Review : Matrices & Vectors — Question 8

Question 8 – Solution

Strategy. Separate a vector into components parallel and perpendicular to the normal. The formula changes the sign of only the normal component.

Step 1: Identify the reflection axis. Here $n^Tn=5$, so $$\boxed{H=\frac 15\begin{pmatrix}3&-4\\-4&-3\end{pmatrix}.}$$ Write $x=x_\perp+x_\parallel$, where $x_\parallel=n(n^Tx)/5$ and $n^Tx_\perp=0$. Then $Hx=x_\perp-x_\parallel$: it reflects across the line $\boxed{x_1+2x_2=0}$, whose direction is $(2,-1)^T$.

Step 2: Verify the metric and inverse identities. The matrix is symmetric, and its numerator squared is $\begin{pmatrix}25&0\\0&25\end{pmatrix}$. Thus $H^2=H^TH=I_2$. For any $x,z$, $(Hx)^T(Hz)=x^TH^THz=x^Tz$, preserving dot products and hence lengths. Also $H^{-1}=H$: reflecting twice restores the original vector.

Step 3: Transform and inspect the triangle. The images are $$HO=O,\qquad Ha=(1,-3)^T,\qquad Hb=(-1,-2)^T.$$ The original ordered edge determinant is $3\cdot 2-1\cdot 1=5$, so its area is $5/2$. The reflected edge determinant is $1(-2)-(-1)(-3)=-5$. Its area remains $5/2$ but its orientation reverses, consistent with $\det H=-1$. The midpoint of $a$ and $Ha$ is $(2,-1)$ on the axis; the displacement $Ha-a=(-2,-4)=-2n$ is perpendicular to that axis.

Step 4: Classify fixed and reversed vectors. From the formula, $Hx=x$ exactly when $n^Tx=0$, giving the axis. For $Hx=-x$, the decomposition gives $x_\perp=0$, so $x$ is a multiple of $n$. The zero vector belongs to both sets. These conclusions follow from the components directly, without a characteristic-polynomial calculation. The plot uses equal axis scales.