Doing one linear map and then another is again a linear map, so it has a matrix of its own. That matrix is called the product of the two, and it is defined so that this is true.

Two currency conversions done one after the other are a single conversion at a combined rate. The combined rate is not invented; it is worked out from the two.

If $B$ maps into the space $A$ maps out of, then applying $B$ and then $A$ is a linear map, and its matrix is written $AB$. Its column j is what the pair does to $\mathbf{e}_j$, which is $A$ applied to column j of $B$. The order is right to left, because that is the order the maps are applied in.

• $AB$ exists only when the columns of $A$ number the same as the rows of $B$.
• If $A$ is $m\times n$ and is , then is .

$B$ turns $\mathbf{x}$ a quarter turn, and then $A$ stretches what it gets to twice the width. The product $AB$ is the single matrix carrying the blue arrow to the green one in one step, which is what it was defined to do. It is written right to left because $B$ acts first.

Unfold the left side as a weighted sum of the columns of $AB$, which are the images under $A$ of the columns of $B$. Because $A$ is linear, those weights can be moved back inside, gathering the columns of $B$ into $B\mathbf{x}$ with waiting outside.

Unfold the product with the vector, using the definition of the columns of $AB$.

The map of $A$ is linear, so a weight may be moved through it in either direction.

Linearity again, this time gathering a sum of images into the image of the sum.

The bracket on the right is $B\mathbf{x}$ by definition, which is the statement.

The two threes meet in the middle and disappear.

$\mathbf{x}$ is nearest to $B$, so $B$ reaches it first.

Nothing about the other columns of $B$ is needed.

$A$ is $2\times3$, $B$ is $3\times4$

$AB$ exists and $BA$ does not, since 4 and 2 do not agree.