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sid9221
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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}<br />
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}<br />
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 ??
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}<br />
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}<br />
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 ??
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