Show that the following set W={(a,b,c,d)|3a+b=c, a+b+2c=2d} is a vector space under the ordinary addition and scalar multiplication of vectors.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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`W` is a subset of `RR^4,` which is a vector space under the ordinary definitions of addition and scalar multiplication, so we only have to verify that `W` is nonempty and closed under these same operations.

Clearly `(0,0,0,0)in W` since `3*0+0=0` and `0+0+2*0=2*0,` so `W` is nonempty. Now if we add two vectors in `W,` we get

`(a,b,c,d)+(e,f,g,h)=(a+e,b+f,c+g,d+h)` and we have to show that `3(a+e)+(b+f)=(c+g)` and `(a+e)+(b+f)+2(c+g)=2(d+h).` The first one is easy, since we already know that



Just add thse equations columnwise and we get what we need. Proving the second equality is just as easy and uses the same idea- that we know `a+b+2c=2d` and `e+f+2g=2h.`

As for scalar multiplication, for `(a,b,c,d)in W` we have

`t(a,b,c,d)=(ta,tb,tc,td),` and `3*ta+tb=t(3*a+b)=tc,` which proves closure here.

We've proved that W is a subspace (and thus a vector space in its own right) of ```RR^4.`

If you've studied kernels of linear transformations between vector spaces and know that the kernel is a subspace, another proof goes:

Define the linear transformation `T:RR^4->RR^4` by

`T(a,b,c,d)=(0,0,c-3a-b,d-(a+b+2c)/2)` and note

`T(a,b,c,d)=(0,0,0,0)` if and only if `c=3a+b` and `a+b+2c=2d.`

Therefore the kernel is the set `W,` making `W` a subspace of `RR^4.`