Is trace(A*A) always positive?

  • Context: Undergrad 
  • Thread starter Thread starter zeebek
  • Start date Start date
  • Tags Tags
    Positive
Click For Summary
SUMMARY

The discussion confirms that for any non-trivial n x n matrix A, the expression trace(A*A) is always non-negative. A counter-example initially proposed was the matrix A = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}, which actually results in A*A being the identity matrix I, yielding a trace of 2. The proof provided demonstrates that A*A is self-adjoint and positive semidefinite, leading to non-negative eigenvalues and thus a non-negative trace. The misunderstanding regarding the definition of A*A was clarified, affirming the original statement's validity.

PREREQUISITES
  • Understanding of matrix operations, specifically matrix multiplication.
  • Familiarity with concepts of self-adjoint and positive semidefinite matrices.
  • Knowledge of eigenvalues and eigenvectors in linear algebra.
  • Basic comprehension of the trace function in matrix theory.
NEXT STEPS
  • Study the properties of self-adjoint matrices in linear algebra.
  • Learn about positive semidefinite matrices and their applications.
  • Explore the implications of eigenvalues in matrix theory.
  • Investigate the trace function and its significance in various mathematical contexts.
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in matrix theory and its applications in various fields such as physics and engineering.

zeebek
Messages
27
Reaction score
0
I have a feeling that for any n x n non-trivial matrix A, trace(A*A) is always positive.
Is it true?
 
Physics news on Phys.org
Nevermind, I found contre example.
 
I'm eager to hear your counter-example, because the statement is true.
 
Let A= \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}
 
@HallsofIvy: is this meant to be a counter-example? With this A, I get A*A=I (the identity matrix), which has trace equal to 2.

My proof: for an arbitrary matrix A, the product A*A is self-adjoint (because (A*A)*=A*A) and positive semidefinite (because (A^*Av,v)=(A^*v,A^*v)=\|A^*v\|^2=\|Av\|^2). Hence A*A has an orthonormal basis of eigenvectors, i.e. is diagonalizable with non-negative eigenvalues. The trace is then the sum of the eigenvalues, which is non-negative.
 
Ah. I has assumed that "A*A" simply meant A times itself. Otherwise, as you say, there is no "counterexample".
 
Now I understand that I was wrong. A*A is indeed positive semidefinite
thanks to everybody
 

Similar threads

Graduate Trace of the inverse of matrix products
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Understanding Hessian for multidimensional function
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Proof about a positive definite matrix
  • · Replies 12 ·
Replies
12
Views
2K
MHB Proving or disproving this matrix V is invertible.
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad What is the definition of trace for n-indexed tensor in group theory?
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad How can I test for positive semi-definiteness in matrices?
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Want to understand how to express the derivative as a matrix
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Comparing Rank and Trace of a Matrix
  • · Replies 1 ·
Replies
1
Views
2K
MHB Could you prove that f(A)>=0 whenever A>0?
  • · Replies 6 ·
Replies
6
Views
1K
Undergrad Number of Binary Operations on a Set with a Special Property
  • · Replies 2 ·
Replies
2
Views
3K