# Thread: Dimension of the column space of a matrix

1. ## 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.

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

I row reduced it to:

$\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?

2. ## recheck

Hi

I get after row-operations:

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

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

3. ## you're right

Thanks mate.
I put
$
a_{33} = 1
$

instead of -1 in the original matrix before I reduced it.

cheers.