Question

Discrete Mathematics I. University Question Hard.
Q4) provided in the attachment. Cheers! :)

Answer 1

(a) Take any point
`(x,y)` in the plane
`\mathbb R^2`. To reflect this point in the
`y`-axis, you need to negate the
`x`-component, so that

`\mathbf T(\vec x)=\begin{bmatrix}-x\\y\end{bmatrix}`

(b) This can be done easily by choosing

`\mathbf T=\begin{bmatrix}-1&0\\0&1\end{bmatrix}`

as

`\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-x\\y\end{bmatrix}`

(c) Computing
`\mathbf T^2` (very easy since it's diagonal) returns the identity matrix
`\mathbf I_2`, which as a linear transformation returns whatever it is applied to. Geometrically, we're applying the same reflection twice, which returns the original vector to its starting value/position in the plane.

(d) If

`\mathbf A=\begin{bmatrix}a&b\\c&d\end{bmatrix}`

then

`\mathbf{TA}=\begin{bmatrix}-a&-b\\c&d\end{bmatrix}`

`\mathbf{AT}=\begin{bmatrix}-a&b\\-c&d\end{bmatrix}`

`\mathbf{TAT}=\begin{bmatrix}a&-b\\-c&d\end{bmatrix}`

Multiplying
`\mathbf A` by
`\mathbf T` on the left is equivalent to negating the first row of
`\mathbf A`, and on the right to negative the first column of
`\mathbf A`. The third product is equivalent to negating the antidiagonal.