# Basis change matrix

1. Dec 6, 2008

### Marin

Hi all!

I want to determine to matrices, S and T, so that for the matrix B,

$$B=\left(\begin{array}{cccc}-1&-2&-5&-3\\1&1&4&2\\4&-3&9&1\\2&-6&0&-4\end{array}\right)$$

it´s true that:

$$SBT=\left(\begin{array}{cc}E_k&0\\0&0\end{array}\right)$$

where E_k is the identity matrix of dimension k x k and k is the rank of B

I successfully calculated that rg(B)=2 and I also have the vectors that solve Bx=0:

$$v_1=\left(\begin{array}{c}-3\\-1\\1\\0\end{array}\right)$$

and

$$v_1=\left(\begin{array}{c}-1\\-1\\0\\1\end{array}\right)$$

I don´t know why, but I cannot express the columns of B as linear combination of v_1 and v_2. I checked my calculations twice...?

I´m pretty sure it´s a basis change problem, but instead of having the two basis I have the transformed matrix.

How is one supposed to do such transformations?

Any help will be much appreciated!

2. Dec 7, 2008

### HallsofIvy

If the rank of B is 2 (image is two dimensional), then you can't write it in the form
$$SBT=\left(\begin{array}{cc}E_k&0\\0&0\end{array}\right)$$
It would have to have rank 3 to be written that way.

Last edited by a moderator: Dec 8, 2008
3. Dec 7, 2008

### Marin

no no, E_k is the identity matrix where k is 2=rgB so SBT is a 4 x 4 matrix in the top left corner of which we write the identity 2 x 2 matrix and zeroes elsewhere

but I ´m still searching for a way to perform this transformation...