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Suppose you have a 2x2 real-valued matrix, $\mathbf{A}$. If you perform a singular value decomposition (SVD), then this can be understood geometrically as a decomposition of $\mathbf{A}$ into a 2-D rotation, scaling and second 2-D rotation of the form:

$$\mathbf{A} = \mathbf{R_1 S R_2}$$

However, if the entries of $\mathbf{A}$ are complex, then what is the geometric meaning of this? Now $\mathbf{R_1}$ and $\mathbf{R_2}$ will also be complex. Are these still rotation matrices? Is there an additional underlying rotation happening in the complex plane? What does it mean if you have a "complex rotation matrix"?

Any help, clarification or references is appreciated.

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  • $\begingroup$ They'll be unitary matrices, which can still be thought of as rotations in a suitable sense. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 15, 2018 at 21:27
  • $\begingroup$ Rotation through what angle though? What does it mean to have a "complex rotation matrix"? The matrix multiplication does not preserve real and imaginary components as separate. For example, if $\mathbf{R_1}$ is a complex rotation matrix and you take some real valued vector, $\mathbf{v}$, then $\mathbf{R_1v}$ will be complex valued as well. I'm just struggling to understand what this actually means geometrically. $\endgroup$
    – Darcy
    Commented Nov 15, 2018 at 21:59
  • $\begingroup$ The eigenvalues of a unitary matrix, like an orthogonal matrix, are unit complex numbers $e^{i \theta}$, so can be thought of in terms of rotations by $\theta$. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 15, 2018 at 22:02
  • $\begingroup$ Dear Darcy, any vector in an $n$-dimensional complex vector space may be regarded as a vector in a $2n$-dimensional real vector space (every complex number gives you two real numbers, namely the real and imaginary part). Please convince yourself that a unitary transformation in an $n$-dimensional complex space is also an ordinary rotation in this $2n$-dimensional space. Be warned though: the converse is not true. Rotation in $\mathbb R^{2n}$ is unitary as an operator on $\mathbb C^{n}$ if and only if it is complex linear. $\endgroup$
    – Blazej
    Commented Jun 27, 2019 at 7:14

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