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

[sid9221]

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

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

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

Row reducing I get

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

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

Now, for the image/range.

I can write the transformation as:

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

So we proceed to find the basis of Span S

$$\begin{bmatrix}<br /> 1 &-1 \\ <br /> 1 &-3 \\ <br /> 2 &-8 \\ <br /> 3 &-27 <br /> \end{bmatrix}$$

Row reducing I get:

$$\begin{bmatrix}<br /> 1 &0 \\ <br /> 0 &1 \\ <br /> 0 &0 \\ <br /> 0 &0 <br /> \end{bmatrix}$$

Which implies S is linearly independent.

So the basis is:$$S=\begin{pmatrix}<br /> 1 \\ <br /> 1 \\ <br /> 2 \\ <br /> 3 <br /> \end{pmatrix}, \begin{pmatrix}<br /> -1 \\ <br /> -3 \\ <br /> -8 \\ <br /> -27 <br /> \end{pmatrix}$$

Now for U1 union U2

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

I have no idea about the addition any ideas ??

 

[sid9221]

Any one ?