# Matrix Relative to B and B' R3 to R3

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1. Apr 21, 2016

### siimplyabi

1. The problem statement, all variables and given/known data
For problems 1 and 2 use [PLAIN]http://T: [Broken] R^3 to R^3, T(<x1,x2,x3>) = <2x1-x2, x2+3x3, x1 - x2+2x3>,[/PLAIN] , T: R^3 to R^3, T(<x1,x2,x3>) = <2x1-x2, x2+3x3, x1 - x2+2x3>, , bases B = { <1,0,1>, <1,1,0>, <0,1,1> } and B' = { <1,1,-2>, <2,1,-1>, <3,1,1> }. Find T ( <3,-1,2> ) by using

The matrix relative to B and B'

2. Relevant equations
How do i correctly use the c1w1+c2w2+c3w3 ..

3. The attempt at a solution
I've looked in the book and other places online and everyone seems to be doing a 2x2 which is straighforward and makes more sense. I am sort of lost with trying to do this one. I know there must be a T(v1) T(v2) and T(v3) but where is v1, v2 and v3 coming from? Any guidance or help with starting the problem would be great. I know that once I get those v1 v2 v3 they become the columns of the matrix..

Last edited by a moderator: May 7, 2017
2. Apr 21, 2016

### Ray Vickson

I guess the notation $\langle x_1,x_2,x_3 \rangle$ means the vector $\vec{v}$ whose components are $x_1, x_2, x_3$ in the usual basis E, consisting of the vectors $\vec{e_1}, \vec{e_2}, \vec{e_3}$, where $\vec{e_1} = \langle 1,0,0 \rangle$, $\vec{e_2} = \langle 0,1,0 \rangle$ and $\vec{e_3} = \langle 0,0,1 \rangle$.

Step 1 is to write the components of $\vec{v}$ in the new basis B given in the problem. Then, given components $v_1, v_2, v_3$ of $\vec{v}$ in the B-basis, you can figure out its components in the E-basis and from that, get its image $T(\vec{v})$ in the E-basis of the range space. Then you can figure out what that becomes in the B'-basis of the range space.

Note that what I am suggesting is that, rather than using canned formulas, you do it step-by-step explicitly. Once you have really grasped what is happening, then you can turn to canned formulas to lessen the work.

Last edited by a moderator: May 7, 2017