Matrix multiplication as composition

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Recap:

• Linear Transformations are functions with vectors as input and vectors as output
• A Linear Transformation is completely determined by where the transformation moves the, basis vectors, $\hat{i}$ and $\hat{j}$ for 2D spaces
• Any vector in that 2D space can be represented as a linear combination of the 2 base vectors
• If we had the original co-ordinates of the vector before the transformation, we can compute the new co-ordinates of the vector, after the transformation. if only we had the co-ordinates of the base vectors

A Matrix represent a specific linear transformation. And multiplying a vector by the matrix is what it means to apply the transformation of the vector

Vector composition

• A combination of two or more linear combinations is called a composition.
• If we make two linear transformations on a 2D space, one after the another, the result will be the same as applying the composition (product of the 2 transformations) on the space.
• $\left[\begin{array}{c}11\\ 00\end{array}\right]\left(\begin{array}{c}\left[\begin{array}{c}0-1\\ 10\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}\right)=\left[\begin{array}{c}1-1\\ 10\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$
• Matrix multiplication is alway a product of 2 transformation
• The matrix multiplication should be read from right to left. i.e. we apply the transformation of the right most matrix first, and then move from right to left.

M_2 = \begin{bmatrix} 0 & 2 \ 1 & 0 \end{bmatrix}, \quad M_1 = \begin{bmatrix} 1 & -2 \ 1 & 0 \end{bmatrix}

M_2 M_1 = \begin{bmatrix} 0 & 2 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \ 1 & 0 \end{bmatrix}

\begin{bmatrix} 2 & ? \ 1 & ? \end{bmatrix}

- The co-ordinates of the $\hat{j}$ after the first transformation is the second column of the matrix $M_{1}$ by definition of the matrix, the first column is where $\hat{i}$ lands and second column is where $\hat{j}$ lands

\text{Second column computation:} \quad \begin{bmatrix} 0 & 2 \ 1 & 0 \end{bmatrix} \begin{bmatrix} -2 \ 0 \end{bmatrix} = -2 \cdot \begin{bmatrix} 0 \ 1 \end{bmatrix} + 0 \cdot \begin{bmatrix} 2 \ 0 \end{bmatrix}

\begin{bmatrix} 0 \ -2 \end{bmatrix}

- When applying the second transformation think of it as finding the position of this new vector in the new transformation. Forget that it is the $\hat{j}$ and think of it just as a vector. - So to find it you use the same linear combination of $x.\hat{i}$ + $y.\hat{j}$, here $\hat{i} \text{ and } \hat{j}$ are the first and second column of the matrix respectively. > The order of the transformation matters > $M_{1}.M_{2}\neq M_{2}.M_{1}$ > Matrix multiplication is associative, because irrespective of (AB)C or A(BC), we always apply the transformations in the order, first C, then B, then A. irrespective of the paranthesis, so Matrix multiplication is associative.

M_2 = \begin{bmatrix} a & b \ c & d \end{bmatrix}, \quad M_1 = \begin{bmatrix} e & f \ g & h \end{bmatrix}

M_2 M_1 = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} e & f \ g & h \end{bmatrix}

\begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix}

\text{First column computation:} \quad \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} e \ g \end{bmatrix}

e \cdot \begin{bmatrix} a \ c \end{bmatrix}

• g \cdot \begin{bmatrix} b \ d \end{bmatrix} = \begin{bmatrix} ae + bg \ ce + dg \end{bmatrix}