Thread: Linear mappings and vector spaces proof

let be a linear mapping from vector spaces U to V and let X be a subspace of U. Show that

is a subspace of V

I can understand why this is, it seems pretty trivial but I am not sure how you would go about proving it.

Thanks for any help

I presume you mean . You don't "prove it"- that is the definion of T(X)

I think you meant

right? (Note the capital X on the LHS.)

I think I'd need a bit more background in order to understand your problem, because this is probably how I would define the set on the LHS. How does your book or professor define the LHS normally?

oh sorry I was meant to prove that is a subspace of V sorry about that

So, you could just show that it's closed under scalar multiplication and vector addition, and that the candidate subspace T(X) contains the zero vector. Then you're done, right? So how does this look for you?

Yeah but as T is defined as a linear mapping then these are satidfied? and as U is a subspace it contains 0 which maps to 0 so I am done? or am I missing something?

Thanks for the help

Originally Posted by hmmmm

Yeah but as T is defined as a linear mapping then these are satidfied? and as U is a subspace it contains 0 which maps to 0 so I am done? or am I missing something?

Thanks for the help

Well, I think you should write out the equations that show this. I agree that the linearity of T gets the job done, but that's precisely what you're asked to show. So, how would you write it out?

Suppose u and v are in T(X). That is, there exist x in X such that u= T(x) and there exist y in X such that v= T(y). Now u+ v= T(x)+ T(y)= T(x+ y).

Suppose u is in T(X) and a is any scalar. Then there exist x in X such that u=T(x). Now au= aT(x)= T(ax).

Do you see how those prove that u+ v and au are in T(X)?