For the subspace given a find a basis and b state the dimens

Solution

{[4s,-3s,-t]:s,t in R}

You can tell that there are two \"dimensions\" in this subspace because you have two parameters. This isn\'t a rigorous explanation, but you see that I think.

So to find a basis, you need two things that are linearly independent. The best way to do this is to think of what I just said.

If each parameter represents a dimension, then you can \"reformulate\" this subspace as simply the space R^2 defined by:

R^2 = { (s,t) : s and t are real numbers }

What is the standard basis?

(1,0)
(0,1)

Use that as your guide:

s=1,t=0 in W gives you the vector:

(4,-3,0)

s=0,t=1 in W gives you the vector:

(0,0,-1)

These must be linearly independent (notice where the zeros are in the third, first, and second coordinates). Thus it is a basis:

B = { (4,-3,0) , (0,0,-1) }

Now you can even do more with this! We wanted to say this \"looks like\" R^2. You can now explicitly construct a mapping:

Take w in W. This vector must be of the form:

w = s(4,-3,0) + t(0,0,-1)

for real numbers s and t, since we have B as a basis. We map this element to the element (s,t) in R^2. We notice that it transforms our basis as follows:

s(4,-3,0) + t(0,0,-1) (s,t)
(4,-3,0) (1,0)
(0,0,-1) (0,1)

You can check easily that the map is a proper linear map from one space to another. Because it maps the basis onto the basis, we know the subspace W is equivalent (also called \"isomorphic\") to R^2.