Subspaces R^n

  • Thread starter KataKoniK
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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.
 
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So basically we got to chose arbitrary (sp?) choose a "s" and "t", so that it satisfies the three main conditions?
 
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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?
 
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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?
 
168
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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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