Reverse Order Law for the Transposition of Matrices

We have covered matrix transposition in a previous lesson, but now that we know about matrix multiplication, we can learn about an important rule when transposing the product of two matrices.

Generally, the following holds for two conformable matrices $A$ and $B$:

As you can see, we had to swap the order of $A$ and $B$ in the multiplication.

Proof

If you’re interested, here is the proof for this.

Let $n$ be the number of columns in matrix $A$ and the number of rows in matrix $B$.

By the definition of matrix transposition and matrix multiplication, we get that for each row $i$ and column $j$

Since $A_{jk}$ and $B_{ki}$ are scalars, their multiplication is commutative, so

By the definition of matrix transposition we have that

Finally, by the definition of matrix multiplication we have that

We have just shown that $\left[ (AB)^T \right]_{ij} = \left[ B^T A^T \right]_{ij}$ for all rows $i$ and columns $j$. That must mean that it also holds for the entire matrix. That is,

Thus, the transpose of a product always reverses the order of the factors.

Multiplying by the Transpose

From what we proved above, another interesting fact about matrices can be deduced.

For any matrix $A_{m \times n}$, the products $A^T A$ and $A A^T$ are always symmetric matrices.

More formally, for any matrix $A$,

and

We can prove this using the reverse order law of transposition that we proved above. According to the reverse order law,

Then,

The same can be said for