Luke1294
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Find a basis for Ker T that contains S = \begin{pmatrix}<br />
1\\<br />
0\\<br />
1\\<br />
0\\<br />
\end{pmatrix}, \begin{pmatrix}<br />
0\\<br />
1\\<br />
0\\<br />
2\\<br />
\end{pmatrix} where T : R^4 -> R^4 is defined by
T\begin{pmatrix}<br /> 1\\<br /> b\\<br /> c\\<br /> d\\<br /> \end{pmatrix} = \begin{pmatrix}<br /> a - b - c\\<br /> a - 2b + c\\<br /> 0\\<br /> 0\\<br /> \end{pmatrix}.
Well, I have found a basis 'B' for Ker (T) to be B ={\begin{pmatrix}<br /> 3\\<br /> 2\\<br /> 1\\<br /> 0\\<br /> \end{pmatrix}, \begin{pmatrix}<br /> 0\\<br /> 0\\<br /> 0\\<br /> 1\\<br /> \end{pmatrix}}.
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?
T\begin{pmatrix}<br /> 1\\<br /> b\\<br /> c\\<br /> d\\<br /> \end{pmatrix} = \begin{pmatrix}<br /> a - b - c\\<br /> a - 2b + c\\<br /> 0\\<br /> 0\\<br /> \end{pmatrix}.
Well, I have found a basis 'B' for Ker (T) to be B ={\begin{pmatrix}<br /> 3\\<br /> 2\\<br /> 1\\<br /> 0\\<br /> \end{pmatrix}, \begin{pmatrix}<br /> 0\\<br /> 0\\<br /> 0\\<br /> 1\\<br /> \end{pmatrix}}.
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?