \operatorname{im} M = 1$, we have $K\cdot \alpha = \im M$. Now let $1 \le i \le n$ and $e_i$ the $i$-th standard unit vector. Now $Me_i \in \im M$, hence there is $\beta_i \in K$ with $Me_i = \beta_i \alpha$. Let $\beta = (\beta_1, \ldots, \beta_n)^\top \in K^n$. Then we have for each $i$ \[ Me_i = \beta_i \cdot\alpha = \alpha\beta^\top e_i \] hence $M = \alpha\beta^\top$.

If we work over a ring, the condition $\rk M = 1$ tells us in the same way that $\im M$ has a generating set of size 1. If moreover our ring $R$ is a PID, we know that $\im M$, being a submodule of the free module $R^n$, is free, hence the same will work, taking as $\alpha$ a basis of $\im M$.

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Thanks, your answer provides a new view to me! –  hxhxhx88 Oct 31 '12 at 8:31

Yes, this is true. It is quite easy to see. For rank $0$ we have $M=0$ and we are done. For rank $1$, we have a non-zero column in $M$. Set $\beta$ to be this column. Now, since $M$ has rank $1$, any other column will be a multiple of $\beta$, so your matrix looks like $$ M=(a_1 \cdot \beta ,\, a_2 \cdot \beta ,\, \ldots ,\, a_n \cdot \beta). $$ You just assemble all the coefficients $a_i$ into a vector $\alpha$, and there you have it, $M=\beta \alpha^T$.

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..oh..it's so easy.I should have thought it over more..thx!! –  hxhxhx88 Oct 31 '12 at 8:31

If the rank is equal to $1$ then there is a column $c \neq 0$ of $M$ such that every column $c_i$ of $M$ is a multiple of $c$. Lets say $$c_i=\alpha_i c, \ i=1,\ldots, n$$ for some $\alpha_i \in \mathbb{F}$. Set $\beta=c^t.$

Then $M=\alpha \beta^t \ \text{where} \ \alpha=(\alpha_1,\ldots,\alpha_n).$

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I got it. thank you! –  hxhxhx88 Oct 31 '12 at 8:32

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