A proof regarding rank

**szpengchao** · February 6th 2009, 06:19 AM

$rank(\alpha+\beta) \leq rank(\alpha) + rank(\beta)$

**Mush** · February 6th 2009, 07:30 AM

Originally Posted by szpengchao

$rank(\alpha+\beta) \leq rank(\alpha) + rank(\beta)$

'rank' is a very broad word and has many different meanings in different topics in mathematics.

What topics are you talking about and what do alpha and beta represent?

**szpengchao** · February 6th 2009, 07:40 AM

they r linear maps from V to W.

**HallsofIvy** · February 6th 2009, 01:47 PM

So the "rank" of $\alpha$ is the dimension of $\alpha(V)$ as a subspace of W. If w is in $(\alpha+ \beta)(V)$ then there exist v in V such that $(\alpha+ \beta)v= \alpha(v)+ \beta(v)$ is in W. Certainly that is true if $\alpha(v)$ and $\alpha(v)$ are in W. Do you see why that means that $(\alpha+ \beta)V$ must be a subspace of the direct sum $\alpha(V)$ and $\beta(V)$ ?

**szpengchao** · February 6th 2009, 02:02 PM

wait. please check my proof:
so it is equivalent to prove:
$Im(\alpha+\beta)\subset Im(\alpha)\bigoplus Im(\beta)$

$\forall w\in Im(\alpha+\beta), \exists \ \ unique \ \ v\in V, \ \ s.t. \ \ (\alpha+\beta)(v)=w \ \ \Rightarrow w\in Im(\alpha)\bigoplus Im(\beta)$

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A proof regarding rank

linear algebra

ok.

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