Transforming Vectors

Question

Let $T$ be the linear transformation from $\mathbb{R}^3$ to $\mathbb R^3$ that reflects every vector about the $xy$-plane and then triples its length.

How do I find the matrix for $T$?

Answer 1

I'd like to give a little exposition on why it is possible in general to completely characterize any linear transformation by simply knowing how it transforms a set of basis vectors of the domain. This part can thus be skipped if you'd just like the answer to your specific question:

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Let $T: \Bbb R^4 \to \Bbb R^3$ be a linear transformation over the field of real numbers. This means that for every $x,y \in \Bbb R^4$ and $\alpha \in \Bbb R$, we have:
$\begin{align}& 1)\ T(x+y) = T(x) + T(y) & \text{This is called additivity.} \\ & 2)\ T(\alpha x) = \alpha T(x)& \text{This is called homogeneity of degree $1$.} \end{align}$

Let's take the standard basis of $\Bbb R^4$ to be our basis and denote them $e_i$ (for instance the first one is the column vector $e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$).

Then let's say that we know exactly how each of those basis vectors is transformed under our transformation, i.e. let $T(e_i)=f_i$. By linearity, we can then find how any vector transforms under $T$. E.g. let $v\in \Bbb R^4$ be given by $v=v_1e_1 + v_2e_2 + v_3e_3+v_4e_4$, where the $v_i$'s are real numbers. Then $T(v)$ $= T(v_1e_1 + v_2e_2 + v_3e_3+v_4e_4)$ $= T(v_1e_1) + T(v_2e_2) + T(v_3e_3)+T(v_4e_4)$ $= v_1T(e_1) + v_2T(e_2) + v_3T(e_3)+v_4T(e_4) = v_1f_1 + v_2f_2 + v_3f_3 + v_4f_4$. But we already know what the $v_i$'s and $f_i$'s are, thus we've completely characterized our transformation by simply knowing how it transforms our basis.

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Now let's look at your matrix. Clearly it'll be a $3\times 3$ matrix because it maps vectors from $\Bbb R^3$ to $\Bbb R^3$. So it'll look like $\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}$. Let's look at how this matrix transforms the basis vectors. Here again, the easiest basis to choose is the standard basis.

So what you need to do is to figure out what the vectors $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ map to under your transformation. I'll let you figure that out, but say that $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ should map to $\begin{pmatrix} p \\ q \\ r \end{pmatrix}$. Then let's just look at what that would mean for our matrix:

$$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} a_{11} \\ a_{21} \\ a_{31}\end{pmatrix}= \begin{pmatrix} p \\ q \\ r \end{pmatrix}$$

So apparently the first column of the matrix $A$ which represents your transformation is just the vector that $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ maps to. And you can easily verify that the other two basis vectors map to the other two columns of your matrix.