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 January 28th, 2017, 10:35 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 Positive Definite Given a quadratic form: $\displaystyle Q=x'Ax=x_{i}A_{ij}x_{j}$, summation convention, A symmetric. Let P diagonalize A: $\displaystyle D=P^{-1}AP$. Let $\displaystyle x=Py$. Then $\displaystyle Q=y'Dy=\lambda _{i}y_{i}^{2}$, which is positive for all y if $\displaystyle \lambda$ are all positive. Question: Why is $\displaystyle x'Ax$ positive for all x?
 January 30th, 2017, 03:35 PM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 Range of values of x'Ax doesn't change under non-singular linear substitution.
January 30th, 2017, 11:31 PM   #3
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Quote:
 Originally Posted by zylo Given a quadratic form: $\displaystyle Q=x'Ax=x_{i}A_{ij}x_{j}$, summation convention, A symmetric. Let P diagonalize A: $\displaystyle D=P^{-1}AP$. Let $\displaystyle x=Py$. Then $\displaystyle Q=y'Dy=\lambda _{i}y_{i}^{2}$, which is positive for all y if $\displaystyle \lambda$ are all positive. Question: Why is $\displaystyle x'Ax$ positive for all x?

It isn't necessarily. However, when it is, this is the definition of a positive definite matrix. If you are assuming that $A$ is positive definite then this follows immediately from what you have written since all eigenvalues are non-negative.

If you aren't assuming this, then here is a simple counterexample which shows not every symmetric matrix is positive definite.

$$\left\( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$

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