# Linear Transformations, Linear Algebra Question

1. May 10, 2015

### raay

Hi can any one give me some hints with this question thanks

A = \begin{pmatrix} 3 & -2 &1 & 0 \\ 1 & 6 & 2 & 1 \\ -3 & 0 & 7 & 1 \end{pmatrix}

be a matrix for T:ℝ4→ℝ3 relative to the basis

B = {v1, v2, v3, v4} and B'= {w1, w2, w3}

v1 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \end{pmatrix}
v2 = \begin{pmatrix} 2 \\ 1 \\ -1 \\ -1 \end{pmatrix}
v3 = \begin{pmatrix} 1 \\ 4 \\ -1 \\ 2 \end{pmatrix}
v4 = \begin{pmatrix} 6 \\ 9 \\ 4 \\ 2 \end{pmatrix}
w1 = \begin{pmatrix} 0 \\ 8 \\ 8 \end{pmatrix}
w2 = \begin{pmatrix} -7 \\ 8 \\ -1 \end{pmatrix}
w3 = \begin{pmatrix} -6 \\ 9 \\ 1 \end{pmatrix}

a- Find [T(v_1)]B' , [T(v_2)]B' , [T(v_3)]B' and [T(v_4)]B'.
b- Find T(v1), T(v2), T(v3) and T(v4).
c- Find a formula for T( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} )

Attempt solution for part (a)
[T(v)]B' = [T]B→B' × [v]B
If this is right I dont know how to use it.
Also i tried drawing a diagram but i think i have to use the diagram to find a formula for T in part (c).

Thanks

Last edited by a moderator: May 11, 2015
2. May 11, 2015

### Fredrik

Staff Emeritus
If I understand your notation correctly, it's correct. I would write $[Tv]_{B'}=[T]_{B',B}[v]_B$. The right-hand side is the product of two matrices, so the next step should be to multiply the matrices.

3. May 11, 2015

### raay

but how do i get [T]B' ,B ? like [T(v1)]B' ,B I know how to do it with polynomials but I have no idea with matrices.

4. May 11, 2015

### Fredrik

Staff Emeritus
$[T]_{B',B}$ is the matrix you included at the start of your post.

5. May 14, 2015

### raay

Please can you explain why it is that ? Thanks

6. May 14, 2015

### Fredrik

Staff Emeritus
There isn't much to explain. By definition $[T]_{B',B}$ denotes the matrix of T with respect to the pair of bases B and B' (B for the domain and B' for the codomain), and that's what you said that A is. So we have $A=[T]_{B',B}$.

The https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ [Broken] may be useful.

Last edited by a moderator: May 7, 2017
7. May 14, 2015

### raay

Oh kkkkkkkkkk got it. Thanks so much.