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**HallsofIvy** To be a subspace, we must have closure under addtion and scalar multiplication- and we can put those together- A is a subspace if and only if, for all u and v in A and scalars a and b, au+ bv is also in A.

Let u= (s, 0, 2s) and v= (t, 0, 2t). au+ bv= (as, 0, 2as)+ (bt, 0, 2bt)= (as+ bt, 0, 2(as+ bt)). Is that also in A?

Is -1(x, y, z) also in B?

Is 0(t+1, t, 3t) in C? (The 0 vector, of course, is in every subspace.)

Let u= (x, y, z) and v= (i, j, k) where 4x+ y- z= 0 and 4i+ j- k= 0 so that they are in D. au+ bv= (ax, ay, az)+ (bi, bj, bk)= (ax+ bi, ay+ bj, az+ bk). For that to be in D, we must have 4(ax+ bi)+ (ay+ bj)- (az+ bk)= 0. Rewrite that as (4ax+ ay- az)+ (4bi+ bj- bk= a(4x+ y- z)+ b(4i+ j- k).