Consider this matrix transformation:
[tex]\[
\left[\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{array}\right]
\][/tex]

What is the geometric effect of this transformation?



Answer :

To understand the geometric effect of the transformation matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
on vectors and areas in a 2-dimensional space, let’s analyze how it transforms some basic vectors and how it affects areas.

### 1. Effect on Unit Vectors

#### a. Unit Vector along the x-axis:
The unit vector along the x-axis is
[tex]$ \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $[/tex]

When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 1 + 0 \cdot 0 \\ 0 \cdot 1 + \frac{1}{2} \cdot 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right] $[/tex]

Therefore, the unit vector along the x-axis is scaled down to
[tex]$\left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right]$[/tex].

#### b. Unit Vector along the y-axis:
The unit vector along the y-axis is
[tex]$ \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $[/tex]

When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 0 + 0 \cdot 1 \\ 0 \cdot 0 + \frac{1}{2} \cdot 1 \end{array}\right] = \left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right] $[/tex]

Therefore, the unit vector along the y-axis is scaled down to
[tex]$\left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right]$[/tex].

### 2. Effect on Area

The area effect of a transformation matrix is determined by its determinant. The determinant of the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is:
[tex]$ \det \left( \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) - 0 \cdot 0 = \frac{1}{4} $[/tex]

This value tells us that any area in the 2D space will be scaled by a factor determined by the determinant of the matrix. In this case, areas are scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex].

### Conclusion

The geometric effect of the transformation represented by the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is to scale both the x and y components of any vector by [tex]\( \frac{1}{2} \)[/tex]. This results in vectors being halved in length along both axes. Additionally, any area in the plane is scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex], as indicated by the determinant of the transformation matrix.