Finding a basis of a vector space

[brru25]

1. The problem statement

Let W = {(x, y, z, t): x + y + 2z - t = 0} be a vector space under R^4. Find a basis of W over R.

2. The attempt at a solution

To me I would think that the vector space itself could its own basis, but I know I'm probably way off. I also tried solving x = t - y - 2z and say that could be a basis but my confidence level on each possibility is not too high.


Thank you ahead of time for your help.

 

[Dick]

The vector space itself spans the vector space. That doesn't mean it's a basis. In some sense a basis is a set containing the smallest number of vectors that span a vector space. Can you name some vectors in W? Writing them as (x,y,z,t), I'll give you one. (1,0,0,1) is in W. Can you give me another? Now how many do you really need so that linearly combinations of them cover all of W?

 

[brru25]

From what I see, (a, 0, 0, a) where a is any real number could be an element of W. Other ones could be (0, a, 0, a), (0, -a, a, a), and (-a, 0, a, a).

How about (0, -2a, a, 0)? Couple that with (a, 0, 0, a), wouldn't that cover all of W?

 

[Dick]

Why not just put a=1? You can make a basis of W with just three vectors.

 

[brru25]

Why not just put a=1? You can make a basis of W with just three vectors.

a = 1 would give us (1, 0, 0, 1) & (0, -2, 1, 0). What would the third vector be? I thought that these two alone could be enough for the basis.

 

[rootX]

(1,1,1,4) is under W. Does (0, -2a, a, 0) and (a, 0, 0, a) cover that?

I was replying to #3.. delayed a bit in replying.

 

[Dick]

(1,-1,0,0) is also in the span.

 

[rootX]

a = 1 would give us (1, 0, 0, 1) & (0, -2, 1, 0). What would the third vector be? I thought that these two alone could be enough for the basis.

It is simple. Pick one vector, pick second, and pick third that is not a combination of the first and second.

 

[Mark44]

If you write your defining equation a certain way, the vectors will just about pop out at you.
x = -y -2z +t
y = y
z = z
t = t

One vector I see here is (-1, 1, 0, 0), as Dick also pointed out. Another is (1, 0, 0, 1), as bruu pointed out. There is one more obvious one.

 

[HallsofIvy]

Once you have x = t - y - 2z, let each variable on the right be 1 while the others are 0.

For example, it t= 1, y= z= 0, x= 1- 0- 0= 1 so <1, 1, 0, 0> is such a vector. If y= 1, t= z= 0, x= 0- 1- 0= -1 so <0, -1, 1, 0> is such a vector. If z=1, t= y= 0, x= 0- 0- 2= -2 so <0, -2, 0, 1> is such a vector.

 

[brru25]

Thank you to everyone for your help, I think I have it down!