Real Symmetric Positive Definite Matrices

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Homework Statement


Let A be a real symmetric positive definite matrix. Show that |aij|<(aii+ajj)/2 for i not equal to j.


Homework Equations





The Attempt at a Solution


I really don't even know where to start with this. I think that aii and ajj must both be > 0 since they are on the diagonal and real symmetric positive definite matrices have positive diagonal elements. Other than this, I honestly have NO ideas. Does anyone else??
 

Answers and Replies

  • #2
Dick
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Let {ei} be the basis vectors. Then ei*Aei, ej*Aej and (ei+ej)*A(ei+ej) are all positive. What does that mean in terms of the matrix entries of A?
 
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  • #3
Tom Mattson
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As with any proof in mathematics, you start by looking at the definitions of the objects in question. Answer the following questions in order. They should help you get started.

1.) What is the definition of a real symmetric matrix?
2.) What is the definition of a positive definite matrix?
3.) Consider a simple case: a 2x2 matrix [itex]A[/itex], given below.

[tex]A = \left[\begin{array}{cc}a_{11} & a_{12}\\a_{21} & a_{22}\end{array}\right][/tex]

Apply the definitions you stated in 1.) and 2.) to say as much as you can about the [itex]a_{ij}[/itex].

4.) Let [itex]\vec{x}^T=\left[x_1,x_2,x_3\right]\in\mathbb{R}^2[/itex]. Apply the definition of a positive definite matrix to [itex]A[/itex] using this vector.

At this point you should be able to deduce the result for the 2x2 case. Hopefully that sheds some light on the subject.
 
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  • #4
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Thank you for all your guidance. Here is where I am at right now.

1. Real Symmetric: A = AT --> aij=aji

2. Positive Definite: xTAx > 0 for x not equal to 0, also aii and ajj > 0

So, I did this out for a 2x2 matrix A: xTAx

In the end I got: a11x12 + a22x22+2a12x1x2 > 0


Then, I get down to: (a11x12+a22x22)/2 > |a12x1x2|

However, we don't want the x's in here, and I still do not really understand how to apply this to a symmetric positive definite matrix in general (a more general case as opposed to this 2x2). Thank you again!
 
  • #5
Dick
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Substitute x1=0 and x2=1, x1=1 and x2=0 and finally x1=1 and x2=1. Put these into the first inequality. Your derivation so far doesn't prove the second one because you don't know a11 and a22 are positive. Look at what the three resulting inequalities tell you. This is basically the contents of my first post.
 
  • #6
Tom Mattson
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2. Positive Definite: xTAx > 0 for x not equal to 0, also aii and ajj > 0
Yes, for any [itex]\itex{x}\in\mathbb{R}^n[/itex] not equal to [itex]\vec{0}[/itex]. That means you can choose the components to be, say, one.

Then, I get down to: (a11x12+a22x22)/2 > |a12x1x2|
What do you get when you set all of the [itex]x_i[/itex] equal to one?

To generalize, try to answer the question in Dick's post. Can you express [itex]\left(\vec{e}_i+\vec{e_j}\right)^TA\left(\vec{e}_i+\vec{e}_j\right)[/itex] in terms of the [itex]a_{ij}[/itex]?
 
  • #7
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Why don't you know a_11 and a_22 are positive? They are diagonal elements on a symmetric positive definite matrix, doesn't that imply that they are positive?
 
  • #8
Dick
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Why don't you know a_11 and a_22 are positive? They are diagonal elements on a symmetric positive definite matrix, doesn't that imply that they are positive?
It does, but you haven't shown it yet. If you have already proved that, then never mind.
 
  • #9
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Yes, it was a fact stated in class, so I won't have to show it. Thank you though :)

Unfortunately, I'm still completely confused how to show this generally.

I tried doing something with (ei+ej)TA(ei+ej)>0. All I got was aii+ajj>0, which we already know is true, so I'm guessing I didn't do that right.
 
  • #10
Dick
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Yes, it was a fact stated in class, so I won't have to show it. Thank you though :)

Unfortunately, I'm still completely confused how to show this generally.

I tried doing something with (ei+ej)TA(ei+ej)>0. All I got was aii+ajj>0, which we already know is true, so I'm guessing I didn't do that right.
You didn't do it right. ei^T A ej and ej^T A ei give you contributions too. What are they?
 
  • #11
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Oh duh, thank you. I'll try that. Thanks so much.
 
  • #12
Dick
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Oh duh, thank you. I'll try that. Thanks so much.
Hint: you'll also need to think about (ei-ej)^T A (ei-ej)>0 to complete the inequality.
 
  • #13
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Okay, let me try this again.



(ei+ej)TA(ei+ej)>0

eiTAei+ejTAei+eiTAej+ejTAej>0

aii+aji+aij+ajj>0

but aji=aij so

aii+ajj+2aij>0

aii+ajj>-2aij

aii+ajj>|-2aij| since the left side is positive anyways

(aii+ajj)/2>|aij|

I think this is okay now. Thank you SO much for all of your patience and help!!
 
  • #14
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Hint: you'll also need to think about (ei-ej)^T A (ei-ej)>0 to complete the inequality.
Thanks. Is it possible though to just take the absolute value of both sides knowing that the left side is already positive?
 
  • #15
Dick
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Nope, doesn't quite work. a>b doesn't imply a>|b|. b could be negative and |b|>a. You need both a>b and a>-b. See my last hint.
 
  • #16
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Thank you SO SO SO much, Dick. I really appreciate you walking me through this.
 

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