Does a Positive Definite Hermitian Matrix Always Have a Positive Determinant?

In summary, if X is a column vector and A is a Hermitian 3 x 3 matrix, then it implies that determinant (A) > 0. This can be proven using quadratic forms or eigenvalue methods.
  • #1
David932
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1. Problem statement : suppose we have a Hermitian 3 x 3 Matrix A and X is any non-zero column vector. If
X(dagger) A X > 0 then it implies that determinant (A) > 0.

I tried to prove this statement and my attempt is attached as an image. Please can anyone guide me in a step by step way to approach this problem. I am not sure and not very clear about my way of approach about the problem.

P.S. I am new to the forum and I apologize if my post has flaws. I tried my best to formulate the problem and my attempt at it clearly but still I am not very proficient in this formatting stuff. I wrote it in MSword too and then posted it but the whole text appeared struck out and then it was removed.
Question1.jpg
 
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  • #2
With the free or cheap app DocScan HD you can make your page look like this:
2q0porc.jpg


In your case I don't know whether that's enough to get anyone to read it. :oldbiggrin:
 
  • #3
Attempted transliteration:

David932 said:
1. Problem statement : suppose we have a Hermitian 3 x 3 Matrix A and X is any non-zero column vector. If
X(dagger) A X > 0 then it implies that determinant (A) > 0.

I tried to prove this statement...

Consider ##F(t) = det( t\ I + (1-t)A) ## defined on ## t \in [0,1] ##

##F(0) = det(A)##
##F(1) = 1 ##

If we manage to prove ##F(t) \ne 0## for each ##t \in [0,1]## then ##F(0)## and ##F(1)## have the same sign, by the intermediate value theorem.

If ##t \in [0,1]## and ##x## is a non-zero real vector then ##x^T( t\ I + (1 - t)A) x = t x^T x + (1-t)x^T A x > 0 ##

which implies ##t\ x + (1-t)A## is nonsingular. This implies ## F(t) \ne 0##.
 
  • #4
You're on the cusp of some very interesting stuff called quadratic forms. I'd suggest approaching this from a few different angles -- it connects quite a few different concepts in Linear Algebra. With that in mind, here are a few other ways to look at the problem:
- - - -

When you get down to it, these are optimization problems that are eigenvalue problems. (And determinant = product of all eigenvalues).

For Hermitian ##\mathbf A##, consider the optimization problem:

##\mathbf x^H \mathbf A \mathbf x##

where ##\mathbf x \neq \mathbf 0##
(where ##^H## indicates conjugate transpose -- your post says Hermitian, so we really should be using ##^H## not ##^T##. Note that ##\mathbf 0## is the zero vector).

How would you minimize this expression? How would you maximize this? (There are basically two approaches, one is use Lagrange multipliers, and the other is diagonalize with unitary basis formed by eigenvectors. Both are worth working through in detail.)

I would spend some time thinking on quadratic forms.

- - - - -
A simpler, but less elegant approach is to not directly mention eigenvalues, and consider this:

##\mathbf x^H \mathbf A \mathbf x##. Assume that ##\mathbf A## has a non trivial nullspace. Then that means ##\mathbf A## is singular and I can find some ##\mathbf x \neq \mathbf 0##, where ##\mathbf A \mathbf x = \mathbf 0##. If this is the case, then ##\mathbf x^H \mathbf A \mathbf x = \mathbf x^H \big(\mathbf A \mathbf x\big) = ?##. What does this tell you?

This approach is simpler, but feels a lot lot less constructive (and satisfying) to me than actually working through the quadratic form optimization.
- - - -
Note there is a sticky on using LaTeX formatting here:

https://www.physicsforums.com/help/latexhelp/

Also you can right click my formatted text and show the math commands to see what I entered to get this result.
- - - -
By the way this was posted in the 'precalculus' forum. So I'm not totally sure how to think about the use of intermediate value theorem (or Lagrange multipliers) here.
 
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Related to Does a Positive Definite Hermitian Matrix Always Have a Positive Determinant?

1. What are matrices and determinants?

Matrices and determinants are mathematical tools used to organize and manipulate data or solve systems of equations. A matrix is a rectangular array of numbers or variables, while a determinant is a numerical value that can be calculated from a matrix.

2. How are matrices and determinants used in real life?

Matrices and determinants are used in a variety of fields, such as computer graphics, economics, and physics. They are especially useful in solving systems of linear equations, which arise in many real-life situations.

3. What is the difference between a matrix and a determinant?

A matrix is a collection of numbers arranged in rows and columns, while a determinant is a single value that can be calculated from a matrix. A determinant is used to represent the properties of a matrix, such as its invertibility and the number of solutions to a system of equations.

4. How do you calculate the determinant of a matrix?

The determinant of a square matrix can be calculated using different methods, such as the cofactor expansion method or the Gaussian elimination method. The size of the matrix will determine which method is more efficient, but both methods follow a set of rules and operations to obtain the determinant value.

5. Can matrices and determinants be multiplied together?

Yes, matrices and determinants can be multiplied together. The result of multiplying two matrices together is another matrix, while the result of multiplying a matrix by a determinant is a new matrix with each element multiplied by the determinant value. However, the dimensions of the matrices must be compatible for multiplication to be possible.

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