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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.

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  • $\begingroup$ It might get a bit clunky, but if you write out the indices, you can show the equality holds for each entry $(i, j)$. $\endgroup$ Commented Feb 8, 2019 at 16:57
  • $\begingroup$ 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. $\endgroup$
    – amd
    Commented Feb 9, 2019 at 1:11

1 Answer 1

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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}

Proof of the little theorem:

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.

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  • $\begingroup$ Thank you. I didn't understand until you made that edit. I'll accept the answer in 4 minutes. $\endgroup$
    – Mossmyr
    Commented Feb 8, 2019 at 17:02

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