Is Rank(TS) Always Less Than or Equal to Rank(T)?

[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 ?

 

[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 ?

 

[daveyinaz]

Im(T) = W when T is onto.

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

 

[adriank]

You were pretty much done, but you did some extra stuff that wasn't necessary.

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.