Subspaces R^n

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Q: Determine whether U is a subspace of R^3.

U = {[0 s t]^T | s and t in R}

A: Yes. U = span {[0 1 0]^T, [0 0 1]}

Can someone explain to me how the heck they come up with that answer? Seems so random.
 
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Let's write column vectors as row vectors, shall we.

U = { (0, s, t); s and t in R } = { (0, s, 0) + (0, 0, t); s and t in R } = { s(0, 1, 0) + t(0, 0, 1); s and t in R }.

See now how one might notice that (0, 1, 0) and (0, 0, 1) form a basis for U?

Of course, this is not necessary to show that U is a subspace of R^3, one can directly use the definition of subspace instead.
 

matt grime

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not random at all - they are just noting that given two (or any set of) vectors, they span a subspace and the things they span in this case is exactly the set U. it is easy to check the axioms if you need to.
 
So basically we got to chose arbitrary (sp?) choose a "s" and "t", so that it satisfies the three main conditions?
 
Actually, how would you do this question?

U = {[r 0 s]^T | r^2 + s^2 = 0, r and s in R}

Determine if U is a subspace of R^3?
 

HallsofIvy

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What are the possible (real) values for r and s so that s2+ y2= 0? What are the possible vectors in this vector space?
 
HallsofIvy said:
What are the possible (real) values for r and s so that s2+ y2= 0? What are the possible vectors in this vector space?
r = 0 and s = 0 correct? Possible vectors? That's what I cannot figure out. Are there many different answers for this?
 

Hurkyl

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Well, there's one element in {[r 0 s]^T | r^2 + s^2 = 0, r and s in R} for each r,s pair satisfying r^2 + s^2 = 0, right?
 
So from what you are saying then yes, there is an element in {[r 0 s]^T | r^2 + s^2 = 0, r and s in R} for each r, s pair satisfying r^2 + s^2 = 0 where r and s = 0?
 

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