# Thread: Kernel of Transformation?

1. ## Kernel of Transformation?

I have the matrix
1st line [1 0 1 -2]
2nd line [1 1 -3 4]
3rd line [3 1 -1 0]
4th line [4 2 -4 4]
5th line [0 1 -4 6]

I need to transform it from R5 to R4, ann then find the base nullity. How do I go about doing this and if possible show a working out. Cheers

2. Originally Posted by LooNiE
I have the matrix
1st line [1 0 1 -2]
2nd line [1 1 -3 4]
3rd line [3 1 -1 0]
4th line [4 2 -4 4]
5th line [0 1 -4 6]

I need to transform it from R5 to R4, ann then find the base nullity. How do I go about doing this and if possible show a working out. Cheers
Let $M$ be that matrix if you define $T: \mathbb{R}_4 \to \mathbb{R}_5$ by $T(\bold{x}) = M\bold{x}$ you get a linear transformation.

The nullity of $T$ is $\text{dim} \left( \{ \bold{x}\in \mathbb{R}_4 |T(\bold{x}) = \bold{0} \} \right)$.

The first step is to turn the system $M\bold{x} = \bold{0}$ to row-reduced echelon form.

Can you do that first?

3. Originally Posted by ThePerfectHacker
Let $M$ be that matrix if you define $T: \mathbb{R}_4 \to \mathbb{R}_5$ by $T(\bold{x}) = M\bold{x}$ you get a linear transformation.

The nullity of $T$ is $\text{dim} \left( \{ \bold{x}\in \mathbb{R}_4 |T(\bold{x}) = \bold{0} \} \right)$.

The first step is to turn the system $M\bold{x} = \bold{0}$ to row-reduced echelon form.

Can you do that first?
Ok i tried reducing it, and this is what i got:
[1 0 1 -2]
[0 1 -4 -2]
[0 0 0 8]
[0 0 0 16]
[0 0 0 8]

4. Originally Posted by LooNiE
Ok i tried reducing it, and this is what i got:
[1 0 1 -2]
[0 1 -4 -2]
[0 0 0 8]
[0 0 0 16]
[0 0 0 8]
When you have the system $M\bold{x} = \bold{0}$ and you are solving for $\bold{x}$.
The matrix you end up with is actually:
$\begin{bmatrix} 1&0&1&-2&0 \\ 1&1&-3&4&0 \\ 3&1&-1&0&0 \\ 4&2&-4&4& 0 \\ 0&1&-4&6&0 \end{bmatrix}$

Now if you done the row-reductions (assuming you were correct!) we get:
$\begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&8&0\\0&0&0&16&0\\0&0&0&8&0 \end{bmatrix}$
You can bring the bottom three rows to:
$\begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

However this matrix is still now in row-reduced echelon form.
We will add the (3rd row $\times$ 2) to 2nd row and 1st row.
$\begin{bmatrix} 1&0&1&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

Finally, add (2nd row $\times -\tfrac{1}{4}$) to 1st row:
$\begin{bmatrix} 1&0&0&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

This tells you that:
$x_1 = 0$
$x_2 - 4x_3 = 0$
$x_4 = 1$.

Thus, $x_1=0,x_4=1,x_2 = 4x_3$ (notice that $x_3$ is "free").
This means if $t\in \mathbb{R}$ then $\bold{x} = \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix}$ solves $M\bold{x} = \bold{0}$, and conversely.

Thus, the nullspace is $\left\{ \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} : t\in \mathbb{R} \right\} = \left\{ t \begin{bmatrix}0\\4\\1\\1 \end{bmatrix} : t\in \mathbb{R}\right\}$

We see the dimension is one, and so the nullity is one.

5. Originally Posted by ThePerfectHacker
When you have the system $M\bold{x} = \bold{0}$ and you are solving for $\bold{x}$.
The matrix you end up with is actually:
$\begin{bmatrix} 1&0&1&-2&0 \\ 1&1&-3&4&0 \\ 3&1&-1&0&0 \\ 4&2&-4&4& 0 \\ 0&1&-4&6&0 \end{bmatrix}$

Now if you done the row-reductions (assuming you were correct!) we get:
$\begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&8&0\\0&0&0&16&0\\0&0&0&8&0 \end{bmatrix}$
You can bring the bottom three rows to:
$\begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

However this matrix is still now in row-reduced echelon form.
We will add the (3rd row $\times$ 2) to 2nd row and 1st row.
$\begin{bmatrix} 1&0&1&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

Finally, add (2nd row $\times -\tfrac{1}{4}$) to 1st row:
$\begin{bmatrix} 1&0&0&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

This tells you that:
$x_1 = 0$
$x_2 - 4x_3 = 0$
$x_4 = 1$.

Thus, $x_1=0,x_4=1,x_2 = 4x_3$ (notice that $x_3$ is "free").
This means if $t\in \mathbb{R}$ then $\bold{x} = \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix}$ solves $M\bold{x} = \bold{0}$, and conversely.

Thus, the nullspace is $\left\{ \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} : t\in \mathbb{R} \right\} = \left\{ t \begin{bmatrix}0\\4\\1\\1 \end{bmatrix} : t\in \mathbb{R}\right\}$

We see the dimension is one, and so the nullity is one.
Ok I checked my working again, and it turns out i was wrong with reducing. The second line is not [0 1 -4 -2] but is [0 1 -4 6]. What does that change?

6. rref(A) =
1 0 1 -2
0 1 -4 6
0 0 0 0
0 0 0 0
0 0 0 0

rank(A)=2, nullity(A)=2

ker(A) = span(
-1 2
4 -6
1 0
0 1