# Are there more matrix transpose properties than this one?

In a recent exercise, our professor wrote that

$$t^TXw = w^T X^T t$$

where $$t,w$$ are vectors and $$X$$ is a matrix.

I tried it for a simple example and the identity seems to hold true. However, I can't find any reference to this online, so it made me curious why this works.

Does it work if $$t$$ and/or $$w$$ are matrices? Is the identity true only because there are 3 factors? Et cetera.

• It might get a bit clunky, but if you write out the indices, you can show the equality holds for each entry $(i, j)$. Commented Feb 8, 2019 at 16:57
• The key to this particular identity is that the value of the product on either side of the identity is a scalar, which is trivially its own transpose.
– amd
Commented Feb 9, 2019 at 1:11

In general, for any two matrices that can be multiplied, we have this little theorem: $$(AB)^T = B^T A^T$$ This has been applied twice to prove the thing your professor wrote. Note that $$t^T X w$$ is a real number, so its transpose is equal to itself.
\begin{align} t^T X w&= (t^T X w)^T\\ &= (t^T (Xw))^T \\ &= (Xw)^T (t^T)^T & \text{by the little theorem, with A = t^T, B = Xw} \\ &= w^T X^T (t^T)^T & \text{by the little theorem, with A = w, B = X} \\ &= w^T X^T t & \text{because transposing twice gives back the original matrix} \\ \end{align}
The $$ij$$ entry of $$AB$$ is $$\sum_k a_{ik} b_{kj}$$; the $$ij$$ entry of $$(AB)^T$$ is therefore $$\sum_k a_{jk} b_{ki}$$
The $$ik$$ entry of $$B^T$$ is $$b_{ki}$$; the $$kj$$ entry of $$A^T$$ is $$a_{jk}$$. So the $$ij$$ entry of $$B^T A^T$$ is $$\sum_k b_{ki} a_{jk},$$ and this is evidently the same thing we computed above. QED.