Question

Find the linear or affine transformations that satisty the desired properties and write it in the form T(x) = Ax + b: The transformation T : mathbb{R}

Transformation properties
ANSWERED
asked 2020-11-02
Find the linear or affine transformations that satisty the desired properties and write it in the form \(T(x) = Ax\ +\ b:\)
The transformation \(T\ :\ \mathbb{R}^{2}\ \rightarrow\ \mathbb{R}^{2}\) sending the origin to itself and a triangle of vertices
\((0,\ 0),\ (1,\ 0),\ (0,\ 1)\) to a triangle of verices (0, 0),
\((\sqrt{\frac{2}{2}},\ \sqrt{\frac{2}{2}}),\ (- \sqrt{\frac{2}{2}},\ \sqrt{2}).\)

Answers (1)

2020-11-03

We have the tranformation as,
\(\vec{T}(t_{1}\ :\ t_{2})=(-1)\frac{det\left(\begin{array}{c}0 & \vec{x} & \vec{y} & \vec{z}\\ t_{1} & a_{1} & b_{1} & c_{1} \\ t_{2} & a_{2} & b_{2} & c_{2} \\ 1 & 1 & 1 & 1\end{array}\right)}{det\left(\begin{array}{c}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ 1 & 1 & 1 \end{array}\right)}\)
Graphical representation:
image
The red triangle is transformed into blue triangle under the transformation T.
Here,
\(\vec{T}(t_{1}\ :\ t_{2})=(-1)\frac{det\left(\begin{array}{c}0 & \vec{x} & \vec{y} & \vec{z}\\ t_{1} & a_{1} & b_{1} & c_{1} \\ t_{2} & a_{2} & b_{2} & c_{2} \\ 1 & 1 & 1 & 1\end{array}\right)}{det\left(\begin{array}{c}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ 1 & 1 & 1 \end{array}\right)}\)
\(=\frac{det\left(\begin{array}{c}0 & \vec{x} & \vec{y} & \vec{z}\\ t_{1} & 0 & 1 & 0 \\ t_{2} & 0 & 0 & 1 \\ 1 & 1 & 1 & 1\end{array}\right)}{det\left(\begin{array}{c}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right)}\)
\(=\ -\frac{\vec{x}t_{1}\ -\ \vec{y}t_{1}\ +\ \vec{x}t_{2}\ -\ \vec{x}\ -\ \vec{z}t_{2}}{1}\)
\(=(\vec{y}\ -\ \vec{x})t_{1}\ +\ (\vec{z}\ -\ \vec{x})t_{2}\ +\ \vec{x}\)
Now, plugging in the transformed entries.
\(=\left(\left(\begin{array}{c}\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)\ -\ \left(\begin{array}{c}0\\ 0\end{array}\right)\right)t_{1}\ +\ \left(\left(\begin{array}{c}\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)\ -\ \left(\begin{array}{c}0\\ 0\end{array}\right)\right)t_{2}\ +\ \left(\begin{array}{c}0\\ 0\end{array}\right)\)
\(=\left(\begin{array}{c}\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)t_{1}\ +\ \left(\begin{array}{c}-\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)t_{2}\)
\(=\left(\begin{array}{c}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right)\left(\begin{array}{c}t_{1}\\ t_{2}\end{array}\right)\)
\(=\left(\begin{array}{c}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}t_{1}\\ t_{2}\end{array}\right)\)
Ansver:
Hence,
\(T(x)=\left(\begin{array}{c}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)x\)
So,
\(A=\left(\begin{array}{c}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\ and\ b=\left(\begin{array}{c}0\\ 0\end{array}\right)\)

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