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

• mitch_1211
In summary, Mitch_1211 found the line joining (3/2, 2, 1) and (3, 4, 2) by forming the vector from one point to the other and then combining that with one of the points.
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

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 ).

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).

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

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.

HallsofIvy said:
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.

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...

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.

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

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.

mitch_1211 said:
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>}

## What is the definition of "basis" and "dimension" in linear algebra?

In linear algebra, a basis is a set of vectors that can be used to express any other vector in a vector space. The dimension of a vector space is the number of vectors in its basis.

## How do you find the basis and dimension of a given vector space?

To find the basis and dimension of a vector space, you need to find a set of linearly independent vectors that span the space. These vectors will form the basis, and the number of vectors in the basis will be the dimension of the space.

## What are the steps for finding the basis and dimension of W = {(a,b,c,0)} where abc are real numbers?

To find the basis and dimension of W = {(a,b,c,0)} where abc are real numbers, follow these steps:

1. Write out the vectors in W as a linear combination of variables, such as (a, b, c, 0) = a(1, 0, 0, 0) + b(0, 1, 0, 0) + c(0, 0, 1, 0).
2. Set up a system of equations, one for each variable, using the coefficients from the linear combination. In this case, the system would be a = 1, b = 0, c = 0.
3. Solve the system of equations to find the values of a, b, and c that make the vectors linearly independent.
4. The vectors (1, 0, 0, 0), (0, 1, 0, 0), and (0, 0, 1, 0) form a basis for W, since they are linearly independent and span W. Therefore, the dimension of W is 3.

## What is the significance of finding the basis and dimension of a vector space?

Finding the basis and dimension of a vector space is important because it allows us to understand the structure and properties of the space. It also helps us to determine the number of linearly independent vectors needed to represent any vector in the space, which is useful for solving problems and performing calculations in linear algebra.

## Are there any other methods for finding the basis and dimension of a vector space?

Yes, there are other methods for finding the basis and dimension of a vector space. Some common methods include Gaussian elimination, row reduction, and using determinants. Each method may be more or less efficient depending on the specific vector space and problem at hand.

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