Finding basis and dimension of W = {(a,b,c,0)} where abc are real numbers

[mitch_1211]

With normal vectors i usually check there is the correct number of vectors i.e 3 for R3 2 for R2 etc and then just check for linear independence but reducing the matrix that results from c1v1+c2v2+..cnvn=0 and determining of unique solution or infinite solutions. There are the right number of vectors, so spanning will automatically follow if the vectors are linearly independent.

The dimensions of R3 is 3 right? DimR2=2, DimP3=4?

In this case is the dimension of W = {(a,b,c,0)} where abc are real numbers 4?

Also how would i go about determining a basis?

Thanks!

Mitch

 

[Bacle]

Try the vectors v1=(1,0,0,0), v2=(0,1,0,0) and V3=(0,0,1,0) , see if they span
(they are clearly LI ).

 

[HallsofIvy]

That is a three dimensional subspace of R⁴.

Note we can write (a, b, c, 0)= (a, 0, 0, 0)+ (0, b, 0, 0)+ (0, 0, c, 0)= a(1, 0, 0, 0)+ b(0, 1, 0, 0)+ c(0, 0, 1, 0).

 

[Bacle]

So, Ivy, what's wrong with my post?

 

[HallsofIvy]

I didn't say there was anything wrong. I was just pointing out to mitch_1211 how he could have seen your answer directly from the given information.

 

[mitch_1211]

Note we can write (a, b, c, 0)= (a, 0, 0, 0)+ (0, b, 0, 0)+ (0, 0, c, 0)= a(1, 0, 0, 0)+ b(0, 1, 0, 0)+ c(0, 0, 1, 0).

Thanks, i figured that what Bacle was referring to. These type of problems are actually quite simple when you realize that the vectors you pull out are the vectors that span the space.

Thanks again to both of you.

 

[mitch_1211]

I actually have one more question,
I have a feeling that the basis for the line joining (3/2 , 2 , 1) and (3 , 4 , 2) is
(3/2 , 2 , 1) but I'm not sure how to show this is the case...

 

[Bacle]

If you mean that as a line in R^3, then it must go thru the origin for it to be

a subspace of R^3. Then any vector in the line, e.g., the segment from 0 to P,

for some point P in the line, would be a basis for the line.

 

[mitch_1211]

so the point (3/2,2,1) would be a basis?

 

[Bacle]

Well, if the line goes thru the origin, then any vector in the line , e.g., the

vector from (0,0,0) to any point would generate the line. Usually we scale

vectors, not points; given a vector v=(a,b,c) thru the origin, then:

(ta,tb,tc) : t in R determines a line thru the origin, and, conversely, any

line thru the origin is ( can be ) determined this way.

 

[HallsofIvy]

I actually have one more question,
I have a feeling that the basis for the line joining (3/2 , 2 , 1) and (3 , 4 , 2) is
(3/2 , 2 , 1) but I'm not sure how to show this is the case...

With the (a, b, c) notation, it is difficult to distinguish between vectors and points. Do you mean the line containing the points (3/2, 2, 1) and (3, 4, 2)? The standard method of finding equations for the line given two points is to form the vector from one point to the other: <3- 3/2, 4- 2, 2- 1>= <3/2, 2, 1> and then combining that with one of the points:
x= 3/2t+ 3/2, y= 2t+ 2, z= t+ 1. A difficulty with that is that a line through two given points does NOT in general pass through the origin and so does NOT form a vector space. It happens here that, because (3, 4, 2)= 2(3/3, 2, 1) that this line does pass through the origin so we can use (0, 0, 0) as a given point and write x= 3/2 t+ 0, y= 2t+ 0, z= t+ 0 or <x, y, z>= <3/2t, 2t, t>= <3/2, 2, 1>t.

That is, {<3/2, 2, 1>} spans the subspace. Of course, so does any multiple, like {<3, 4, 2>}