Board 8 > Math help [Linear Algebra]

(+) Yeah! (+)

I suck at math, give me a hand?

T: R^n ---> R^m is a linear transformation and S a subspace of R^n. Demonstrate that T(S) = {T(s), s in S} is a subspace of R^m.

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Surskit
.-#Elements of Water#-.

I took linear algebra and hated it.
And you just reminded me why.
I think I just have a difficult time thinking in terms of sets and spaces, though.

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~Peaf~
"Later that day, I f***ed a panda." - PSO

So you have to show that:

T(S) is a subset of R^m (show that T(s) is in R^m for each s in S)
If u and v are in T(S), u+v is in T(S)
If u is in T(S) and c is any scalar, cu is in T(S).

Using the fact that T is linear and that S is a subspace of R^n, these should be fairly straightforward to prove.

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http://img255.imageshack.us/img255/2636/ivotedphoenixyi0.png
No I'm not a damn furry. Looney Tunes are different. - Guiga

Isn't this one of the basic theorems of linear algebra? If you have a subspace, you can linearly transform it into a subspace of another space.

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_foolmo_
'It's easy to get yourself in trouble if you start quoting people who don't like you in your signature' - Mods

Yeah, I know those conditions, I just am not sure how to proceed, though! Like, how do I demonstrate this is a subspace of R^m specifically?

sorry i just really suck at math, and the more abstract it gets, the worse I am

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Surskit
.-#Elements of Water#-.

You need to show that the 3 properties of a subspace hold. I'll use u and v as arbitrary elements in S.

1. 0 is in {T(s), s in S}
0 is in S because S is a subspace, and T(0) = T(0 u) = 0T(u) [because T is a linear transformation] = 0

2. for any two elements in {T(s), s in S}, any linear combination of the two is in {T(s), s in S}.
Without loss of generality, you may let the two elements be T(u) and T(v). Then any linear combination of the two is of the form aT(u) + bT(v) for some a and b. By properties of a linear transformation, aT(u) + bT(v) = T(a u) + T(b v) = T(a u + b v). a u + bv is a linear combination of u and v, so it's in S. Thus, T(a u + b v) is in {T(s), s in S}.

3. for any element in {T(s) [call in T(u) for u in S], s in S} and c is a scalar, cT(u) is in {T(s), s in S}.
cT(u) = T(c u). c u is in S because u is in S, c is a scalar, and S is a vector space. Thus, cT(u) is in {T(s), s in S}.

no offense but it's mostly pretty straightforward stuff that follows from definitions.

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GotD Contest Champ

No offense taken! I can't stress enough how little understanding I have of this.

But many thanks, this helps a lot. :D

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Surskit
.-#Elements of Water#-.