Find Matrix for Linear Transformation Given Criteria

Question

Suppose the linear transformation $T$ does the following:

• maps $R^2$ to the plane given by $x-2y+3z=0$,
• preserves angles and distances, and
• maps $(2,4)$ to $(-3\sqrt{2},0,\sqrt{2})$.

I have two questions. First, how might we construct a matrix for $T$? I'm sure that there are multiple ways to do this. Is there a "standard" way that would make sense to someone studying linear algebra for this first time?

Second, is $T$ unique? If not, what additional criteria would make it unique?

This is not a homework question. I am a high school teacher trying to create a challenge problem for some advanced students. Appreciate any responses.

Answer 1

We can look at this transformation as a rotation that has two conditions:

• The image of $u_1=[0, 0, 1]^T$ is $v_1 =\dfrac{1}{\sqrt{14}}[1, -2, 3]^T $
• The image of $u_2=[2, 4, 0]^T$ is $v_2=[-3 \sqrt{2}, 0, \sqrt{2}]^T $

From these two pairs, we can generate a third pair as follows:

• The image of $u_1 \times u_2$ is $v_1 \times v_2$

Hence, we now have the following matrix equation

$ [v_1, v_2, v_3] = R [u_1, u_2, u_3] $

so that

$ R = [v_1, v_2, v_3] [u_1, u_2, u_3]^{-1} $

By writing a small code to implement this, you get

$ R = \begin{bmatrix} -0.27308 && -0.92412 && 0.267261 \\ 0.755929 && -0.37796 && -0.53452 \\ 0.594979 && 0.056064 && 0.801784 \end{bmatrix} $

The linear transformatin matrix from $\mathbb{R}^2 $ to $\mathbb{R}^3 $ is the first two columns of $R$, hence,

$ A = \begin{bmatrix} -0.27308 && -0.92412 \\ 0.755929 && -0.37796 \\ 0.594979 && 0.056064 \end{bmatrix} $

If we now repeat the above steps, but with the image of $u_1$ being $-v_1$, then the rotation matrix will be

$ R = \begin{bmatrix} -0.57545 && -0.77294 && -0.26726 \\ -0.75593 && 0.377964 && 0.534522 \\ -0.31214 && 0.509621 && -0.80178 \end{bmatrix} $

So that the transformation matrix is the first two columns and this gives us

$ A = \begin{bmatrix} -0.57545 && -0.77294 \\ -0.75593 && 0.377964 \\ -0.31214 && 0.509621 \end{bmatrix} $