Splitting a vector into a rowspace component and a nullspace component

[starcoast]

Homework Statement

Find a basis for the orthogonal complement of the row space of A:
A =
[1 0 2
1 1 4]

Split x = (3,3,3) into a row space component x_r and a nullspace component x_n.

The Attempt at a Solution

For the first part of the problem I took A to RREF
R =
[1 0 2
0 1 2]
and then solved to find a basis for the nullspace, x * (-2,-2,1), which should be the basis for the orthogonal complement of the row space.

I THINK that's right but maybe it's not because I'm stuck on the second part, splitting x = (3,3,3). Do I try to find a vector in Row(A) and a vector in Nul(A) that, when added, produce the vector (3,3,3)? How do I solve for that?

Help is very much appreciated!

Also, what's the best way to represent a matrix on these message boards?

 

[starcoast]

nevermind, I figured it out

 

[Deveno]

an observation:

x(-2,-2,1) isn't the basis for Null(A), it's the entire space. "a" basis, is any particular vector, using a non-zero value for x. x = 1 works well, so one basis is:

{(-2,-2,1)}.

to find x_n and x_r, we're looking for a,b, and c with:

(3,3,3) = a(-2,-2,1) + (b(1,0,2) + c(1,1,4))

the first term is in Null(A) and the second sum of two terms in in Row(A).

this isn't that hard, we have:

(3,3,3) = (b-2a+c,c-2a,a+2b+4c), or, if you prefer:

\begin{bmatrix}-2&1&1\\-2&0&1\\1&2&4\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\3\end{bmatrix}

solve this for a,b and c.

****