Mathematics Asked by As soon as possible on January 7, 2022

I have a differential equation $X’=AX$ where $Ainmathcal M_n(Bbb R)$. 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

$$0=varphi'(t)=2langle X'(t),X(t)rangle=2langle AX(t),X(t)rangle=2langle X(t),A^TX(t)rangle$$

How can I complete my answer? Any other suggestion?

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 $t$.

Answered by Robert Lewis on January 7, 2022

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.

Answered by Cesareo on January 7, 2022

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 $x=e_i+e_j$.

Answered by Lutz Lehmann on January 7, 2022

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