Spectral Theorem

A symmetric matrix $A \in \R^{n \times n}$ has orthogonal eigenvectors and real eigenvalues. Moreover, it can be orthogonally diagonalized into an orthogonal matrix $Q$, containing its eigenvectors, and a diagonal matrix $\Lambda$, containing its eigenvalues.

$$ A = Q\Lambda Q^T $$


Sign Definite Matrices

Symmetric matrices $A \in \R^{n \times n}$ are characterized by the sign of their eigenvalues which, equivalently, corresponds to the sign of the product $x^T A x$.

  1. Positive Definite $(\succ 0)$. $\sigma(A) > 0 \equiv x^T Ax \geq 0 \text{ with } x^TAx =0 \iff x=0$
  2. Positive Semidefinite $(\succeq 0 )$. $\sigma(A) \geq 0 \equiv x^TAx \geq 0$
  3. Negative Definite $(\prec 0)$. $\sigma (A) < 0 \equiv x^T Ax \leq 0 \text{ with } x^TAx =0 \iff x=0$
  4. Negative Semidefinite $(\preceq 0)$. $\sigma(A) \leq 0 \equiv x^TAx \leq 0$

Multivariate Taylor Expansion

A smooth function $f : \R^n \rightarrow \R^m$ can be approximated to an arbitrary degree via its Taylor expansion centered at a point $x_0 \in \R^n$. In this course, we typically look at first and second order approximations, which only require $1$ and $2$ times continuous differentiability.

$$ f(x) \approx f(x_0) + \nabla f(x_0)^T(x-x_0) + O(d^2) \\

f(x) \approx f(x_0) +\nabla f(x_0)^T(x-x_0) + \frac{1}{2}(x-x_0)^T\nabla^2 f(x_0)(x-x_0)+O(d^3) $$

<aside> <img src="/icons/castle_yellow.svg" alt="/icons/castle_yellow.svg" width="40px" />

Mid-Value Theorem. States that, for any $x, x_0 \in \R^n$ there exist points $u, v \in \R^n$ on the straight line segment from $x$ to $x_0$ which make the first and second order Taylor approximations exact.

image.png

$$ f(x) = f(x_0) + \nabla f(u)^T(x-x_0) $$

$$ f(x) = f(x_0 ) + \nabla f(x_0)^T(x-x_0) + \frac{1}{2}(x-x_0)^T \nabla^2f(v)(x-x_0) $$

The point $u$ can be interpreted as the point for which the total change in $f$ between $x$ and $x_0$ is exactly a linear term in $u$. Analogously, the point $v$ can be interpreted as the point for which the total change in $f$ between $x$ and $x_0$ can be written as a linear approximation and a quadratic term in $v$.

</aside>


Hessians