Showing That $(AB)x = A(Bx)$: A Clue!

[evinda]

Hello! (Wave)

I want to show that if $A$ and $B$ are two $n \times n$ matrices and $x \in \mathbb{R}^n$, then $(AB) x=A(B x)$.

What do we deduce from the above equality for the relation between the composition of the functions $x \to Bx, y \to Ay$ and the multiplication of matrices?Could you give me a hint how we could show that $(AB) x=A(B x)$?

 

[Ackbach]

So,
$$(Bx)_i=\sum_{j=1}^n B_{ij}x_j,$$
and
$$(AB)_{ki}=\sum_{l=1}^n A_{kl}B_{li}.$$
Can you continue?

 

[evinda]

So,
$$(Bx)_i=\sum_{j=1}^n B_{ij}x_j,$$
and
$$(AB)_{ki}=\sum_{l=1}^n A_{kl}B_{li}.$$

So $k$ is the row and $i$ the column, right?

So do I have to find now the form of $(AB)_{ki} x_i$ ?

 

[Deveno]

The $i$-th coordinate of $Bx$ is:

$\sum\limits_j b_{ij}x_j$ <----this sum is a scalar, we wind up with $i$ different scalar entries of $Bx$.

Notice that for this to even make sense, the "$j$" 's must match up: $B$ must have the same number of COLUMNS, as $x$ has ROWS ($x$ is taken to be a column-vector, so it only has 1 column, considered as a matrix).

Now we "hit this with $A$". We use $A = (a_{ki})$, since, again, for this to be well-defined, we must have that $A$ has $i$ columns, as $Bx$ has $i$ entries (rows).

Now, $A(Bx)$ will have $k$ entries, the $k$-th entry will be:

$[A(Bx)]_k = \sum\limits_i a_{ki}(Bx)_i = \sum\limits_i a_{ki}\left(\sum\limits_j b_{ij}x_j \right)$.

Using the distributive law, we can rewrite the above as:

$= \sum\limits_i \sum\limits_j a_{ki}b_{ij}x_j$

and using the distributive law AGAIN, to collect all the $x_j$ terms, for each $j$:

$= \sum\limits_j \left(\sum\limits_i a_{ki}b_{ij}\right)x_j$

Now, the stuff in the parentheses is the $k,j$-th entry of $AB$ (by the definition of matrix multiplication), so we have:

$[A(Bx)]_k = [(AB)x]_k$, for each $k$.

It easier to see what is going on with specific values for $k,i,j$: so let's say $A$ is a 3x2 matrix, and $B$ is a 2x2 matrix.

So we start with a 2-vector: $(x_1,x_2)$.

Then $Bx = (b_{11}x_1 + b_{12}x_2, b_{21}x_1 + b_{22}x_2)$.<---a different 2-vector now.

Now $A(B(x)) =
(a_{11}(b_{11}x_1 + b_{12}x_2) + a_{12}(b_{21}x_1 + b_{22}x_2),
a_{21}(b_{11}x_1 + b_{12}x_2) +a_{22}(b_{21}x_1 + b_{22}x_2),
a_{31}(b_{11}x_1 + b_{12}x_2) + a_{32}(b_{21}x_1 + b_{22}x_2))$

$=((a_{11}b_{11} + a_{12}b_{21})x_1 + (a_{11}b_{12} + a_{12}b_{22})x_2,
(a_{21}b_{11} + a_{22}b_{21})x_1 + (a_{21}b_{12} + a_{22}b_{22})x_2,
(a_{31}b_{11} + a_{32}b_{21})x_1 + (a_{31}b_{12} + a_{32}b_{22})x_2)$

Note how all the "grouped terms" are entries of the matrix $AB$, for example:

$(a_{11}b_{11} + a_{12}b_{21}$ is the $1,1$-entry of $AB$, and $a_{11}b_{12} + a_{12}b_{22}$ is the $1,2$-entry of $AB$, and so on,

so that we have the above is $(AB)x$.