The easy spectral theorem

This is the one we use normally: If $A$ is a symmetric matrix, then there exists an orthonormal eigenbasis for the space of $A$.

More Nuanced Spectral Theorems

Complex Spectral Theorem (1)

A linear operator $T$ on a complex, finite-dimensional vector space has a real, orthonormal eigehbasis iff $T$ is self-adjoint. (i.e essentially symmetric)

Real spectral theorem (2)

A linear operator $T$ on a real finite dimensional vector space has an orthonormal basis of eigenvectors iff $T$ is self-adjoint.

Normal spectral theorem (3)

A linear operator on a complex finite-dimenisional vector space has an orthonormal basis of eigenvectors iff $T$ is normal ($TT^* = T^*T$) [in the real domain, this is automatically true because $T^TT$ is symmetric.

Spectral decomposition, left and right

let $X$ be the matrix of right eigenvalues, and $Y$ be the matrix of left eigenvalues (both are column matrices). the following is true, assuming that $A$ has $n$ distinct eigenvalues (A is NOT NECESSAIRLY symmetric)

1. $AX = XD$ (previous definition)

2. $Y^TA = DY^T$

3. $Y^TX = I_n$

4. $A = XDX^{-1} = XDY^T = \sum \lambda_ix_iy_i^T$

The last point is very interesting. The spectral decomposition of a matrix can be thought of as a weighted sum of the inner products of the left and right inner products

Spectral theorem, non-abstractly

$A \in R^{n\times n}$ is symmetric iff $A$ is orthogonally diagonalizable. What does this mean/ WEll, it means that there exists some orthogonal matrix $Q$ and diagonal matrix $D$ with REAL entries where $A = QDQ^T$.

• here, we note that because $Q$ is orthogonal, $Q^T = Q^{-1}$

• Conversely, if $A = QDQ^T$ then we know that $A$ is symmetric (proof by taking $A^T$ and simplifying). Therefore, these are equivalent statements

The columns of $u_i$ of $Q$ form an orthogonal eigenbasis

This actually yields a profound result: with symmetric matrices, $Q$ contains the right eigenvectors as the columns and $Q^T$ contains the left eigenvalues as the colums!