derivative vanishes

Question

I'm seeing from the Professor's notes that the that the derivative of the Euclidean scalar product squared $||y(x)||^2 = \langle y(x), y(x)\rangle$ simply vanishes. That is $\frac{d}{dx} ||y(x)||^2=0$. I've got no clue why. Here $y(x)$ is the solution of the DGL $y(x)'=Ay(x)$. $A$ is a $n\times n$-matrix in this case and $y(x)$ is a function from $\mathbb{R}$ to $\mathbb{R}^n$. Anyone can help me?

Answer 1

$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}\newcommand{\Transp}[1]{#1^{\mathsf{T}}}$If $y' = Ay$, then \begin{align*} \frac{d}{dt} \Brak{y(t), y(t)} &= \Brak{y'(t), y(t)} + \Brak{y(t), y'(t)} \\ &= 2\Brak{y(t), y'(t)} \\ &= 2\Brak{y(t), Ay(t)} = 2\Brak{\Transp{A}y(t), y(t)}. \end{align*} If $\Transp{A} = -A$, the last expression is equal to $-2\Brak{y(t), Ay(t)}$, implying $$ \frac{d}{dt} \Brak{y(t), y(t)} = 0. $$

Conversely, if $y' = Ay$ and $$ 0 = \frac{d}{dt}\bigg|_{t = 0} \Brak{y(t), y(t)} = 2\Brak{y(0), Ay(0)} = 2\Brak{y(0), \Transp{A}y(0)} $$ for every initial condition $y(0) = y_{0}$, then $$ 0 = \Transp{y_{0}}\bigl[\tfrac{1}{2}(A + \Transp{A})\bigr] y_{0}\quad\text{for every $y_{0}$.} $$ By the spectral theorem, $\tfrac{1}{2}(A + \Transp{A})$ (the symmetric part of $A$) has real eigenvectors; letting $y_{0}$ run through an eigenbasis shows $\frac{1}{2}(A + \Transp{A}) = 0$, i.e., $A = -\Transp{A}$.