1. ## Linear Algebra

My text is not helping at all and the wikipedia explanations of the concepts: nullspace and range are way over my head.

Given A is a 3x3 matrix
0 1 0
0 0 1
0 0 0

and B is also a 3x3 matrix
0 0 0
1 0 0
0 1 0

specify the nullspace of A and B, and the range of A and B

Thanks

2. Originally Posted by jblorien
My text is not helping at all and the wikipedia explanations of the concepts: nullspace and range are way over my head.

Given A is a 3x3 matrix
0 1 0
0 0 1
0 0 0

and B is also a 3x3 matrix
0 0 0
1 0 0
0 1 0

specify the nullspace of A and B, and the range of A and B

Thanks
The range is the span of the column vectors

$a \cdot \begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix} + b \cdot \begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix} + c \cdot \begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix}

$

These vectors span the xy plane so that is your range.

The null space is the solution to Ax=0

$\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \\
\end{bmatrix} \begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}

$

using the augmented matrix we get

$\begin{bmatrix}
0 && 1 && 0 && 0 \\
0 && 0 && 1 && 0\\
0 && 0 && 0 && 0\\
\end{bmatrix}$

So by parameterizing our solution we

$x=t$
$y=0$
$z=0$

$\begin{bmatrix}
t \\
0 \\
0 \\
\end{bmatrix} = t \cdot \begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix}

$

So the null space is all of the points on the x axis.

3. Note that elementry row operations will change the column space of a matrix.

the span of the columns is

$a \cdot \begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix}
+b \cdot \begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}
+c \cdot \begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}

$

This is the yz plane.

Note that elementry row operation will NOT change the null space.

solving the augmented matrix

$\begin{bmatrix}
0 && 0 && 0 && 0\\
1 && 0 && 0 && 0\\
0 && 1 && 0 && 0\\
\end{bmatrix}$

in reduced row form

$\begin{bmatrix}
1 && 0 && 0 && 0\\
0 && 1 && 0 && 0\\
0 && 0 && 0 && 0\\
\end{bmatrix}$

so parameterizing the solution

$x=0$
$y=0$
$z=t$

so we get...

$\begin{bmatrix}
0 \\
0 \\
t \\
\end{bmatrix} =t \cdot \begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}

$

So the null space is all points on the z-axis.