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

crabchef

New member
Feb 18, 2012
3
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!
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
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.

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.
 
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crabchef

New member
Feb 18, 2012
3
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 hw 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)