# Thread: rank and null space

1. ## rank and null space

just need to know how to solve a rank and null space of an 3x5 matrix. any help would be helpful thanks.

2. Originally Posted by fatboy
just need to know how to solve a rank and null space of an 3x5 matrix. any help would be helpful thanks.
The rank of a matrix is the dimension of its row or column space (both will be the same). Now, you could find a basis for the row space, and then the number of elements in the basis will be the rank of the matrix. To find a basis for the row space, you can use the fact that the nonzero rows of a matrix in row-echelon that is row-equivalent to another matrix A will actually form a basis for the row space of A.

So, for example:

$A = \left[\begin{array}{ccc}
2 & -3 & 1\\5 & 10 & 6\\8 & -7 & 5
\end{array}\right]$

Put A in row echelon form to get:

$B = \left[\begin{array}{ccc}
1 & -\frac32 & \frac12\\ 0 & 1 & \frac15\\0 & 0 & 0
\end{array}\right]$

Now, since $B$ has two nonzero rows, you can conclude that $\text{rank}(A) = \text{dim}(\text{rowspace}(A)) = 2$.

However, once you know the nullspace of a matrix, you usually don't need to go through the above process to find the rank.

To find the nullspace of A, simply solve the system $A\textbf{x} = \textbf{0}$.

For example, using the same matrix:

$A = \left[\begin{array}{ccc}
2 & -3 & 1\\5 & 10 & 6\\8 & -7 & 5
\end{array}\right]$

Augment this matrix with the 0 column vector and reduce:

$\left[
\begin{array}{ccc|c}
2 & -3 & 1 & 0\\5 & 10 & 6 & 0\\8 & -7 & 5 & 0
\end{array}
\right]
\rightarrow
\left[
\begin{array}{ccc|c}
1 & 0 & \frac45 & 0\cr 0 & 1 & \frac15 & 0\cr 0 & 0 & 0 & 0
\end{array}\right]
$

So,

$
\text{nullspace}(A) = N(A) =
\left[\begin{array}{c}
-\frac45t\\-\frac15t\\t
\end{array}\right] =
t\left[\begin{array}{c}
-\frac45\\-\frac15\\1
\end{array}\right]
$

The dimension of the nullspace is called the nullity. Knowing the nullity of a matrix allows you to find the rank very easily, without finding the row or column space: For any $m\times n$ matrix, $\text{rank}(A) + \text{nullity}(A) = n$.

For our matrix, we see that $n = 3,\text{ and nullity}(A) = 1\Rightarrow\text{rank}(A) = 3 - 1 = 2$, which agrees with what we found earlier.

I hope that helped!