Positive-Definite Matrices

Energy-Based Definition

In Linear Algebra, a matrix an $n \times n$ matrix is Positive-definite matrix (PDM) if

  • $\mathbf v^T A \mathbf v > 0$ for all $\mathbf v \in \mathbb R^n$
  • This is the energy based definition


Why energy?

  • because $\mathbf v^T A \mathbf v$ or $\frac{1}{2} \mathbf v^T A \mathbf v$ is called the energy of the system $A$


Positive Semi-Definite Matrices

  • A matrix is semi-positive definite if
  • $\mathbf v^T A \mathbf v \geqslant 0$ for all $\mathbf v \ne \mathbf 0 \in \mathbb R^n$
  • so some eigenvectors can be 0


Motivating Example

  • Let [math]A = \begin{bmatrix} 2 & 6 \\ 6 & 18 \\ \end{bmatrix}[/math]
  • then for any $\mathbf x = (x_1, x_2)$ we want to check
  • [math]\big[x_1 \ x_2 \big] \begin{bmatrix} 2 & 6 \\ 6 & 18 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = 2 \, x_1^2 + 12 \, x_1 x_2 + 18 \, x_2^2[/math]
  • note that this is not a linear anymore:
  • we have an equation $a x_1^2 + 2b \, x_1 x_2 + c \, x_2^2$
  • this is a Quadratic Form
  • we want to know if this quantity is always positive or not
  • are there such $x_1, x_2$ that $a x_1^2 + 2b \, x_1 x_2 + c \, x_2^2 < 0$?

So we have a function $f(\mathbf x) = \mathbf x^T A \, \mathbf x$ and we want to check if it's always positive for any $\mathbf x$


Another example

  • let [math]A_1 = \begin{bmatrix} 2 & 6 \\ 6 & 7 \\ \end{bmatrix}[/math]
  • then $f(\mathbf x) = \mathbf x^T A_1 \, \mathbf x = 2 x_1^2 + 12 x_1 x_2 + 7 x_2^2$
  • there exists $\mathbf x$ such that $f(\mathbf x) < 0$, e.g. $(1, -1)$
  • in this system, there's a Saddle Point - a max for one direction and min for another
  • 8065cdad5e2c4642bc8a9b74feb907d9.png


Consider an alternative:

  • [math]A_2 = \begin{bmatrix} 2 & 6 \\ 6 & 20 \\ \end{bmatrix}[/math]
  • $f(\mathbf x) = \mathbf x^T A_2 \, \mathbf x = 2 x_1^2 + 12 x_1 x_2 + 20 x_2^2$
  • here squares always overwhelm $12 x_1 x_2$
  • 07acbca564ff4962993cf450951146bf.png


We say that $A_1$ is indefinite, and $A_2$ is positive-definite


Source: [1]


Finding Minima

Recall from Calculus:

  • 1st Derivative is needed for finding extremum, but you don't know if it's min or max
  • so you have to look for the 2nd derivative to learn if it's positive or negative
  • you want to find $\cfrac{du}{dx} = 0$ and $\cfrac{d^2 \, u}{d \, x^2} > 0$

Consider $A_2$ again:

  • $f(\mathbf x) = \mathbf x^T A_1 \, \mathbf x = 2 x_1^2 + 12 x_1 x_2 + 20 x_2^2$
  • Let's complete the square: $2 \, (x_1 + 3 \, x_2)^2 + 2 \, x_2^2$
  • now it's easy to see that this function is indeed always positive: we completed the square and there are no negative terms

What about $A_1$?

  • $f(\mathbf x) = \mathbf x^T A_1 \, \mathbf x = 2 x_1^2 + 12 x_1 x_2 + 7 x_2^2$
  • let's try to complete the square: $2 \, (x_1 + 3 \, x_2)^2 - 11 \, x_2^2$
  • we have a minus!


Matrix vs Function

Let's have a look again at $A_2$:

  • $f(\mathbf x) = \mathbf x^T A_1 \, \mathbf x = 2 x_1^2 + 12 x_1 x_2 + 20 x_2^2 = 2 \, (x_1 + 3 \, x_2)^2 + 2 \, x_2^2$
  • the numbers in the completed square form come from Gaussian Elimination!
  • Let's do $A = LU$ transformation:
    • [math]L = \begin{bmatrix} 1 & 0 \\ 3 & 1 \\ \end{bmatrix}, U = \begin{bmatrix} \boxed 2 & 6 \\ 0 & \boxed 2 \\ \end{bmatrix}[/math]
  • multipliers before squares come from pivots of $U$:
    • $\boxed{2} \, (x_1 + 3 \, x_2)^2 + \boxed 2 \, x_2^2$
  • coefficients inside each square come from $L$
    • $2 \, (1 \, x_1 + 3 \, x_2)^2 + 2 \, (0\, x_1 + 1 \, x_2)^2$
  • so positive pivots of $U$ are good


Derivative Matrix

So a matrix of second derivatives (Hessian Matrix) is

  • [math]\begin{bmatrix} \cfrac{\partial x_1^2}{\partial^2 x_1} & \cfrac{\partial x_1 \partial x_2}{\partial x_1 \partial x_2} \\ \cfrac{\partial x_2 \partial x_1}{\partial x_2 \partial x_1} & \cfrac{\partial x_2^2}{\partial^2 x_2} \\ \end{bmatrix}[/math]
  • we want it to be positive-definite
  • then the function $f(\mathbf x) = \mathbf x^T A \, \mathbf x$ is positive-definite


Checking for Positiveness

So, how to check for positive definitiveness?

  • using the definition: check that $\mathbf v^T A \mathbf v > 0$ for all $\mathbf v$
  • check that all eigenvalues are positive
  • or that all pivots of $L$ in $A = LU$ are positive
  • or that all Determinants and sub-determinants are positive


Checking using positiveness of eigenvalues:

  • if for all $\mathbf v$, $\mathbf v^T A \, \mathbf v > 0$,
  • $A \mathbf v = \lambda \mathbf v$, multiply by $\mathbf v^T$ on the left
  • $\mathbf v^T A \, \mathbf v = \lambda \mathbf v^T \mathbf v$
  • $\mathbf v^T A \, \mathbf v = \lambda \| \mathbf v \|^2$
  • $\| \mathbf v \|^2$ is always positive, so it means that if $\lambda > 0$, then so is $\mathbf v^T A \, \mathbf v$
  • therefore we can check if all eigenvalues are positive


Properties

Sum

If $A$ and $B$ are both PDM

  • then so is $A + B$
  • because the energies add when we add matrices:
  • $\mathbf v^T A \, \mathbf v + \mathbf v^T B \, \mathbf v$


Inverse

  • if $A$ is PDM, the inverse is also PDM
  • eigenvalues of the inverse are $\lambda^*_i = \frac{1}{\lambda_i}$
  • so eigenvalues are also positive
  • but careful with semi-positive definite matrices: they do not have an inverse!


$R^T R$ and $R R^T$ Matrices

They are always semi-positive definite


Sources

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