# Linear algebra: Transformations

1. Aug 3, 2008

### Niles

1. The problem statement, all variables and given/known data
A linear transformation L : R2 -> R3 is defined by:

$$L({\bf{x}}) = \left( {x_2 ,x_1 + x_2 ,x_1 - x_2 } \right)^T$$

I wish to find the matrix representation of L with respect to the orderes bases [u1, u2] and [b1, b2, b3], where

u1 = (1,2)
u2 = (3,1)

and

b1 = (1,0,0)
b2 = (1,1,0)
b3 = (1,1,1).

3. The attempt at a solution
Ok, I what I want to do is to find the matrix representation of L with respect to U and the standard basis E (I call this matrix A), and then find the matrix representation of L with respect to E and B (I call this matrix X). Then I will multiply these two matrices:

$$$A = \left( {\begin{array}{*{20}c} 2 \hfill & 1 \hfill \\ 3 \hfill & 4 \hfill \\ { - 1} \hfill & 2 \hfill \\ \end{array}} \right)$$$

and

$$$X = \left( {\begin{array}{*{20}c} { - 1} \hfill & 0 \hfill \\ 0 \hfill & 2 \hfill \\ 1 \hfill & { - 1} \hfill \\ \end{array}} \right)$$$.

I believe that the matrix I am being asked for is X*A. But this wont work because of the dimensions. What am I missing here?

Sincerely,
Niles.

2. Aug 3, 2008

### TeTeC

If we define $_u(x)$ to be the coordinates of the vector x in the basis $u = \{u_{1},u_{2}\}$ (with x in $\matbb{R}^{2}$) and $_{b}(L(x))$ the coordinates of the vector L(x) in the basis $b = \{b_{1},b_{2},b_{3}\}$, then we have:

$$_b(L(x))=_{b}(L)_{u}\thinspace_{u}(x)$$

where $_{b}(L)_{u}$ is the matrix in which the j-th column is given by $_b(L({u}_{j}))$ (the coordinates of $L({u}_{j})$ in the basis $b = \{b_{1},b_{2},b_{3}\}$)

The matrix representation of L is the matrix $_{b}(L)_{u}$

The first column would be calculated this way:

$$L({u}_{1}) = (2,3,-1)^{T} = -b_{1} + 4b_{2} - b_{3}$$

(As you can check). And

$$_{b}(L({u}_{1})) = (-1,4,-1)^{T}$$

gives you the first column.

3. Aug 4, 2008

The way I was thinking of doing this is first finding L in the basis of (1,0), (0,1)-> (1,0,0),(0,1,0),(0,0,1) since this is easily done and then putting an transform matrix on each side of it. To the right of L you would want a matrix that goes from $${u_1,u_2}$$ to (1,0),(0,1) and on the left side a matrix that transforms from (1,0,0),(0,1,0),(0,0,1) to $${b_1,b_2,b_3}$$. The L matrix should be 3 x 2. The right side transform matrix should be 2 x 2 and the left side transform matrix should be 3 x 3. Just my 2 cents as to how I think of approaching a problem like this.

4. Aug 4, 2008

### Defennder

You can approach the problem this way:

Let $x_1,x_2$ be the coefficients of basis vectors u1, u2 and $c_1,c_2,c_3$ be the coefficients of the basis vectors b1,b2,b3. So by your notation, we have $$A(x_1\vec{u_1} + x_2\vec{u_2}) = c_1\vec{b_1} + c_2\vec{b_2} + c_3\vec{b_3}$$.

So the matrix multiplication on the left can be seen as

$$\left( \begin{array}{ccc} A\vec{u_1}&A\vec{u_2} \end{array} \right) \left(\begin{array}{c}x_1\\x_2 \end{array}\left) = (\vec{b_1} \ \vec{b_2} \ \vec{b_3}) \left( \begin{array}{c}c_1\\c_2\\c_3 \end{array} \right)$$

where Au1, Au2 and bn are column vectors in the matrix. Now, you should be able to find c1,c2,c3.