Prove that $$c(AB)= (cA)B= A(cB)$$ where $A,B$ are matrices with $AB$ is defined and $c$ is a scalar.
I don't know where to start, so I guess we start there. This is a theorem that is used in my textbook, and my professor stated that we should practice proving some of these for the test. I can see that this is true but I can't put it into a proof. Obviously it's true just thinking about it, but how do you even start to prove this.
In order to do this properly, you're going to need proper definitions for scalar and matrix multiplication.
Start with scalar multiplication first, as it's easier. If $A$ is an $n \times m$ matrix, and $c$ is a scalar, then $$(cA)_{ij} := c(A)_{ij}.$$ This is to say, the new matrix which we are calling $cA$, has an entry in its $i$th row and $j$th column equal to the entry of $A$ in the same position, just multiplied by $c$. More briefly, we form $cA$ by multiplying each entry by $c$.
A little more complicated is the definition of matrix multiplication. If $A$ is $n \times m$ and $B$ is $m \times k$, then we define $$(AB)_{ij} := \sum_{t = 1}^m (A)_{it} (B)_{tj}.$$ You should convince yourself that this is indeed the definition of matrix multiplication that you're familiar with: each entry is the dot product between the appropriate row of $A$ and the appropriate column of $B$.
Now we have what we need to actually do a proof. We have, \begin{align*} (c(AB))_{ij} &= c(AB)_{ij} & \ldots \text{ by definition of scalar multiplication} \\ &= c \sum_{t = 1}^m (A)_{it} (B)_{tj} & \ldots \text{ by definition of matrix multiplication} \\ &= \sum_{t = 1}^m c(A)_{it} (B)_{tj} & \ldots \text{ by the distributive property of real numbers} \\ &= \sum_{t = 1}^m (cA)_{it} (B)_{tj} & \ldots \text{ by definition of scalar multiplication} \\ &= ((cA)B)_{ij} & \ldots \text{ by definition of matrix multiplication.} \end{align*} Since they are equal in each entry, we therefore have $$c(AB) = (cA)B$$ as required. Try showing they are equal to $A(cB)$ yourself!