Which is the geometric interpretation of the maps?

Question

We consider the matrices \begin{equation*}A=\begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}, B:=\begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix} , \ \ \ S:=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\end{equation*}

I want to give the geometric interpretation of the maps $\vec{x}\mapsto B\cdot \vec{x}$, $\vec{x}\mapsto S\cdot \vec{x}$, $\vec{x}\mapsto A^{-1}\cdot \vec{x}$ and $\vec{x}\mapsto B^{-1}\cdot \vec{x}$.

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Let's consider the map $\vec{x}\mapsto B\cdot \vec{x}$.

We have that: \begin{equation*}B\cdot \vec{x}=\begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}2x_1 \\ x_1+x_2\end{pmatrix}\end{equation*}

Now we consider the map $\vec{x}\mapsto S\cdot \vec{x}$.

We have that: \begin{equation*}S\cdot \vec{x}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}-x_2 \\ x_1\end{pmatrix}\end{equation*}

To give the geometeic interprtation do we have to check the image of the unit square under these maps?

Answer 1

For the first example stated in simplest terms

$$ (x,\quad y)\rightarrow ( 2x, \quad x+y ) $$ is a transformation function just like geometrical form changes like reflection, magnification, sliding, shear etc.

Applying the above Rule of transformation

$$ (0,\quad0)\rightarrow (0,\quad0) $$ $$ (0,\quad 1)\rightarrow (0,\quad 1) $$

$$ (1,\quad 0)\rightarrow (2,\quad 1) $$ $$ (1,\quad 1)\rightarrow (2, \quad 2) $$

It maps a square to a parallelogram with rotation of elements parallel to x-axis by a shear angle $\tan^{-1}\dfrac12$ and a linear magnification of $ \dfrac{\sqrt 3}{2},$

etc..

Answer 2

"To check the image of the unit square under these maps" implies to check the change of basis. Indeed, the standard basis for $\mathbb R^2$ is: $$E_2=\{e_1,e_2\}=\left\{{1\choose 0},{0\choose 1}\right\}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$$ Hence, the transformed basis is: $$S\cdot E_2=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}=S=\left\{{0\choose 1},{-1\choose 0}\right\}=\{s_1,s_2\}$$ Hence, the given vector $\vec x={x_1\choose x_2}$ in base $E_2$ will be scaled in $S$: $$\vec{y}=S\cdot E_2\cdot \vec x=S\cdot \vec x=x_1{0\choose 1}+x_2{-1\choose 0}={-x_2\choose x_1},$$ which is the rotation by $90^\circ$ counterclockwise:

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For $B=\begin{pmatrix}2&0\\ 1&1\end{pmatrix}$: $$\vec{y}=B\cdot E_2\cdot \vec x=B\cdot \vec x=x_1{2\choose 1}+x_2{0\choose 1}={2x_1\choose x_1+x_2},$$

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which is the vertical shear mapping.

See also this for the examples of linear transformation matrices.