Rank

Hey there,

If anyone could give me a hint that would be awesome.

Quote:

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Let V be a finite-dimensional vector space and let $\displaystyle S,T \in \mathcal{L}(V)$.

prove rank(S+T)<= rankS+rankT

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All I have is:

Take any x from V.

Rank (S+T) implies dim((S+T)(x)) = dim(S(x)+T(x))

But I have no clue what to do from here. I know it's really trivial, but I'm stuck.

Hints?

Quote:

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Originally Posted by manygrams

Hey there,

If anyone could give me a hint that would be awesome.



All I have is:

Take any x from V.

Rank (S+T) implies dim((S+T)(x)) = dim(S(x)+T(x))

But I have no clue what to do from here. I know it's really trivial, but I'm stuck.

Hints?

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Hint:
Note first that $\displaystyle \left(S+T)(V)\subseteq S(V)+T(V)$. To see this merely note that if $\displaystyle (S+T)(v)\in (S+T)(V)$ then $\displaystyle (S+T)(v)=S(v)+T(v)\in S(V)+T(V)$.

Solution:

Spoiler:

Thus, $\displaystyle \dim\left((S+T)(V)\right)\leqslant \dim\left(S(V)+T(V)\right)=\dim S(V)+\dim T(V)-\dim\left(S(V)\cap T(V)\right)\leqslant \dim S(V)+\dim T(V)$.