- #1
Luke1294
- 57
- 0
Find a basis for Ker T that contains S = [tex]\begin{pmatrix}
1\\
0\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
1\\
0\\
2\\
\end{pmatrix}[/tex] where [tex]T : R^4 -> R^4[/tex] is defined by
[tex]T\begin{pmatrix}
1\\
b\\
c\\
d\\
\end{pmatrix} = \begin{pmatrix}
a - b - c\\
a - 2b + c\\
0\\
0\\
\end{pmatrix}[/tex].
Well, I have found a basis 'B' for Ker (T) to be B ={[tex]\begin{pmatrix}
3\\
2\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
0\\
0\\
1\\
\end{pmatrix}[/tex]}.
I noticed that the two sets of vectors S and B are linearly independent of one another. Does this mean that there is no basis for Ker(T) that contains S, as a basis must be a minimal spanning set? Or have I gone astray somewhere?
1\\
0\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
1\\
0\\
2\\
\end{pmatrix}[/tex] where [tex]T : R^4 -> R^4[/tex] is defined by
[tex]T\begin{pmatrix}
1\\
b\\
c\\
d\\
\end{pmatrix} = \begin{pmatrix}
a - b - c\\
a - 2b + c\\
0\\
0\\
\end{pmatrix}[/tex].
Well, I have found a basis 'B' for Ker (T) to be B ={[tex]\begin{pmatrix}
3\\
2\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
0\\
0\\
1\\
\end{pmatrix}[/tex]}.
I noticed that the two sets of vectors S and B are linearly independent of one another. Does this mean that there is no basis for Ker(T) that contains S, as a basis must be a minimal spanning set? Or have I gone astray somewhere?