Yes, it looks correct. Good job!

[beetle2]

Homework Statement

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

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

Homework Equations

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?

 

[lanedance]

comments as below, tried to follow your work

Homework Statement

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

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

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
$$T^e=$$
$$\[ \left( \begin{array}{ccc}1 & 1 &1 \\1 & 1 & 0\\ 1 & 0 & 0\\\end{array} \right)^{-1}\]$$

Homework Equations

The Attempt at a Solution

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

looks ok here, based on the definition 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
$$u^f = (A^{-1}T^eA)v^f = B v^f$$

Than,

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

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...

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


Does this look allright?