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 January 28th, 2017, 09:35 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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, 02:35 PM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Range of values of x'Ax doesn't change under non-singular linear substitution. January 30th, 2017, 10:31 PM   #3
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 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)$$ Tags definite, positive Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post calypso Applied Math 0 January 16th, 2017 07:07 AM wannabe1 Linear Algebra 1 April 26th, 2010 07:00 AM BSActress Linear Algebra 2 August 23rd, 2009 03:39 PM BSActress Linear Algebra 0 August 5th, 2009 08:06 PM angelz429 Linear Algebra 1 May 7th, 2008 07:42 AM

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