Geometry of Skew-Hermitian matrices

Question

In the question I will be referring to finite dimensional complex vector spaces. I know that the eigenvalues of an Skew-Hermitian matrix $S$ are purely imaginary, but what is their geometric effect on the vector space? I know that Hermitian matrices H have an orthonormal basis of eigenvectors which span the whole vector space and their eigenvalues are purely real. Their effect is as far as I understand that Hermitian matrices $H$ just scale the vector space in the direction of these orthogonal eigenvectors by the eigenvalues and have no rotational effect. Is the effect of Skew-Hermitian matrices then just purely rotational?

Answer 1

That's not quite correct.

It's helpful to first establish that because we are in a complex vector space, the one-dimensional subspaces are two dimensional spaces relative to multiplication by real numbers, and so is perhaps best visualized as a two-dimensional space. Within the span of any single non-zero vector, multiplication by $e^{i \theta}$ corresponds to a rotation (within this "one-dimensional" span) by an angle of $\theta$.

Skew-Hermitian matrices have purely imaginary eigenvalues, and just like Hermitian matrices, they can be unitarily diagonalized. So, for any skew-Hermitian matrix $A$, there is an orthonormal basis $v_1,\dots,v_n$ such that $A v_j = \pm i\lambda_j v_j$ for some $\lambda_j \geq 0$. So, we have broken up $\Bbb C^n$ as the direct sum of $n$ mutually orthogonal "1-dimensional" vector spaces, each the span of an eigenvector $v_j$. Within the span of $v_j$, the action of $A$ can be thought of as a multiplication by $\pm i$ (a rotation by $90^\circ$) followed by scaling by factor $\lambda_j$.

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Another helpful perspective comes from the study of Lie Groups. The unitary matrices may be thought of as corresponding to rotations within $\Bbb C^n$. Unitary matrices are also unitarily diagonalizable; for any unitary matrix $M$, there exists an orthonormal basis $v_1,\dots,v_n$ such that $Mv_j = e^{i \theta_j}v_j$ for some $\theta_j \in \Bbb R$. In other words, within each of $n$ mutually orthogonal "1-dimensional" vector spaces, the action of $M$ is a rotation by angle $\theta$.

The skew-Hermitian matrices, in a sense, correspond to "infinitesimal rotations". If $A$ is skew-Hermitian, then $(I + \frac An)^n$ (for positive integer $n$) is approximately a rotation. In the limit as $n \to \infty$, we find that $$ \lim_{n \to \infty}\left(I + \frac An\right)^n = e^A $$ is a unitary matrix. In fact, for every $M$, there exists a Hermitian matrix $A$ that "generates" $M$ in this sense.

Another helpful perspective: we can extend our description of $M$ to describe an entire family of rotations (called a "one parameter subgroup"). For any $t \in \Bbb R$, we'll say that $M(t)$ is the matrix for which $M(t) v_j = e^{it\theta_j} v_j$. Notably, $M(0) = I$ and $M(1) = M$ (the first unitary matrix we described). The "derivative" of this function of $t$ at $t = 0$ yields another transformation, and this transformation corresponds to a skew-Hermitian matrix. In particular, for each $j$, $$ \left.\frac d{dt} e^{it \theta_j}v_j\right|_{t = 0} = i\theta_j v_j. $$ In this way, the Hermitian matrices correspond to the "tangent space" of the set of unitary matrices at the point $I$.