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

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- Feb 6th 2009, 06:19 AMszpengchaoA proof regarding rank
$\displaystyle rank(\alpha+\beta) \leq rank(\alpha) + rank(\beta) $

- Feb 6th 2009, 07:30 AMMush
- Feb 6th 2009, 07:40 AMszpengchaolinear algebra
they r linear maps from V to W.

- Feb 6th 2009, 01:47 PMHallsofIvy
So the "rank" of $\displaystyle \alpha$ is the dimension of $\displaystyle \alpha(V)$ as a subspace of W. If w is in $\displaystyle (\alpha+ \beta)(V)$ then there exist v in V such that $\displaystyle (\alpha+ \beta)v= \alpha(v)+ \beta(v)$ is in W. Certainly that is true if $\displaystyle \alpha(v)$ and $\displaystyle \alpha(v)$ are in W. Do you see why that means that $\displaystyle (\alpha+ \beta)V$ must be a subspace of the direct sum $\displaystyle \alpha(V)$ and $\displaystyle \beta(V)$?

- Feb 6th 2009, 02:02 PMszpengchaook.
wait. please check my proof:

so it is equivalent to prove:

$\displaystyle Im(\alpha+\beta)\subset Im(\alpha)\bigoplus Im(\beta) $

$\displaystyle \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) $