# Homework Help: Finding a basis in ImT using Gaussian Elimination

1. May 19, 2012

### JoshMaths

1. The problem statement, all variables and given/known data

$$\begin{pmatrix} -1&3&0\\ 2&0&-1\\ 0&-6&1 \end{pmatrix}$$

Finding the ImT basis of this

3. The attempt at a solution

I got it down to

$$\begin{pmatrix} 1&0&-1/2\\ 0&1&1/6\\ 0&0&1 \end{pmatrix}$$

I know that by the principle of having pivots as the only non-zero entities in their respective columns this makes that column one of the basis vectors. So answer is

[-1,2,0] [3,0,-6]

What i don't understand is why (In the r-echelon form) i cannot subtract -1/2(Row 3) from Row 1 and Subtract 1/6(Row 3) from Row 2 to give the Guass-Jordan form or Identity form which would imply that the entire first matrix was a basis for itself right? Meaning the basis contains three vectors instead of the actual two is contains in the correct answer.

Thanks, and i hope the question has come out clearly, just say if clarification is needed.

Josh

Last edited: May 19, 2012
2. May 19, 2012

### genericusrnme

What is 'ImT basis'?
The image of some basis under T?

3. May 19, 2012

### JoshMaths

No it's just the basis over T i guess. The generic basis that spans T(v) where v is an arbitrary vector and the matrix for T is the above.