# Rank

• December 1st 2010, 03:27 PM
manygrams
Rank
Hey there,

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

Quote:

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

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

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?
• December 1st 2010, 04:17 PM
Drexel28
Quote:

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?

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

Solution:

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