# Matrix coefficients of vector dot product

while reading a proof for conservation of energy of a certain system involving skew symmetric matrices (AT = -A), I came across this mathematical peculiarity that I cannot explain. Within the proof is the following statement:

$$2A\vec{x}\cdot\vec{x} = (A + A^T)\vec{x}\cdot\vec{x}$$

I tried this out with a sample skew symmetric matrix A, and indeed it works. But by inspection 2A cannot always equal A + AT (unless A = 0)! So my question is, am I violating some kind of matrix or dot product rule by trying to equate 2A and A + AT, and if so, what rule?

Sorry if this question comes across as ill-posed, I do not have a particularly sophisticated math background and would benefit most from simple answers, however possible.

• If $A$ is skew-symmetric, shouldn't $A + A^T = 0$? Oct 2, 2020 at 3:29
• Yes it is, hence my confusion that 2A = A + A' (at first glance) Oct 2, 2020 at 15:10

The problem with equating $$2A$$ and $$A + A^T$$ is that you essentially try to take the inverse of a function which you have not checked to be invertible. In particular, let $$f_x(A) = \langle Ax, x \rangle$$ and note that your equation can be rewritten as $$f_x(2A, x) = f_x(A + A^T)$$. Now, to conclude equality in the first argument (in general), you (essentially) need this function to be invertible -- and you have already discovered that it is not.
Also, while we are at it, we may also prove the formula you cite. As the scalar product is linear and symmetric (at least on the real numbers), we find \begin{align} \langle (A + A^T)x, x \rangle &= \langle Ax, x \rangle + \langle A^T x, x \rangle \\ &= \langle Ax, x \rangle + \langle x, A^T x \rangle \\ &= (Ax)^T x + x^T (A^T x) \\ &= (Ax)^T x + (Ax)^T x \\ &= 2\langle Ax, x \rangle. \end{align}
• Wow this was a very clear and elegant explanation, thank you. Just to clarify one thing for myself, you are saying that to conclude equality between $$f(2A,x) = f(A+A^T,x)$$ I would essentially have to prove $$1 = f^{-1}(2A,x)f(A+A^T,x)$$ which I can't do because f(2A,x) is not necessarily invertible, correct? Oct 2, 2020 at 15:22
• To get from $f_x(2A) = f_x(A + A^T)$ to $2A = A + A^T$ we apply the inverse of $f_x$ to both sides. It seems you write $f_x^{-1}(A)$ to denote $1/f_x(A)$, which is not what we want to do here (and also that notation is pretty universely disliked). Oct 2, 2020 at 17:58