4

Given $$A=LDU$$ where

  • $A$ is a real symmetric matrix
  • $L$ is a lower unitriangular matrix
  • $D$ is a diagonal matrix
  • $U$ is an upper unitriangular matrix

can we say that $$A>0 \iff D>0$$ ?

Edit:

My thinking is that $(LD^{1/2})(D^{1/2}U)$ is (probably?) the Cholesky decomposition, and $D^{1/2}$ exists iff $D>0$.

Museful
  • 869

1 Answers1

3

First of all, if $A = LDU$ is symmetric, we must have $U=L^T$ by uniqueness of the decomposition ($LDU = A=A^T = U^T D L^T$). so then $A=LDL^T$.

So if $D>0$ then by for all $x\neq 0$ denoting $y=L^Tx$ (why is $y \neq 0$?) yields $$x^T A x = x^T L D L^T x = (L^T x)^T D L^T x = y^T D y > 0$$

And if $A>0$ then for all $x \neq 0 $ denoting $y = (L^T)^{-1}x $ yields $$x^T D x = (L^T y)^T D (L^T y)=y^T LDL^T y = y^T A y > 0$$

Theorem
  • 3,048