# If the solutions of $X'=AX$ have a constant norm then $A$ is skew symmetric.

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?

## 3 Answers

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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