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

`3a+b=c` 

`3e+f=g.`

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.`

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