Matrix operations

First we have scalar product, i.e. the product of a scalar and a matrix, and the sum of two matrices. These are straightforward because they operate elementwise.

• scalar-matrix product: define $c\cdot A$, where $c$ is a scalar and $A$ is an $m\times n$ matrix, to be the $m\times n$ matrix entry $i,j$ equal to $c\cdot a_{i,j}$. In other words:
  If $c$ is a scalar and $A$ is an $m\times n$ matrix, the scalar-matrix product $c\cdot A$ is defined as $$ c \cdot \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} = \begin{bmatrix} c\cdot a_{1,1} & c\cdot a_{1,2} & \cdots & c\cdot a_{1,n}\\ c\cdot a_{2,1} & c\cdot a_{2,2} & \cdots & c\cdot a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ c\cdot a_{m,1} & c\cdot a_{m,2} & \cdots & c\cdot a_{m,n} \end{bmatrix}. $$

  Example: $ 5\cdot \begin{bmatrix} 3 & -1\\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 5\cdot 3 & 5\cdot -1\\ 5\cdot 2 & 5\cdot 4 \end{bmatrix} = \begin{bmatrix} 15 & -5\\ 10 & 20 \end{bmatrix} $

• matrix-matrix sum: define $A + B$, where $A$ and $B$ are both $m\times n$ matrices, to be the $m\times n$ matrix entry $i,j$ equal to $a_{i,j} + b_{i,j}$. In other words:
  If $A$ and $B$ are $m\times n$ matrices, the matrix sum $A+B$ is defined as $$ \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} + \begin{bmatrix} b_{1,1} & b_{1,2} & \cdots & b_{1,n}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m,1} & b_{m,2} & \cdots & b_{m,n} \end{bmatrix} = \begin{bmatrix} a_{1,1}+b_{1,1} & a_{1,2}+b_{1,2} & \cdots & a_{1,n}+b_{1,n}\\ a_{2,1}+b_{2,1} & a_{2,2}+b_{2,2} & \cdots & a_{2,n}+b_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m,1}+b_{m,1} & a_{m,2}+b_{m,2} & \cdots & a_{m,n}+b_{m,n} \end{bmatrix}. $$
  Example: $ \begin{bmatrix} 3 & -1\\ 2 & 4 \end{bmatrix} + \begin{bmatrix} -5 & 0\\ 6 & -3 \end{bmatrix} = \begin{bmatrix} 3+-5 & -1+0\\ 2+6 & 4+-3 \end{bmatrix} = \begin{bmatrix} -2 & -1\\ 8 & 1 \end{bmatrix} $

Observe that we can always view an $m\times n$ matrix as a vector with $mn$ elements and, as long as we remember the dimension, we can bounce back and forth between the matrix view and the vector view. The reason this observation is important is because addition and scalar multiplication are the same in either view. (Hopefully this is easy to see.) That means that the associativity rules for scalar multiplication and sum that we have for vectors carries over to matrices, and we get that for free without having to do new proofs!

ACTIVITY ← we might skip for time
Prove the following theorem

• matrix-matrix product: Matrix multiplication is where we really have to face something new. The basic intuition is that $A \cdot B$ will be defined as doing matrix-vector product of $A$ with each of the column vectors of $B$, and the resulting column vectors are collected into a matrix, which is the final result of the matrix-matrix product. But this means the number of elements in a row of $A$ has to equal the number of elements in a column of $B$, otherwise the dot product isn't defined.
  If $A$ is $m\times n$ matrix, and $B$ is $n\times k$ matrix (note: the number of columns in $A$ is the same as the number of rows in $B$), the matrix product $A\cdot B$ is the matrix of dimension $m\times k$ defined as $$ \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \cdot \begin{bmatrix} b_{1,1} & b_{1,2} & \cdots & b_{1,k}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,k}\\ \vdots & \vdots & \ddots & \vdots\\ b_{n,1} & b_{n,2} & \cdots & b_{n,k} \end{bmatrix} = \begin{bmatrix} c_{1,1} & c_{1,2} & \cdots & c_{1,k}\\ c_{2,1} & c_{2,2} & \cdots & c_{2,k}\\ \vdots & \vdots & \ddots & \vdots\\ c_{m,1} & c_{m,2} & \cdots & c_{m,k} \end{bmatrix} \text{, where } c_{i,j} = \begin{bmatrix}a_{i,1} & \cdots & a_{i,n}\end{bmatrix}\cdot \begin{bmatrix} b_{1,j}\\ \vdots\\ b_{n,j} \end{bmatrix} $$

  

  Example: $ \begin{bmatrix} 3 & -1\\ 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} -5 & 0\\ 6 & -3 \end{bmatrix} = \begin{bmatrix} %% 1,1 \begin{bmatrix}3&-1\end{bmatrix} \cdot \begin{bmatrix}-5\\ 6\end{bmatrix}\hspace{12pt} %% & %% 1,2 \begin{bmatrix}3&-1\end{bmatrix} \cdot \begin{bmatrix}0\\ -3\end{bmatrix}\\ %% &\\ %% 2,1 \begin{bmatrix}2&4\end{bmatrix} \cdot \begin{bmatrix}-5\\ 6\end{bmatrix} %% & %% 2,2 \begin{bmatrix}2&4\end{bmatrix} \cdot \begin{bmatrix}0\\ -3\end{bmatrix} %% \end{bmatrix} = \begin{bmatrix} -21&3\\ 14&-12 \end{bmatrix} $

ACTIVITY
In-class activity on matrix multiplication. Note: this is really important stuff!

Is matrix multiplication commutative? In class I asked you to figure this out and prove either yes or no. The answer is no!, and to disprove all we need to do is find a counter example: just one pair $A$ and $B$ such that $AB \neq BA$. In fact, almost any pair you chose works. Some work because the dimensions do not match up properly when you switch the order. But I suggest you write down two random 2x2 matrixes and try multiplying them in both orders. You will almost certainly see different results! So matrix multuplication is not commutative, not even if we restrict ourselves to square matrices.

Is matrix multiplication associative? The answer is yes!, but the proof is tedious and not very informative. If you write out the symbolic expressions for the $i,j$th element of $(AB)C$ and $A(BC)$ you'll see that they are equal. I won't make you do it because of the tedium.

The ring of nxn matrices