M is real anti-symmetric matrix, prove that exp(M) is isometry

Question

M is nxn real anti-symmetric matrix.I need to prove that exp(M) is isometry.

Could anyone give me any hint , I don't have any approach to this question. thank you

Answer 1

One-line sketch: $$(e^M u)^t(e^M v)=u^t[(e^M)^te^M]v = u^t[e^{M^t+M}]v=u^tv$$ Where I've used the property that if $M$ is anti-symmetric then $M^t+M=0$.

Answer 2

Presumably we're dealing here with matrices operating on a vector space $V = \Bbb R^N$ or $V = \Bbb C^N$, where $N$ is the size of $M$. Then, whether the inner product $\langle \cdot, \cdot \rangle$ is Euclidean or Hermitian respectively, we have that, for $x, y \in V$,

$\langle e^M x, e^M y \rangle = \langle (e^M)^T e^M x, y \rangle, \tag{1}$

since $M$ is real. Furthermore,

$(e^M)^T = (\sum_0^\infty \dfrac{M^j}{j!})^T = \sum_0^\infty \dfrac{(M^T)^j}{j!} = e^{M^T} \tag {2}$

by virtue of the convergence of the series in (2) and the fact that $(M^j)^T = (M^T)^j$ for all integers $j \ge 0$ implies $(p(M))^T = p(M^T)$ for polynomials $p(x)$. Thus we may write

$\langle (e^M)^T e^M x, y \rangle = \langle e^{M^T} e^M x, y \rangle, \tag{3}$

and since $M$ is skew-symmetric, that is $M^T = -M$, (3) becomes

$\langle (e^M)^T e^M x, y \rangle = \langle e^{-M} e^M x, y \rangle. \tag{4}$

Since $M$ commutes with $-M$, $[M, -M] = 0$, the identity

$e^{-M} e^M = e^{-M + M} = e^0 = I \tag{5}$

binds. Here we use the fact that commuting matrices $A, B$, obeying as the do $[A, B] = 0$, aslo satisfy $e^A e^B = e^{A + B}$; a detailed discussion of this is avialable in my answer to this question. In any event, combining (1), (4) and (5) we find

$\langle e^M x, e^M y \rangle = \langle Ix, y \rangle = \langle x, y \rangle, \tag{6}$

which shows that $e^M$ preserves inner products. Since $\Vert x \Vert = \langle x, x \rangle^{\frac{1}{2}}$, the desired result follows immediately. QED.

Moral of the Story: When $M = -M$, $e^M$ is orthogonal.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!