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# Definition

MathJax TeX Test Page A is symmetric iff there exists an orthogonal matrix Q such that $Q^TAQ = Q^{-1}AQ$ is diagonal.

An equivalent way of writing it, is $[A]_{\mathcal{E}}$ is symmetric iff there exists an orthogonal basis for $\mathbb{R}^n$ consisting of eigenvectors of A.

The reason they are equivalent can be recalled from the change-of-basis formula. $$[A]_\mathcal{E} = \underset{E \leftarrow B}{P}[A]_\mathcal{B}\underset{E \leftarrow B}{P}^{-1}$$ If we rewrite the first definition in a slightly different way, we get $$A = QDQ^{-1} = QDQ^T$$ In order for a change-of-basis transformation to be orthonormal, there must be an orthonormal basis! Thus, they are equivalent.

# The Following Are Equivalent (TFAE)

$$\text{1. T is symmetric}$$ $$\text{2. There exists an orthonormal basis such that } [T]_\mathcal{B} \text{ is symmetric}$$ $$\text{3. For every orthonormal basis C in } \mathbb{R}^n, [T]_\mathcal{C} \text{ is symmetric}$$ We can change the order of our orthonormal basis, it doesn't matter. It will still be symmetric. $$\text{4. } T(x) \cdot y = x \cdot T(y)$$ This one is easy to prove. $Ax \cdot y = (xA)^Ty = x^T A^Ty = x \cdot A^Ty = x \cdot Ay$. This is only true if $A^T = A$

# Lagrange’s Revenge

If A is symmetric, then A has a real eigenvalue.

In case it wasn't clear, not all matrices have real eigenvalues. $$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} \text{ has eigenvalues of } \pm i$$ This is mainly important in the proof for the Spectral Theorem, because we prove the Spectral Theorem with induction, and we use Lagrange's Revenge to choose an arbitrary eigenvector, which we wouldn't be able to do without justification.