# Prove a(AB)=(aA)B=A(aB) ..

• Sep 28th 2010, 08:49 PM
alice8675309
Prove a(AB)=(aA)B=A(aB) ..
Prove a(AB)=(aA)B=A(aB) for any scalar a. At first I started out like:

a(AB)=(aA)(aB) and quickly noticed that that was very wrong. And then I kept trying to play around with it in different ways and I seem to get tangled and I know this is such an easy concept.
• Sep 29th 2010, 04:14 AM
HappyJoe
I suppose $\displaystyle A$ and $\displaystyle B$ are matrices.

You need to apply the definition of matrix product. Note that two matrices are equal, precisely when each of the corresponding entries are equal. Let me change the name of the scalar to $\displaystyle c$.

Let $\displaystyle a_{ij}$ be the $\displaystyle (i,j)$-entry of $\displaystyle A$ and let $\displaystyle b_{ij}$ be the $\displaystyle (i,j)$-entry of $\displaystyle B$. Suppose also that $\displaystyle A$ is an $\displaystyle m \times n$ matrix, and that $\displaystyle B$ is an $\displaystyle n \times p$ matrix.

Then the $\displaystyle (i,j)$-entry of $\displaystyle AB$ is "the $\displaystyle i$'th row of $\displaystyle A$ times the $\displaystyle j$'th column of $\displaystyle B$". So this $\displaystyle (i,j)$-entry of $\displaystyle AB$ is:

$\displaystyle \sum_{k=1}^n a_{ik}b_{kj}$,

and so the $\displaystyle (i,j)$-entry of $\displaystyle c(AB)$ is

$\displaystyle c\sum_{k=1}^n a_{ik}b_{kj} = \sum_{k=1}^n ca_{ik}b_{kj},$

where the first expression is the entry that follows by applying the definition of the product directly, and where the equality follows from bringing the scalar $\displaystyle c$ into each term of the sum.

Consider then the product $\displaystyle (cA)B$. We need to check that the $\displaystyle (i,j)$-entry of this product is same expression as for $\displaystyle c(AB)$. But $\displaystyle cA$ is the matrix, where the $\displaystyle (i,j)$-entry is $\displaystyle ca_{ij}$. The product $\displaystyle (cA)B$ has as its $\displaystyle (i,j)$-entry the $\displaystyle i$'th row of this matrix multiplied by the $\displaystyle j$'th column of $\displaystyle B$, so

$\displaystyle \sum_{k=1}^n ca_{ik}b_{kj}.$

But this is exactly the same expression as above.

Similarly you may show that A(cB) has the correct $\displaystyle (i,j)$-entries.