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Let $(a_{ij})$ be a skew-symmetric $3 \times 3$ matrix (i.e., $a_{ij}=-a_{ji}$ for all $i, j$).

Let $v_1$, $v_2$ and $v_3$ be smooth functions of a parameter $s$ satisfying the differential equations $$v'_i= \sum_{j=1}^3 a_{ij} v_j$$

for i = $1, 2$ and $3$, and suppose that for some parameter value $s_0$ the vectors $v_1(s_0)$, $v_2(s_0)$ and $v_3(s_0)$ are orthonormal. Show that the vectors $v_1(s), v_2(s)$ and $v_3(s)$ are orthonormal for all values of $s$.

I am facing difficulties. Any hints how to show that the vectors $v_1(s), v_2(s)$ and $v_3(s)$ are orthonormal for all values of $s$?

EDIT Nr.1:

$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) = \frac{d}{ds}(v_i(s)^t v_j(s) )=(v_i(s)^t)' v_j(s)+v_i(s)^t (v_j(s))' \\= \sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)+v_i(s)^t \sum_{k=1}^3 a_{jk}v_k(s) $

Any ideas how I can continue?

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EDIT Nr.2:

We have $$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) = \frac{d}{ds}(v_i(s)^t v_j(s) )=(v_i(s)^t)' v_j(s)+v_i(s)^t (v_j(s))' \\= \sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)+v_i(s)^t \sum_{k=1}^3 a_{jk}v_k(s) =\sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)-v_i(s)^t \sum_{k=1}^3 a_{kj}v_k(s)$$

It is $$v_k(s)^tv_j(s)=v_k(s)v_j(s)^t$$ right?

Therefore we have $$\frac{d}{ds}(\left< v_i(s), v_j(s) \right>) =\sum_{k=1}^3 a_{ik} v_k(s)^t v_j(s)-v_i(s)^t \sum_{k=1}^3 a_{kj}v_k(s)=\sum_{k=1}^3 [a_{ik} v_k(s) v_j(s)^t-a_{kj} v_i(s)^t v_k(s)]=\sum_{k=1}^3 v_k(s)[a_{ik} v_j(s)^t-a_{kj} v_i(s)^t]$$

What can we do next?

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    $\begingroup$ Can you write down the solution to the system of differential equations you are given in terms of the matrix involved? $\endgroup$ – uniquesolution Oct 13 '15 at 17:21
  • $\begingroup$ I added what I did so far. Can you look at it and tell if it is correct and what we can do next? @uniquesolution $\endgroup$ – user175343 Oct 13 '15 at 18:46
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    $\begingroup$ Your argument uses subscripts to denote both vector components and indices on your triple of vector-valued functions, so unfortunately much of what you write doesn't make sense. Instead, start by differentiating the inner product $f_{ij}(s) := \langle v_{i}(s), v_{j}(s)\rangle$ to get $f'_{ij}(s) = 2\langle v_{i}(s), v'_{j}(s)\rangle$, then write the latter (perhaps using superscripts as indices) as$$2\sum_{k,\ell=1}^{3} a_{k\ell} v_{i}^{k}(s) v_{j}^{\ell}(s)$$and use skew-symmetry to deduce $f_{ij}$ is constant. $\endgroup$ – Andrew D. Hwang Oct 14 '15 at 11:49
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    $\begingroup$ That's the product rule for dot products, except in haste I gave the rule for the special case $i = j$; the correct general formula is$$\frac{d}{ds}\langle v_{i}(s), v_{j}(s)\rangle = \langle v_{i}'(s), v_{j}(s)\rangle + \langle v_{i}(s), v_{j}'(s)\rangle.$$ $\endgroup$ – Andrew D. Hwang Oct 15 '15 at 17:47
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    $\begingroup$ You're still using subscripts as vector components. :) Omitting dependence on $s$, the derivative comes out to$$v_{i}^{t}A^{t}v_{j} + v_{i}^{t}Av_{j} = \sum_{k,\ell} v_{i}^{k}(A_{\ell k} + A_{k\ell})v_{j}^{\ell}.$$ $\endgroup$ – Andrew D. Hwang Oct 15 '15 at 21:47
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Let us assume that $v_i \colon \mathbb{R} \rightarrow \mathbb{R}^n$. Then, define $v \colon \mathbb{R} \rightarrow M_{3 \times n}(\mathbb{R})$ by $v(s) = (v_1(s), v_2(s), v_3(s))^t$. The system of equations for $v$ is equivalent to the linear system $v'(s) = Av(s)$ with constant coefficients. Define also $g \colon \mathbb{R} \rightarrow M_3(\mathbb{R})$ by $g(s) = v(s) v(s)^t$. The matrix $g(s)$ is then the Gram matrix for the vectors $v_1(s), v_2(s), v_3(s)$. You want to show that if $g(s_0) = I$, then $g(s) \equiv I$.

If you are familiar with matrix exponential, you can write the solution to $v'(s) = Av(s)$ as $v(s) = e^{A(s-s_0)} v(s_0)$ and then

$$ g(s) = v(s) v(s)^t = e^{A(s - s_0)} v(s_0) v(s_0)^t (e^{A(s - s_0)})^t = e^{A(s - s_0)} g(s_0) e^{A^t(s - s_0)} $$.

If $g(s_0) = I$ and $A$ is anti-symmetric, we see that $g(s) = e^{A(s - s_0)}e^{-A(s - s_0)} = I$ for all $s \in \mathbb{R}$.

If you don't want to use matrix exponentials, you can calculate

$$ g'(s) = v'(s) v(s)^t + v(s) (v'(s))^t = Av(s)v(s)^t + v(s)v(s)^tA^t = Ag(s) - g(s)A = [A, g(s)]. $$

You can check that $g(s) \equiv I$ satisfies this equation and so, if you know that $g(s_0) = I$ for some $s_0$ then by uniqueness of solutions you must have $g(s) \equiv I$.

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  • $\begingroup$ I added what I did so far. Can you look at it and tell if it is correct and what we can do next? $\endgroup$ – user175343 Oct 13 '15 at 18:46
  • $\begingroup$ I added more details. $\endgroup$ – levap Oct 14 '15 at 11:19

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