Chelsea Lamb

2022-09-26

If the matrix of a linear transformation $:{\mathbb{R}}^{N}\to {\mathbb{R}}^{N}$ 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?

asijikisi67

If ${\mathbb{R}}^{n}$ is endowed with an inner product $⟨\phantom{\rule{thinmathspace}{0ex}}\cdot \phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}\cdot \phantom{\rule{thinmathspace}{0ex}}⟩$ and the matrix $A$ of $T$ is symmetric with respect to an orthonormal basis, then we have the important property that
$⟨y,T\left(x\right)⟩=⟨y,Ax⟩=yAx={y}^{T}{A}^{T}x=\left(Ay{\right)}^{T}x=⟨Ay,x⟩=⟨T\left(y\right),x⟩;$
in this case we say that $T$ itself is symmetric. 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
$\lambda ⟨x,y⟩=⟨\lambda x,y⟩=⟨T\left(x\right),y⟩=⟨x,T\left(y\right)⟩=⟨x,\mu y⟩=\mu ⟨x,y⟩.$
In particular, if $\lambda \ne \mu$ then $⟨x,y⟩=0$, that is, the eigenspaces of $T$ are all orthogonal.

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