# A proof regarding rank

• Feb 6th 2009, 07:19 AM
szpengchao
A proof regarding rank
$rank(\alpha+\beta) \leq rank(\alpha) + rank(\beta)$
• Feb 6th 2009, 08:30 AM
Mush
Quote:

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?
• Feb 6th 2009, 08:40 AM
szpengchao
linear algebra
they r linear maps from V to W.
• Feb 6th 2009, 02:47 PM
HallsofIvy
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 $\beta(V)$?
• Feb 6th 2009, 03:02 PM
szpengchao
ok.
$Im(\alpha+\beta)\subset Im(\alpha)\bigoplus Im(\beta)$
$\forall w\in Im(\alpha+\beta), \exists \ \ unique \ \ v\in V, \ \ s.t. \ \ (\alpha+\beta)(v)=w \ \ \Rightarrow w\in Im(\alpha)\bigoplus Im(\beta)$