# Basis of the kernel

1. Apr 12, 2008

### Luke1294

Find a basis for Ker T that contains S = $$\begin{pmatrix} 1\\ 0\\ 1\\ 0\\ \end{pmatrix}$$, $$\begin{pmatrix} 0\\ 1\\ 0\\ 2\\ \end{pmatrix}$$ where $$T : R^4 -> R^4$$ is defined by

$$T\begin{pmatrix} 1\\ b\\ c\\ d\\ \end{pmatrix} = \begin{pmatrix} a - b - c\\ a - 2b + c\\ 0\\ 0\\ \end{pmatrix}$$.

Well, I have found a basis 'B' for Ker (T) to be B ={$$\begin{pmatrix} 3\\ 2\\ 1\\ 0\\ \end{pmatrix}$$, $$\begin{pmatrix} 0\\ 0\\ 0\\ 1\\ \end{pmatrix}$$}.

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?

2. Apr 12, 2008

### HallsofIvy

Perhaps you should reread the problem. "Find a basis for the kernel of T that includes <1, 0, 1, 0> and <0, 1, 0, 2>" makes no sense as it is easy to see that those two vectors are NOT in the kernel of T and so cannot be in any basis for that kernel.

3. Apr 12, 2008

### Luke1294

Alright, that is what I had thought but just wanted to verify it with someone. Thank you very much!

4. Apr 13, 2008

### HallsofIvy

For future reference: A problem that says "Find a basis for vector space V that include vectors v1 and v2" , assuming v1 and v2 are in V and are independent, means that you are to find a basis that includes those two vectors and possibly more.

Last edited by a moderator: Apr 14, 2008