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A projection of real vectors is given by $proj_x y = (y^T \cdot x) \frac{x}{x^T x}$

I am not familiar with complex-valued spaces, but could I generalize the projection of complex valued vectors as

$proj_x y = (y^H \cdot x) \frac{x}{x^H x}$, where H is the Hermition operation?

Thanks

Ted Shifrin
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user3284182
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  • Yes, that's exactly how it's done. The $y^H \cdot x$ is an inner product, and this formula works for any inner product space. – Zoe Allen Mar 11 '23 at 04:27
  • @ZoeAllen Thanks for your answer. The inner product order matters in this case, right? – user3284182 Mar 12 '23 at 04:37
  • Oops you're right order does matter, and the order is the wrong way round here. The Hermitian conjugate should be applied to the $x$. – Zoe Allen Mar 12 '23 at 04:44

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Just a slightly longer comment, In order to understand why the order matters, we need to recall that a projection, can be characterized by a projection matrix, i.e. a matrix $P$ such that $P=P^H$ and $P^2=P$. in your case, the projection matrix is $P=\frac{1}{x^H x} x x^H$. It is a projection matrix, and also that its span is the span of $\{ x \}$. Now multiplying $P\cdot y=\frac{x^H y}{x^Hx}x$.

By the way, I have a tendency to consider inner products in the nominator of expression as $1\times 1$ matrices, this typically gives me more intuition about the expression I am working with. If you do this then $y^T x$ cannot be multiplied on the left by $x$, only on the right, therefore the correct expression is $x y^T x$ which indeed feels wrong for a projection.

P. Quinton
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