# Ideas on Linear Algebra proof

1. Jan 29, 2009

### garyljc

I came across a question saying
Let S: U -> V and T:V -> W be linear maps, where U V and W are vectors spaces over the same field K
Show that Rank(TS) =< Rank(T)

This is my attempt
the im(TS) is a subspace of W
and so is the im(T)

am I missing out something ?

2. Jan 29, 2009

### garyljc

This is my second attempot oon it
im(TS) is a subspace of im(T)
and im(T) is a subspace of W
therefore rank(TS) =< rank (T)
is it correct ?

but isn't im(T) = W ?

3. Jan 30, 2009

### daveyinaz

$$Im(T) = W$$ when $$T$$ is onto.

Your second attempt is the right direction, you might want to explain more of the details.

4. Feb 1, 2009

Assuming the subspaces involved are finite-dimensional, $\operatorname{rank} TS \le \operatorname{rank} T$ means, by definition, $\dim(\operatorname{im} TS) \le \dim(\operatorname{im} T)$. This happens exactly when $\operatorname{im} TS \subseteq \operatorname{im} T$, which you know is true.