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I frequently see people explaining Hermitian matrices as a generalization of real symmetric matrices e.g. Wikipedia, Math StackExchange. I understand that all real symmetric matrices are Hermitian matrices, but it seems like there's really two changes between real symmetric matrices and Hermitian matrices: (1) permit complex numbers, and (2) the transpose must equal the element-wise complex conjugate. Why was this second step included in the definition of Hermitian matrices? My guess would be that it has something to do with physicists wanting to ensure particular sequences of linear transformations produce real-valued outputs; is this correct?

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    $\begingroup$ Welcome to Mathematics Stack Exchange. I think it's not just physics; e.g., taking the complex conjugate along with transposing a vector to get the dot product ensures that the length of a vector is a non-negative real number $\endgroup$ – J. W. Tanner Sep 5 '19 at 2:01
  • $\begingroup$ You should look up what it means for a linear operator to be self-adjoint with respect to a bilinear form on a vector space. If you take $\mathbb{R}^n$ with the dot product, you find that a matrix is self-adjoint when it is equal to its transpose. Do the same thing with $\mathbb{C}^n$ with the standard inner product, and a self-adjoint matrix is one which is equal to its transpose conjugate. The fact that the notion of “self-adjoint” is defined not only for matrices, but for linear operators (ie does not depend on a choice of basis) may be surprising. $\endgroup$ – Joppy Sep 5 '19 at 2:19
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The "dot product" in $\Bbb R^n$ has the property that $$ d(x) = \sqrt{x \cdot x} = \sqrt{x^t x} $$ is a metric. In particular, $x \cdot x = \sum_i x_i^2$ is always a nonnegative real, so we can take a square root. You can fancy this up and look at $$ \sqrt{x^t M x} $$ for some matrix $M$, or (generalizing a little) look at a product defined by $$ (x, y) \mapsto x^t M y. $$

If you want that to be symmetric (which is a nice thing for generalized inner products), then $M$ has to be symmetric.

If you try to do the same thing for a complex vector $z$, with complex number entries $z_i$, and define $$ z \cdot z = \sum_j z_j^2, $$ then the resulting sum is usually not real. (Example: if each $z_j$ is $\sqrt{-1}$, then...)

But if you say that $$ z \cdot z = \sum_j z_j \overline{z}_j, $$ then you DO get something that's a nonnegative real, and can mimic all the stuff you did for $\Bbb R^n$. But when it comes to the matrix $M$, you need (to get symmetric of your generalized inner product) not that $M^t = M$, but that $\overline{M}^t = M$...so that's where the generalization comes from.

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    $\begingroup$ So the goal is to preserve the concept of a distance that exists in real spaces? $\endgroup$ – Rylan Schaeffer Sep 5 '19 at 2:11
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    $\begingroup$ The dot product (or inner product) is what gives us a concept of length and distance in $\mathbb{R}^n$ since the length of a vector is $|v| = \sqrt{v\cdot v}$ and the distance between $v, w$ is $|v-w|$. Therefore, we can translate much of our geometric intuition in $\mathbb{R}^n$ over to these inner product spaces (and so naturally they are very useful in physics). Hermitian matrices, and in general self-adjoint maps are those maps from the space to itself that satisfy a nice symmetry condition relative to the dot product as seen in my answer. This symmetry gives them their importance. $\endgroup$ – Andrew Sep 5 '19 at 2:35
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Yes; in physics, observables such as the position or momentum of a particle are modeled with linear operators on a Hilbert space (a vector space over $\mathbb{R}$ or $\mathbb{C}$ with a dot product and other nice properties). In order for this interpretation to make sense, the eigenvalues (which correspond to probabilities or measurements) must be real-valued.

Another way to see why this might be the correct definition is from looking at the dot-product between vectors $v$ and $w$. In $\mathbb{R}^n$, this is given by $$v\cdot w = \sum_{i=1}^nv_iw_i $$ but in the complex vector space $\mathbb{C}^n$ it is given by $$v\cdot w = \sum_{i=1}^nv_i\overline{w_i}; $$ $A$ being Hermitian is equivalent to the property that $$Av\cdot w = v\cdot Aw$$ for any vectors $v, w$. The $A = \overline{A^T}$ definition guarantees this to be true, and the presence of the complex conjugate therefore comes from the dot product. For general linear operators (not just matrices), we say the function is self-adjoint or Hermitian (https://en.wikipedia.org/wiki/Self-adjoint_operator).

The reason we think of them as like the real numbers is we can think of the operation $A \mapsto \overline{A^T}$ as a type of inversion, just like complex conjugation is an inversion on the complex plane. The real numbers are exactly those that satisfy $\overline{x} = x$, i.e. they are equal to their image under the inversion. Hermitian matrices are exactly those that satisfy the same property for the adjoint inversion operation.

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  • $\begingroup$ Thanks for the answer Andrew! I'm going to give the first responder credit for answering the question, but your answer was also a nice perspective! $\endgroup$ – Rylan Schaeffer Sep 5 '19 at 14:48

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