Norm-preserving implicit numerical method

Question

Consider the implicit single-step method $$x_1 = x_0 + h A(x_0 + x_1)$$ where matrix $A$ is skew-symmetric. Show that this method preserves the norm, i.e., $\|x_1\| = \|x_0\|$.

I don't really know how to go about it.

Answer 1

With

$y = x + hA(x +y), \tag 1$

we have:

$y = x + hAx + hAy, \tag 2$

$y - hAy = x + hAx, \tag 3$

$(I - hA)y = (I + hA)x; \tag 4$

now the matrices $(I \pm hA)$ are invertible, for with $z \ne 0$,

$\langle (I - hA)z, (I - hA)z \rangle = \langle z, (I - hA)^T(I - hA)z \rangle$ $= \langle z, (I - hA^T)(I - hA)z \rangle = \langle z, (I + hA)(I - hA)z \rangle$ $= \langle z, (I - h^2A^2)z \rangle = \langle z, (I + h^2A^TA)z \rangle = \langle z, z \rangle + h^2\langle z, A^TAz \rangle$ $= \langle z, z \rangle + h^2\langle Az, Az \rangle \ge \langle z, z \rangle > 0, \tag 5$

which implies

$\forall z \ne 0 \; (I - hA)z \ne 0 \Longrightarrow \exists (I - hA)^{-1}, \tag 6$

with a similar calculation for $I + hA$; thus it follows from (4) that

$y = (I - hA)^{-1}(I + hA)x. \tag 7$

I claim the matrix

$O = (I - hA)^{-1}(I + hA) \tag 8$

is in fact orthogonal; indeed,

$O^T = [(I - hA)^{-1}(I + hA)]^T = (I + hA)^T((I - hA)^{-1})^T$ $= (I + hA^T)(I - hA^T)^{-1} = (I - hA)(I + hA)^{-1}, \tag 9$

from which

$OO^T = (I - hA)^{-1}(I + hA)(I - hA)(I + hA)^{-1}$ $= (I - hA)^{-1}(I - hA)(I + hA)(I + hA)^{-1} = I. \tag{10}$

Since $O$ is orthogonal and

$y = Ox, \tag{11}$

we conclude that

$\Vert y \Vert = \Vert Ox \Vert = \Vert x \Vert; \tag{12}$

thus the iterative scheme

$x_{n + 1} = x_n + hA(x_n + x_{n + 1}) \tag{13}$

is inherently norm-preserving as well. $OE\Delta$.

Answer 2

Hint: Since $A$ is skew-symmetric, you can unitarily diagonalize $A$ over $\mathbb{C}$ and have purely imaginary diagonals. Hence you just need to show $\frac{1+hi\theta}{1-hi\theta}$ ($\theta\in\mathbb{R}$) has absolute value $1$, which is obvious.