# Matrix dimension of a vector in R4

1. Jul 25, 2012

### Nicksto

1. The problem statement, all variables and given/known data
Let W be the subspace of R4 defined by W={x:V^TX=0}. Calculate dim(w) where
V=(1 2 -3 -1)^T
note: V^T means V Transpose, sorry I don't know how to do transpose sign in here.

2. Relevant equations

3. The attempt at a solution
I tries to do it (1 2 -3 -1)(x1 x2 x3 x4)^T=0
x1+2(x2)-3(x3)-x4=0
So does it means dim(w)=1 ?

2. Jul 25, 2012

### LCKurtz

No. I will use a,b,c,d instead of the subscripted x's to save typing. You have $a+2b-3c-d=0$. Solving for $a$ gives $a=-2b+3c+d$. So$$\left(\begin{array}{c} a \\ b \\ c \\d \end{array}\right) = \left(\begin{array}{c} -2b+3c+d \\ b \\ c \\d \end{array}\right) = b\left(\begin{array}{c} -2 \\ 1 \\ 0 \\0 \end{array}\right) + c\left(\begin{array}{c} 3 \\ 0 \\ 1 \\0 \end{array}\right) + d\left(\begin{array}{c} 1 \\ 0 \\ 0 \\1 \end{array}\right)$$
Does that give you a hint about the dimension?

Last edited: Jul 25, 2012
3. Jul 25, 2012

### Nicksto

It's 3 right? thx man