# Linear Transformation

1. Jul 19, 2009

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

Write the standard matrix representation for T1 and use it to find [T1(1,-3,0)]E.

2. Relevant equations

$$T_1\left(x_1,x_2,x_3\right)=\left(x_3,-x_1,x_3\right)$$

3. The attempt at a solution

I just wanted to check to see if I am doing this right. Thanks in advance!

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

$$\left[T_1(1,-3,0)\right]_E=A\left( \begin{array}{c} 1 \\ -3 \\ 0 \end{array} \right)=\left( \begin{array}{ccc} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right).\left( \begin{array}{c} 1 \\ -3 \\ 0 \end{array} \right)=\left( \begin{array}{c} 0 \\ -1 \\ 0 \end{array} \right)$$

2. Jul 19, 2009

### Office_Shredder

Staff Emeritus
Your A is transposed from what it should be.

3. Jul 19, 2009

### HallsofIvy

Staff Emeritus
No, Officeshredder,
Applying T1 to each basis vector in turn gives the columns.

T1(1, 0, 0)= (0, -1, 0)
T1(0, 1, 0)= (0, 0, 0)
T1(0, 0, 1)= (1, 0, 1)

So T1 is represented by
$$\begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$

And, of course, T1(1,-3,0)= (0,-1,0) as said.

4. Jul 19, 2009

Thank you both for your input!

5. Jul 19, 2009

### HallsofIvy

Staff Emeritus
Office shredder may be using a different convention than you and I:

$$T_1(1,-3,0)= \begin{bmatrix}1 & -3 & 0\end{bmatrix}\begin{bmatrix}1 & -1 & 0\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}$$

6. Jul 19, 2009