Dear Mohammad,

Since Tr(A) is the sum of eigenvalues, E(A)=Tr(A)=Tr(B)=E(B).

However, A>B>0 usually signifies that A and B are positive definite. In such case are eigenvalues positive and also Tr(A)>0. It cannot be true that Tr(A)=0.

Dear Mohammad,

Since Tr(A) is the sum of eigenvalues, E(A)=Tr(A)=Tr(B)=E(B).

However, A>B>0 usually signifies that A and B are positive definite. In such case are eigenvalues positive and also Tr(A)>0. It cannot be true that Tr(A)=0.

Dear Ari, maybe you mention similar matrices? Such matrices have the same eigenvalues.

https://math.stackexchange.com/questions/8339/similar-matrices-have-the-same-eigenvalues-with-the-same-geometric-multiplicity

Dear Viera,

Thank you so much. But I want to explain that if A>B>=0, means that all the entries of A and B, are positive.

but the eigenvalues can be (+ve) or (-ve), and since the energy defined to be the summation of the absolute values of the eigenvalues.

Thus it is possible to find a matrix (A) with tr(A)=0, but E(A)>0.

Regards.

Ari Ben-Judah,

Thank you my dear, for explanation.

Regards

Dear Mohammad,

Tr(A) is defined as the sum of diagonal elements (do we both mention the trace of A?) but whatever is A, the sum of eigenvalues equals Tr(A) as well.

Note: If A>B in the sense you explained, then necessarily for each diagonal element aii>bii. Consequently, Tr(A)>Tr(B). The assumption Tr(A)=Tr(B) does not work.

Dear Viera,

I got your idea, but in case tr(A)=tr(B)=0, it will be possible.

An example of such A and B is,

take (A) to be the adjacency matrix (I. Gutman, 1978)

and (B) is the sum eccentricity matrix as in the folloing

Article The Sum-Eccentricity Energy Of A Graph

Dear Mohammad,

If A>B>0, means that all the entries of A and B, are positive, then tr(A) ≠ 0 and tr(B) ≠ 0.

Regards.

Dear Wiwat and Viera,

Thank you, and sorry for my missing, I edit the question and hope that you got what I mean.

Regards.

Dear Mohammad,

If A>B>=0, means that all the entries of A and B, are positive (I think : A and B are non-negative entries), and tr(A) = tr(B) = 0, is you mean the matrices has all diagonal entries are zeros?

Regards.

If a positive matrix has zero trace then it is identically zero. So I think the energy of both matrices is obviousely zero. Am I missing some thing?

Dear Mohammad,

tr(A)=tr(B) & tr(matrix)=E(matrix) gives necessarily E(A)=E(B). The relation between elements of A, B will not influence this conclusion.

Dear Wiwat,

you can say that all entries in the diagonal are zero, and the other entries are grater than or equal to zero.

Dear Ali,

Note that tr(A) is the summation of the eigenvalues of (A), while E(A) (the energy of a graph associated by (A)) is the summation of the absolute values of the eigenvalues of (A).

Dear Mohammad,

In the above conditions, and in fact (due to Viera) E(A) = tr(A)= tr(B) = E(B) = 0. That is E(A) = E(B) = 0. We have E(A) is not greater than E(B).

Dear Mohammad, are A,B symmetric?

Dear Ali, the matrix {[0,1],[1,0]} has eigenvalues +1,-1 but the trace is 0.

Dear followers of this question,

it is not true that tr(A)=E(A). Trace is the sum of eigenvalues while energy is the sum of singular values. In the case of a symmetric matrix A, its energy is equal to the sum of absolute values of eigenvalues. (There were typing errors in the first formulation of Question.)

It's the reason why I was asking "Are A,B symmetric?"

In my example above is the matrix symmetric with trace=0, eigenvalues 1 and - 1 and energy |1|+|-1|=2.

Really I would like to thank you all for your action and your healthy answers.

Mohammad Issa Sowaity

Mohammad and all,

This looks like a simple exercise for Linear Algebra students, not very suitable as a legitimate homework help for RG. The answer is E(A) = E(B). HINT: This follows directly from the Cayley-Hamilton theorem.

Cheers, Ed Gerck