a question in basis transformation

transgalactic

when i was told to build the T in the B basis

http://img151.imageshack.us/my.php?image=img8322kt1.jpg

are these the right steps???

HallsofIvy

You could check it your self, In terms of the standard basis, T_tr(1, 1, 1)= (3, 3, 7), T_tr(1, 0, 0)= (2, 1, 1), and T_tr(0, 0, 1)= (0, 0, 5). In terms of the B basis, those results would be (3, 3, 7)= a(1, 1, 1,)+ b(1, 0, 0)= c(0, 0, 1)= (a+ b, a, a+ c) so we have a+ b= 3, a= 3, a+ c= 7 which gives a= 3, b= 0, c= 4 or <3, 0, 4> (I am using "< >" for vectors written in the B basis). Similarly, (2, 1, 1) gives a= 1, b= 1, c= 0 or <1, 1, 0> and (0, 0, 5) gives < 0, 0, 5>. If you try to do everything in the "B" basis: multiply your "T_B" matrix by <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> you get the first, second, and third columns, respectively. And they are NOT the same.

You error was when you formed the "S^-1" transformation matrix: you used the B basis vectors as rows and they should be columns. Use
[tex]\left(\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 1\end{array}\right)[/tex]
instead and you should be alright.

transgalactic

i am confused about the vector apeareance
i what form should i put it in the matrix
as a row

i what form should i put it in the matrix
as a column


???

also i was tald that in transformation we put the vectors as rows

for example:
in this sort of question i was to find the basis of V

http://img301.imageshack.us/my.php?i...mg83241re6.jpg

first i thought that when a vector is signed as (x,y,z)
we flip him verticaly
and when its
(x)
(y)
(z)
then it should flip it horisontaly
but apparently thats not how it works

how it works??
how do i write the given vectors in the metrix
and in what form and in what cases??