# Bases of a Linear transformation (Kernel, Image and Union ?

1. May 20, 2012

### sid9221

Bases of a Linear transformation (Kernel, Image and Union !?!

http://dl.dropbox.com/u/33103477/1linear%20tran.png [Broken]

For the kernel/null space
$$\begin{bmatrix} 3 & 1 & 2 & -1\\ 2 & 4 & 1 & -1 \end{bmatrix} = [0]_v$$

Row reducing I get

$$\begin{bmatrix} 1 & 0 & \frac{7}{9} & \frac{-2}{9}\\ 0 & 1 & \frac{-1}{10} & \frac{-1}{10} \end{bmatrix} = [0]_v$$

So the basis of the kernel U1 is:
$$\begin{pmatrix} \frac{-7}{9}\\ \frac{1}{10}\\ 1\\ 0 \end{pmatrix}, \begin{pmatrix} \frac{2}{9}\\ \frac{1}{10}\\ 0\\ 1 \end{pmatrix}$$

Now, for the image/range.

I can write the transformation as:

$$S=\begin{pmatrix} 1 \\ 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} -1 \\ -3 \\ -8 \\ -27 \end{pmatrix}$$

So we proceed to find the basis of Span S

$$\begin{bmatrix} 1 &-1 \\ 1 &-3 \\ 2 &-8 \\ 3 &-27 \end{bmatrix}$$

Row reducing I get:

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

Which implies S is linearly independant.

So the basis is:

$$S=\begin{pmatrix} 1 \\ 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} -1 \\ -3 \\ -8 \\ -27 \end{pmatrix}$$

Now for U1 union U2

Should I just put both the basis together. (After checking if they are all independant ?)