# Dimension of the column space of a matrix

• Apr 17th 2009, 01:47 AM
Bucephalus
Dimension of the column space of a matrix
I have this problem that I am doing in a book, and well, the answer in the back says the dimension of the column space and the row space is two.

$\displaystyle \left( \begin{matrix} 1 & 1 & 0\\ 1 & 3 & 1\\ 3 & 1 & -1\\ \end{matrix} \right)$

I row reduced it to:

$\displaystyle \left( \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \right)$

So I thought this would mean that they are independent columns and the space it spans has a dimension of 3.

Am I missing something here?

• Apr 17th 2009, 02:03 AM
Twig
recheck
Hi

I get after row-operations:

$\displaystyle \left[\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{matrix}\right]$

Which shows that the matrix has two pivot columns and thus rank = 2.
Equivalently, dim Nul (A) = number of free variables, and
n = dim Nul(A) + rank (A)

And we see here that we have one free variable, so rank (A) = n-dim Nul(A) = 2
• Apr 17th 2009, 02:11 AM
Bucephalus
you're right
Thanks mate.
I put
$\displaystyle a_{33} = 1$
instead of -1 in the original matrix before I reduced it.

cheers.