# Change of basis Matrix

1. Nov 1, 2009

### beetle2

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

Let {e1,e2,e3} be a basis for the vector space V, and $T:V \rightarrow V$ a linear transformation.

let f1 ;= e1 f2;=e1+e2 f3;=e1+e2+e3

Find the Matrix B of T with respect to {f1,f2,f3} given that the matrix with respect to {e1,e2,e3} is

$$\left( \begin{array}{ccc}1 & 1 &1 \\1 & 1 & 0\\ 1 & 0 & 0\\\end{array} \right)^{-1}$$

2. Relevant equations

3. The attempt at a solution

Let A be the matrax of the basis {f1,f2,f3}
A= $$\left( \begin{array}{ccc}1 & 1 &1 \\0 & 1 & 1\\ 0 & 0 & 1\\\end{array} \right)$$
Than,

B= $$\left( \begin{array}{ccc}1 & 1 &1 \\0 & 1 & 1\\ 0 & 0 & 1\\\end{array} \right)^{-1}$$$$\left( \begin{array}{ccc}1 & 1 &1 \\1 & 1 & 0\\ 1 & 0 & 0\\\end{array} \right)$$$$\left( \begin{array}{ccc}1 & 1 &1 \\0 & 1 & 1\\ 0 & 0 & 1\\\end{array} \right)$$

B=$$\left( \begin{array}{ccc}0 & 0 &1 \\0 & 1 & 1\\ 1 & 1 & 1\\\end{array} \right)$$

Does this look allright?

2. Nov 2, 2009

### lanedance

there is an inverse sign, that you don't seem to carry later on

so based on the question we have the linear transform, in the e basis given by say:

$$u^e = T^e.v^e$$

with
$$T^e=$$
$$$\left( \begin{array}{ccc}1 & 1 &1 \\1 & 1 & 0\\ 1 & 0 & 0\\\end{array} \right)^{-1}$$$
looks ok here, based on the defintion of the basis, the matrix A transforms a vector from the f basis to the e basis, so
$$u^e = A.u^f$$

and using the inverse
$$u^f = A^{-1}.u^e$$

now looking at the transformation relation
$$u^f = A^{-1}u^e = A^{-1}T^e.v^e = A^{-1}T^e(Av^f) = (A^{-1}T^eA)v^f$$

the matrix B in the f basis is
$$u^f = (A^{-1}T^eA)v^f = B v^f$$

now unless i'm following wrong, it looks like you've dropped an inverse sign, and i'm not too sure how you simplify to get to the next line in any case...