Linear algebra: Transformations

[Niles]

Homework Statement

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)

andb₁ = (1,0,0)
b₂ = (1,1,0)
b₃ = (1,1,1).

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:

$$\[<br /> A = \left( {\begin{array}{*{20}c}<br /> 2 \hfill & 1 \hfill \\<br /> 3 \hfill & 4 \hfill \\<br /> { - 1} \hfill & 2 \hfill \\<br /> \end{array}} \right)<br /> \]$$

and

$$\[<br /> X = \left( {\begin{array}{*{20}c}<br /> { - 1} \hfill & 0 \hfill \\<br /> 0 \hfill & 2 \hfill \\<br /> 1 \hfill & { - 1} \hfill \\<br /> \end{array}} \right)<br /> \]$$.

I believe that the matrix I am being asked for is X*A. But this won't work because of the dimensions. What am I missing here?Niles.

 

[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.

 

[badphysicist]

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.

 

[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 b_n are column vectors in the matrix. Now, you should be able to find c₁,c₂,c₃.