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I have a differential equation $X'=AX$ where $A\in\mathcal M_n(\Bbb R)$. The question is to prove that if all the solutions have a constant norm then $A$ is skew-symmetric matrix.

What I have tried so far: let $\varphi(t)=\Vert X(t)\Vert^2$ where $X$ is a solution. Then $$0=\varphi'(t)=2\langle X'(t),X(t)\rangle=2\langle AX(t),X(t)\rangle=2\langle X(t),A^TX(t)\rangle$$ How can I complete my answer? Any other suggestion?

As soon as possible
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3 Answers3

1

Now conclude from $x^TAx=0$ for all $x$ that $A_{ii}=0$ for all $i$ by setting $x=e_i$ and then $A_{ij}+A_{ji}=0$ by setting $x=e_i+e_j$.

Lutz Lehmann
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0

Think that

$$ \dot X = X \wedge \vec v = A X \mbox{ with } A \ \ \mbox{skew symmetric} $$

then

$$ < X, \dot X > = < X, X \wedge \vec v > = 0 \Rightarrow ||X|| = C_0 $$

NOTE

The external product can be generalized to $n$ dimensions.

Cesareo
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  • The question was about the other direction, $A$ is without assumption, the claim is the structure of $A$. And your external product does not make sense for $n>2$, as $a∧b$ has $\binom{n}2>n$ components $a_ib_j-a_jb_i$, $i<j$.. – Lutz Lehmann May 19 '18 at 22:43
  • In $R^3$ we have $(x_1,x_2,x_3) \wedge (v_1,v_2,v_3) = (v_3x_2-x_2 x_3,v_1 x_3-v_3x_1,v_2x_1-v_1x_2) = A x$ with $A$ skew symmetric. See https://math.stackexchange.com/questions/2371022/cross-product-in-higher-dimensions/2371039 – Cesareo May 19 '18 at 22:55
0

The desired result is a consequence of the following

Observation: A symmetric matrix

$B \in \mathcal M_n(\Bbb R) \tag 1$

such that

$\langle y, By \rangle = 0 \tag 2$

for all vectors $y$ vanishes, i.e.,

$B = 0.\tag 3$

Proof of Observation:

Suppose that

$B \in \mathcal M_n(\Bbb R) \tag{4}$

is symmetric,

$B^T = B; \tag{5}$

then there exists an orthogonal matrix

$O \in \mathcal M_n(\Bbb R), \tag{6}$

that is,

$O^TO = OO^T = I, \tag{7}$

which diagonalizes $B$:

$OBO^T = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag{8}$

where the $\mu_i$, $1 \le i \le n = \text{size}(A)$ are the (real) eigenvalues of $B$. Now consider the $n \times 1$ (column) vectors

$e_i = (\delta_{ij})_{j = 1}^n, \tag{9}$

that is, $i$-th row entry of $e_i$ is $1$ and all other entries are $0$; we easily see that (8) implies

$(OBO^T)e_i = \mu_i e_i, \tag{10}$

and thus,

$\langle O^Te_i, BO^Te_i \rangle = \langle e_i, OBO^Te_i \rangle = \langle e_i, \mu_i e_i \rangle = \mu_i \langle e_i, e_i \rangle = \mu_i; \tag{11}$

setting

$y_i = O^T e_i, \tag{12}$

we have

$\langle y_i, By_i \rangle = \mu_i; \tag{13}$

now if for all vectors $y$,

$\langle y, By \rangle = 0, \tag{14}$

then in accord with (13),

$\mu_i = 0, \; 1 \le i \le n, \tag{15}$

and thus, via (8)

$B = O^T \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) O = 0. \tag{16}$

Thus a symmetric matrix such that

$\langle y, By \rangle = 0 \tag{17}$

for all vectors $y$ must itself be the $0$ matrix. End: Proof of Observation.

We apply this Observation to the problem at hand as follows:

with

$\langle x, x \rangle = \text{constant}, \tag{18}$

we have

$\langle x, x \rangle' = 0, \tag{19}$

whence
$\langle \dot x, x \rangle + \langle x, \dot x \rangle = \langle x, x \rangle' = 0; \tag{20}$

given that

$\dot x = Ax, \tag{21}$

we write (20) in the form

$\langle Ax, x \rangle + \langle x, Ax \rangle = 0; \tag{22}$

whence

$\langle x, A^Tx \rangle + \langle x, Ax \rangle = 0, \tag{23}$

or

$\langle x, A^Tx + Ax \rangle = 0, \tag{24}$

or

$\langle x, (A^T + A)x \rangle = 0; \tag{25}$

now $A + A^T$ is symmetric:

$(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T, \tag{26}$

and since $x$ may be any solution to (21), it may be assumed that $x(t)$ may be any vector at time $t$, and hence (25) yields

$\langle x(t), (A^T(t) + A(t))x(t) \rangle = 0; \tag{27}$

we now invoke the above Observation and conclude that

$A^T(t) + A(t) = 0, \tag{28}$

that is,

$A^T(t) = -A(t) \tag{29}$

for all values of $t$.

Robert Lewis
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