Derivative of matrix product: is it true that $\frac{d}{dt}(A^TA) = 2A^T \frac{dA}{dt}$?

Question

$A$ is a square matrix. All elements of $A$ depend on a parameter $t$, that is, $a_{ij}=a_{ij}(t)$. Let $S(A):=A^TA$, and take the derivative of $S$ w.r.t. $t$: $\displaystyle \frac{dS}{dt}$

Now, pretty clearly

$\displaystyle \frac{dS}{dt} = \frac{dA^T}{dt}A + A^T\frac{dA}{dt}$

But now, can this also be written $\displaystyle \frac{dS}{dt} = 2 A^T \frac{dA}{dt}$ ?

A math text I am working though right now -- if I am reading it right -- implies that this is the case, but I haven't been able to prove it myself.

Thanks.

Answer 1

Even when one assume $A(t)$ is symmetric, the alternate form is still incorrect. For a counter-example, consider

$$A(t) = \begin{bmatrix}0 & 1 & 0\\1 & 0 & t\\0 & t & 0\end{bmatrix}$$

We have

$$\frac{d}{dt} S(t) = \begin{bmatrix}0 & 0 & 1\\0 & 2t & 0\\1 & 0 & 2t\end{bmatrix} \ne \begin{bmatrix}0 & 0 & 2\\0 & 2t & 0\\0 & 0 & 2t\end{bmatrix} = 2A(t)\frac{d}{dt}A(t)$$

Answer 2

Short answer : no.

Think about the $ij$ entry of $A^T A$; it's

$$ s_{ij} = \sum_k a_{ki} a_{kj} $$ Take the derivative with respect to $t$ (using primes to denote that) to get $$ s'_{ij} = \sum_k a_{ki}' a_{kj} + \sum_k a_{ki} a_{kj}' $$ The claim is that this is just $2 \sum_k a_{ki}' a_{kj}$, after some index-shuffling, and that's true if the matrix is symmetric, and not necessarily true otherwise.

Details: let's write that out in the case of a $3 \times 3$ matrix, with $i = 1$ and $j = 2$. We have

\begin{align} s'_{1,2} &= \sum_k a_{k1}' a_{k2} + \sum_k a_{k1} a_{k2}'\\ &= (a_{1,1}' a_{1,2} +a_{2,1}' a_{2,2} +a_{3,1}' a_{3,2}) + (a_{1,1} a_{1,2}' +a_{2,1} a_{2,2}' +a_{3,1} a_{3,2}') \\ \end{align} Those two don't look equal, do they? Let's try a concrete example.

\begin{align} A &= \begin{bmatrix} 1 & t \\ 0 & 2\end{bmatrix} \\ A^t A &= \begin{bmatrix} 1 & 0 \\ t & 2\end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 2\end{bmatrix} = \begin{bmatrix} 1 & t \\ t & t^2 +4\end{bmatrix} \\ (A^t A)' &= \begin{bmatrix} 0 & 1 \\ 1 & 2t\end{bmatrix} \\ A'^t A &= \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 2\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & t\end{bmatrix} \\ A^t A' &= \begin{bmatrix} 1 & 0 \\ t & 2\end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & t\end{bmatrix} \\ A'^t A + A^t A' &= \begin{bmatrix} 0 & 1 \\ 1 & 2t\end{bmatrix} \end{align}

Clearly the last expression is different from $2A^t A'$. So the formula is not correct.

(My apologies for the glib response earlier; I hope my working out the details makes up for it.)

On the other hand, if the matrix is symmetric, I'm pretty sure everything works out OK. :)

{Actually not -- see comments. The example of $s_{12}'$ makes that pretty clear: $a_{3,1}'$ appears in the left group of three, but neither it nor $a_{1,3}'$ appears in the right group of three.}