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 &amp; 1 &amp; 2 &amp; -1\\ <br /> 2 &amp; 4 &amp; 1 &amp; -1<br /> \end{bmatrix} = [0]_v

Row reducing I get

\begin{bmatrix}<br /> 1 &amp; 0 &amp; \frac{7}{9} &amp; \frac{-2}{9}\\ <br /> 0 &amp; 1 &amp; \frac{-1}{10} &amp; \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}<br />

So we proceed to find the basis of Span S

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

Row reducing I get:

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

Which implies S is linearly independant.

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}<br />

Now for U1 union U2

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

I have no idea about the addition any ideas ??

 

[sid9221]

Any one ?