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If the matrix of a linear transformation T$\colon \mathbb{R}^{N} \rightarrow \mathbb{R}^{N}$ with respect to some basis is symmetric, what does it say about the transformation? Is there a way to geometrically interpret the transformation in a nice/simple way?

I have tried to think about this and look it up on the web and in a few books, but I haven't found anything. It's just that the matrix being symmetric looks like such a nice property, there must be something special about the transformation too? Thanks for any help.

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If $\mathbb{R}^n$ is endowed with an inner product $\langle\,\cdot\, , \,\cdot\,\rangle$ and the matrix $A$ of $T$ is symmetric with respect to an orthonormal basis, then we have the important property that $$\langle y, T(x) \rangle = \langle y, Ax \rangle = y Ax = y^TA^Tx = (Ay)^T x = \langle Ay , x\rangle = \langle T(y), x \rangle;$$ in this case we say that $T$ itself is symmetric (w.r.t. the inner product). There's too much to say about why these are important in a single post, but let me point out two useful facts:

(1) By the Spectral Theorem, $T$ is orthogonally diagonalizable, that is, $T$ is conjugate by an orthogonal transformation to a diagonal transformation.

(2) Suppose $x, y$ are eigenvectors of $T$. If they correspond respectively to distinct eigenvalues $\lambda, \mu$, then we have

$$\lambda \langle x, y \rangle = \langle \lambda x, y \rangle = \langle T(x), y \rangle = \langle x, T(y) \rangle = \langle x, \mu y \rangle = \mu \langle x, y \rangle.$$ In particular, if $\lambda \neq \mu$ then $\langle x, y \rangle = 0$, that is, the eigenspaces of $T$ are all orthogonal!

Finally, let me note that there is a comparably important complex version of this, in which we ask for a complex transformation w.r.t. a Hermitian inner product to be Hermitian, i.e., to ask $A^* = A$, where ${}^*$ denotes the conjugate transpose. We call such transformations self-adjoint, and they are of fundamental importance to, for example, quantum mechanics.

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  • $\begingroup$ I had come across the term 'self-adjoint' while looking for an answer, but didn't quite understand it then. I guess I'll look it up again. Thanks for mentioning (2), hadn't thought of it in that direction. $\endgroup$
    – nemo
    Oct 22, 2014 at 13:06
  • $\begingroup$ Until you encounter complex vector spaces, you can rather safely just think of self-adjoint transformations as the appropriate complex versions of symmetric transformations. $\endgroup$ Oct 22, 2014 at 13:09

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