Multiplying matrices and numbers¶

We can multiply vectors and numbers like this: $$ 2\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \cdot 3 \\ 2 \cdot 4 \end{bmatrix} $$ Let's extend this to matrices of any size in the obvious way. For example, $$ 10 \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix} = \begin{bmatrix} 10 \cdot 1 & 10 \cdot 3 & 10 \cdot 5 \\ 10 \cdot 2 & 10 \cdot 4 & 10 \cdot 6 \end{bmatrix}. $$ Sometimes the number is written on the right, and it means the same thing: $$ \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}10 = \begin{bmatrix} 10 \cdot 1 & 10 \cdot 3 & 10 \cdot 5 \\ 10 \cdot 2 & 10 \cdot 4 & 10 \cdot 6 \end{bmatrix}. $$

Matrix-matrix and number-matrix multiplication¶

Recall that each number in $AB$ is a dot product of a row from $A$ and a column from $B$. For example: $$ \begin{bmatrix} \red1&\red3 \\ \magenta2&\magenta4 \end{bmatrix} \begin{bmatrix} \blue5&\green7 \\ \blue6&\green8 \end{bmatrix} = \begin{bmatrix} \red1 \cdot \blue5 + \red3 \cdot \blue6 & \red1 \cdot \green7 + \red3 \cdot \green8 \\ \magenta2 \cdot \blue5 + \magenta4 \cdot \blue6 & \magenta2 \cdot \green7 + \magenta4 \cdot \green8 \end{bmatrix} $$ If the first matrix is multiplied by $c$, that multiplies the resulting matrix by $c$. $$ \begin{align} \begin{bmatrix} \red1c&\red3c \\ \magenta2c&\magenta4c \end{bmatrix} \begin{bmatrix} \blue5&\green7 \\ \blue6&\green8 \end{bmatrix} &= \begin{bmatrix} \red1c \cdot \blue5 + \red3c \cdot \blue6 & \red1c \cdot \green7 + \red3c \cdot \green8 \\ \magenta2c \cdot \blue5 + \magenta4c \cdot \blue6 & \magenta2c \cdot \green7 + \magenta4c \cdot \green8 \end{bmatrix} \\ &= \begin{bmatrix} c(\red1 \cdot \blue5 + \red3 \cdot \blue6) & c(\red1 \cdot \green7 + \red3 \cdot \green8) \\ c(\magenta2 \cdot \blue5 + \magenta4 \cdot \blue6) & c(\magenta2 \cdot \green7 + \magenta4 \cdot \green8) \end{bmatrix} \\ &= c\begin{bmatrix} \red1 \cdot \blue5 + \red3 \cdot \blue6 & \red1 \cdot \green7 + \red3 \cdot \green8 \\ \magenta2 \cdot \blue5 + \magenta4 \cdot \blue6 & \magenta2 \cdot \green7 + \magenta4 \cdot \green8 \end{bmatrix} \end{align} $$ Multiplying the second matrix by $c$ does the same thing, so we get the following result:

This is different from how matrix multiplication works in general: even though $AB$ is not necessarily $BA$, you can put the $c$ wherever you want, and the result will get multiplied by $c$ regardless of where you inserted the $c$.

Because matrix-vector multiplication can be calculated as a matrix-matrix multiplication, we can also use the above result with a vector $\vec v$ instead of $B$, and we get $(cA)\vec v = A(c \vec v) = c(A\vec v)$. For example, the matrix $$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$ rotates by 90 degrees counter-clockwise (explained here). Because $(5A)\vec v = 5(A \vec v)$, the matrix $$ 5A = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} $$ rotates by 90 degrees and also makes the vector $5$ times longer. We would also get the same result by starting with a vector that is 5 times longer, as in $A(5\vec v)$.

Subtracting matrices¶

For matrices $A$ and $B$, we can now define $-B=(-1)B$, and $A-B=A + (-B)$. These operations behave like you would expect: just flip the sign of each number in the matrix or subtract the corresponding elements. For example, $$ - {\begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}} = \begin{bmatrix} (-1)5 & (-1)7 \\ (-1)6 & (-1)8 \end{bmatrix} = \begin{bmatrix} -5 & -7 \\ -6 & -8 \end{bmatrix} $$ and $$ \begin{align} \red{\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}} - \blue{\begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}} &= \red{\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}} + \begin{bmatrix} -\blue 5 & -\blue 7 \\ -\blue 6 & -\blue8 \end{bmatrix} \\ &= \begin{bmatrix} \red1+(-\blue5) & \red3+(-\blue7) \\ \red2+(-\blue6) & \red4+(-\blue8) \end{bmatrix} \\ &= \begin{bmatrix} \red1-\blue5 & \red3-\blue7 \\ \red2-\blue6 & \red4-\blue8 \end{bmatrix}. \end{align} $$ Subtraction also works with matrix multiplication like you would expect: we have $$ (A-B)C = (A + (-1)B)C = AC + (-1)BC = AC - BC, $$ and we similarly get $$ A(B-C) = AB-AC. $$