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January 28th, 2017, 10:35 AM   #1
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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?
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January 30th, 2017, 03:35 PM   #2
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Range of values of x'Ax doesn't change under non-singular linear substitution.
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January 30th, 2017, 11:31 PM   #3
SDK
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Quote:
Originally Posted by zylo View Post
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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