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

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- Feb 16th 2011, 11:33 AMhmmmmLinear 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 - Feb 16th 2011, 11:38 AMHallsofIvy
I presume you mean . You don't "prove it"- that is the

**definion**of T(X) - Feb 16th 2011, 11:40 AMAckbeet
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? - Feb 16th 2011, 11:57 AMhmmmm
oh sorry I was meant to prove that is a subspace of V sorry about that

- Feb 16th 2011, 12:00 PMAckbeet
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?

- Feb 16th 2011, 12:06 PMhmmmm
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 - Feb 16th 2011, 12:16 PMAckbeet
- Feb 17th 2011, 05:08 AMHallsofIvy
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)?