Exercise 1.3.2

Question

Let a linear transformation in $\mathbb{R}^2$ be in the line $x_1 = x_2$. Find its matrix.

Jing Huang · Accepted Answer

extbf{Solution 1.} Reflection is a linear transformation. It is completely defined on the standard basis.  And $\mathbf{e}_1 = [1 \: 0]^{\mathsf{T}} \xRightarrow{T} \mathbf{r}_1 = [0 \: 1]^{\mathsf{T}}$, $\mathbf{e}_2 = [0 \: 1]^{\mathsf{T}} \xRightarrow{T} \mathbf{r}_2 = [1 \: 0]^{\mathsf{T}}$. So the matrix is the combination of the two transformed standard basis as its first and second column. i.e.
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\begin{equation*}
    T=
    [\mathbf{r}_1 \: \mathbf{r}_2] =
    \begin{bmatrix}
    0 & 1 \
    1 & 0
    \end{bmatrix}.
\end{equation*}

Jing Huang · Answer

extbf{Solution 2. (A more general method)} Let $\alpha$ be the angle between the $x$-axis and the line. The reflection can be achieved through following steps: first, rotate the line around the origin ($z$-axis in 3D space) $-\alpha$ so the line aligns with the $x$-axis (This line happens to pass through the origin, if not, translation is needed in advance to make the line pass through the origin and we need to use \emph{homogeneous coordinates} since translation is not a linear transformation if represented in standard coordinates). Secondly, perform reflection about the $x$-axis. Lastly, we need to rotate the current frame back to its original location or perform other corresponding inverse transformation. So 
\begin{equation*}
    T = Rotz(-\alpha)\cdot Ref \cdot Rotz(\alpha).
\end{equation*}
That is
\begin{equation*}
  \begin{split}
    T &= 
    \begin{bmatrix}
    \cos(-\alpha) & -\sin(-\alpha) \
    \sin(-\alpha) & \cos(-\alpha)
    \end{bmatrix}
    \begin{bmatrix}
    1 & 0 \
    0 & -1
    \end{bmatrix}
    \begin{bmatrix}
    \cos(\alpha) & -\sin(\alpha) \
    \sin(\alpha) & \cos(\alpha)
    \end{bmatrix} \
    & =
    \begin{bmatrix}
    \cos(-\frac{\pi}{4}) & -\sin(-\frac{\pi}{4}) \
    \sin(-\frac{\pi}{4}) & \cos(-\frac{\pi}{4})
    \end{bmatrix}
    \begin{bmatrix}
    1 & 0 \
    0 & -1
    \end{bmatrix}
    \begin{bmatrix}
    \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \
    \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4})
    \end{bmatrix} \
    & =
    \begin{bmatrix}
    0 & 1 \
    1 & 0
    \end{bmatrix}.
  \end{split}
\end{equation*}

Exercise 1.3.2

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Exercise 1.3.2

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