If A is nnd, then show that I + A is pd

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Discussion Overview

The discussion revolves around the properties of non-negative definite (nnd) and positive definite (pd) matrices, specifically addressing the relationship between an nnd matrix A and the matrix I + A. Participants explore the implications of these definitions and the conditions under which I + A can be considered positive definite.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant expresses confusion about the concepts of pd and psd matrices, questioning what these properties reveal about the entries of matrix A and how adding the identity matrix affects definiteness.
  • Another participant challenges the assertion that all entries of a non-negative definite matrix must be nonnegative, suggesting that having a zero in any position other than the diagonal does not necessarily invalidate the nnd condition.
  • A different participant clarifies that the definition of nnd is based on the condition $x'Ax \geq 0$ for every nonzero vector x, and argues that this leads to the conclusion that I + A is positive definite.
  • There is a reiteration that the definition of nnd does not imply all entries of the matrix must be nonnegative, indicating a disagreement on this point.

Areas of Agreement / Disagreement

Participants do not reach consensus on whether all entries of a non-negative definite matrix must be nonnegative. The discussion includes competing views on the implications of matrix properties and the conditions under which I + A is positive definite.

Contextual Notes

The discussion highlights potential limitations in understanding the definitions of nnd and pd matrices, particularly regarding the implications of matrix entries and the conditions for definiteness. There are unresolved aspects concerning the interpretation of matrix properties.

insixyears
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nnd = non negative definite matrix
pd = positive definite I'm having a tough time grappling with the concept of pd, psd matrices in general. My understanding basically just boils down to this, basically after multiplying everything out using the matrix formula x'Ax, you will get some sort of polynomial. If you get a polynomial where everything is squared, you're in good shape because it's impossible for the equation to be negative. A question that I have is, what does knowing that a matrix A is psd or pd tell us about the entries of A? Doesn't it only tell us how the entries of A interact (ie, when they are added up, you get something >= 0). If that's true, how would adding the identity matrix give you anymore information about the definiteness of the matrix?Thanks!
 
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crabchef said:
nnd = non negative definite matrix
pd = positive definite I'm having a tough time grappling with the concept of pd, psd matrices in general. My understanding basically just boils down to this, basically after multiplying everything out using the matrix formula x'Ax, you will get some sort of polynomial. If you get a polynomial where everything is squared, you're in good shape because it's impossible for the equation to be negative. A question that I have is, what does knowing that a matrix A is psd or pd tell us about the entries of A? Doesn't it only tell us how the entries of A interact (ie, when they are added up, you get something >= 0). If that's true, how would adding the identity matrix give you anymore information about the definiteness of the matrix?Thanks!

This is clearly not true. For A to be nonnegative definite, all its entries are nonnegative. What if it has a 0 in any element other than on the main diagonal?
Adding the identity matrix won't change that...
 
crabchef said:
nnd = non negative definite matrix
pd = positive definite I'm having a tough time grappling with the concept of pd, psd matrices in general. My understanding basically just boils down to this, basically after multiplying everything out using the matrix formula x'Ax, you will get some sort of polynomial. If you get a polynomial where everything is squared, you're in good shape because it's impossible for the equation to be negative. A question that I have is, what does knowing that a matrix A is psd or pd tell us about the entries of A? Doesn't it only tell us how the entries of A interact (ie, when they are added up, you get something >= 0). If that's true, how would adding the identity matrix give you anymore information about the definiteness of the matrix?
The definition of nnd is that $x'Ax\geq0$ for every nonzero vector $x$. If that condition holds, then $x'(I+A)x = x'x + x'Ax >0$ (because $x'x>0$), and hence $I+A$ is pd.

Prove It said:
For A to be nonnegative definite, all its entries are nonnegative.
That is not true. The definition of nnd is as given by the OP. It does not imply that all the entries of the matrix are nonnegative.
 
Last edited:
Thanks for the help! As with many problems...AFTER seeing the answer it seems so simple. This gives me one more technique to approach my homework problems, but unfortunately I'm already stuck on the next one (going to keep working on it for a bit longer before I post the question)
 

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