Ch 3.2. MatMul: Composition of linear transformations

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1. Matrix multiplication as composition of linear transformations

• To perform multiple linear transformations (in order) on a vector $\text{v}$, you can pre-multiply $\text{v}$ by the matrices representing each transformation.
  • This must be written in reverse order (i.e. the first transformation is written on the right) — see below.
• This is equivalent to multiplying the matrices representing each transformation together, to obtain a composition matrix,
  • Then multiplying the resultant composition matrix by $\text{v}$.

1.1. Example:

Say you wish to apply two transformations to $\text{v}$. First, a rotation, then a shear (in that order).

• Let’s call ${\text{M}}_1$ the $\text{Rotation}$ matrix, and
• ${\text{M}}_2$ the $\text{Shear}$ matrix

We can apply both transformations as follows:

$$\begin{array}{c}\underset{\text{2. Shear}}{\underbrace{\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}}\left(\underset{\text{1. Rotation}}{\underbrace{\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}}\left[\begin{array}{c}x\\ y\end{array}\right]\right)=\underset{\text{Composition}}{\underbrace{\left[\begin{array}{cc}1& -1\\ 1& 0\end{array}\right]}}\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}$$

Note this reads from right to left: ${\text{M}}_2({\text{M}}_1\text{v})$ means first apply ${\text{M}}_1$ to $\text{v}$, then apply ${\text{M}}_2$ to the result of that.

Note that the composition matrix ${\text{M}}_2{\text{M}}_1$ (RHS) is the matrix that represents the combined transformation of the two individual transformations (LHS):

$$\begin{array}{c}\underset{\text{2. Shear}}{\underbrace{\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}}\underset{\text{1. Rotation}}{\underbrace{\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}}=\underset{\text{Composition}}{\underbrace{\left[\begin{array}{cc}1& -1\\ 1& 0\end{array}\right]}}\end{array}$$

1.2. Geometric intuition

1.2.1. Rotation matrix ${\text{M}}_1$:

${\text{M}}_1$ ($\text{Rotation}$ matrix) tells you where the basis vectors $\hat{i}=\left[\begin{array}{c}1\\ 0\end{array}\right]$ and $\hat{j}=\left[\begin{array}{c}0\\ 1\end{array}\right]$ end up after the first transformation (Rotation).

$$\begin{array}{c}{\text{M}}_1=\underset{\text{1. Rotation}}{\underbrace{\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}}\end{array}$$

Specifically:

• Column 1 of ${\text{M}}_1$ states that the first basis vector $\hat{i}$ is “rotated” and its new location is as follows: $\left[\begin{array}{c}1\\ 0\end{array}\right]\to \left[\begin{array}{c}0\\ 1\end{array}\right]$
• Column 2 of ${\text{M}}_1$ states that the second basis vector $\hat{j}$ is “rotated” and its new location is as follows: $\left[\begin{array}{c}0\\ 1\end{array}\right]\to \left[\begin{array}{c}-1\\ 0\end{array}\right]$

1.2.2. Shear matrix ${\text{M}}_2$:

Next, ${\text{M}}_2$ ($\text{Shear}$ matrix) states where these “new” (rotated) basis vectors end up after the second (shear) transformation.

The rotated $\hat{i}$ is “sheared”:

$$\begin{array}{c}\begin{array}{rl}& \underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}}\underset{\begin{array}{c}\text{Rotated i^}\\ \\ {\text{i.e. col}}_1({\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}0\\ 1\end{array}\right]}}=0\left[\begin{array}{c}1\\ 0\end{array}\right]+1\left[\begin{array}{c}1\\ 1\end{array}\right]=\underset{\begin{array}{c}\text{Rotated, then}\\ \text{Sheared i^}\\ \\ {\text{i.e. col}}_1({\text{M}}_2{\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}1\\ 1\end{array}\right]}}\end{array}\end{array}$$

Similarly, the rotated $\hat{j}$ is “sheared” as follows:

$$\begin{array}{c}\begin{array}{rl}& \underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}}\underset{\begin{array}{c}\text{Rotated j^}\\ \\ {\text{i.e. col}}_2({\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}-1\\ 0\end{array}\right]}}=-1\left[\begin{array}{c}1\\ 0\end{array}\right]+0\left[\begin{array}{c}1\\ 1\end{array}\right]=\underset{\begin{array}{c}\text{Rotated, then}\\ \text{Sheared j^}\\ \\ {\text{i.e. col}}_2({\text{M}}_2{\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}-1\\ 0\end{array}\right]}}\end{array}\end{array}$$

The resultant matrix (i.e. the composition of the two transformations) is ${\text{M}}_2{\text{M}}_1$:

$$\begin{array}{c}\begin{array}{rl}{\hat{i}}_{\text{R,S}}=& \underset{\begin{array}{c}\text{Rotated, then}\\ \text{Sheared i^}\\ \\ {\text{i.e. col}}_1({\text{M}}_2{\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}1\\ 1\end{array}\right]}}{\hat{j}}_{\text{R,S}}=\underset{\begin{array}{c}\text{Rotated, then}\\ \text{Sheared j^}\\ \\ {\text{i.e. col}}_2({\text{M}}_2{\text{M}}_1)\end{array}}{\underbrace{\left[\begin{array}{c}-1\\ 0\end{array}\right]}}\\ & \\ {\text{M}}_2{\text{M}}_1=& ({\hat{i}}_{\text{R,S}}|{\hat{j}}_{\text{R,S}})=\underset{\begin{array}{c}\text{Composition}\\ {\text{M}}_2{\text{M}}_1\end{array}}{\underbrace{\left[\begin{array}{cc}1& -1\\ 1& 0\end{array}\right]}}\end{array}\end{array}$$

2. Another example

Say we are applying two transformations ${\text{M}}_1$ then ${\text{M}}_2$ (in that order) to a vector $\text{v}=\left[\begin{array}{c}x\\ y\end{array}\right]$. We can determine the composition matrix ${\text{M}}_2{\text{M}}_1$ as follows:

$$\begin{array}{rl}\underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}0& 2\\ 1& 0\end{array}\right]}}\left(\underset{{\text{M}}_1}{\underbrace{\left[\begin{array}{cc}1& -2\\ 1& 0\end{array}\right]}}\left[\begin{array}{c}x\\ y\end{array}\right]\right)& =\underset{\text{Composition}}{\underbrace{\left[\begin{array}{cc}?& ?\\ ?& ?\end{array}\right]}}\left[\begin{array}{c}x\\ y\end{array}\right]\\ & \\ \underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}0& 2\\ 1& 0\end{array}\right]}}\underset{{\text{M}}_1}{\underbrace{\left[\begin{array}{cc}1& -2\\ 1& 0\end{array}\right]}}& =\underset{\text{Composition}}{\underbrace{\left[\begin{array}{cc}?& ?\\ ?& ?\end{array}\right]}}\end{array}$$

Let’s follow the geometric intuition:

• The columns of ${\text{M}}_1$ tell us where the basis vectors $\hat{i}$ and $\hat{j}$ end up after the first transformation.
• Multiplying ${\text{M}}_2$ by these “new” basis vectors (columns of ${\text{M}}_1$) will tell us where they end up after the second transformation.

$$\begin{array}{rl}\underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}0& 2\\ 1& 0\end{array}\right]}}\underset{{\text{col}}_1({\text{M}}_1)}{\underbrace{\left[\begin{array}{c}1\\ 1\end{array}\right]}}& =1\left[\begin{array}{c}0\\ 1\end{array}\right]+1\left[\begin{array}{c}2\\ 0\end{array}\right]=\underset{{\text{col}}_1({\text{M}}_2{\text{M}}_1)}{\underbrace{\left[\begin{array}{c}2\\ 1\end{array}\right]}}\\ & \\ \underset{{\text{M}}_2}{\underbrace{\left[\begin{array}{cc}0& 2\\ 1& 0\end{array}\right]}}\underset{{\text{col}}_2({\text{M}}_1)}{\underbrace{\left[\begin{array}{c}-2\\ 0\end{array}\right]}}& =-2\left[\begin{array}{c}0\\ 1\end{array}\right]+0\left[\begin{array}{c}2\\ 0\end{array}\right]=\underset{{\text{col}}_2({\text{M}}_2{\text{M}}_1)}{\underbrace{\left[\begin{array}{c}0\\ -2\end{array}\right]}}\end{array}$$

Finally, augment the columns to form the composition matrix ${\text{M}}_2{\text{M}}_1$:

$$\begin{array}{rl}{\text{M}}_2{\text{M}}_1=& ({\text{col}}_1({\text{M}}_2{\text{M}}_1)|{\text{col}}_2({\text{M}}_2{\text{M}}_1))=\underset{\begin{array}{c}\text{Composition}\\ {\text{mat. M}}_2{\text{M}}_1\end{array}}{\underbrace{\left[\begin{array}{cc}2& 0\\ 1& -2\end{array}\right]}}\end{array}$$