Understanding Commutativity and Eigenvalues in the Product of Hermitian Matrices

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Homework Help Overview

The discussion revolves around the properties of Hermitian matrices, specifically focusing on the conditions under which the product of two Hermitian matrices remains Hermitian. The original poster expresses confusion regarding the implications of commutativity and the nature of eigenvalues in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the commutativity of Hermitian matrices and the nature of their product's eigenvalues. Questions arise about the application of the Operator 2 Norm and its relevance to the problem. There is also discussion about the Frobenius norm and its limitations in this context.

Discussion Status

Some participants have provided hints regarding the use of the Operator 2 Norm and its properties, while others express uncertainty about its definition and application. The conversation reflects a mix of attempts to clarify concepts and explore mathematical reasoning without reaching a consensus on the specific steps needed to resolve the original poster's confusion.

Contextual Notes

Participants note the potential complexity of the problem, particularly regarding the understanding of singular values and norms. There is mention of external resources that may aid in comprehension, but no definitive conclusions are drawn about the problem's resolution.

LagrangeEuler
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Homework Statement
For two Hermitian matrices ##A## and ##B## with eigenvalues larger then ##1##, show that
##AB## has eigenvalues ##|\lambda|>1##.
Relevant Equations
Any hermitian matrix ##A## could be written as
[tex]A=\sum_k \lambda_k|k \rangle \langle k|[/tex]
where ##|k\rangle \langle k|## is orthogonal projector ##P_k##.
Product of two Hermitian matrix ##A## and ##B## is Hermitian matrix only if matrices commute ##[A,B]=0##. If that is not a case matrix ##C=AB## could have complex eigenvalues. If
A=\sum_k \lambda_k|k \rangle \langle k|
B=\sum_l \lambda_l|l \rangle \langle l|
AB=\sum_{k,l}\lambda_k\lambda_l|k \rangle \langle k|l\rangle \langle l|
Now I am confused what to do. Definitely, ##\langle k|l \rangle \leq 1 ##. Could you help me?
 
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hint: the Operator 2 Norm is submultiplicative -- find a way to use this.
 
Thanks. So
||AB||\leq ||A|| \cdot ||B||
this is submultiplicativity? But which norm?
 
LagrangeEuler said:
Thanks. So
||AB||\leq ||A|| \cdot ||B||
this is submultiplicativity? But which norm?
The Operator 2 norm-- I said this in post #2. A very quick google search should tell you that the operator 2 norm is given by ##\big \Vert C \big\Vert_2##
 
Sorry, but I am not sure what is Operator 2 norm.
 
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Thanks, but I still do not understand. Let's take matrix from the example
<br /> I_2=\begin{bmatrix}<br /> 1 \\[0.3em]<br /> 0 \\[0.3em]<br /> 0 \\[0.3em]<br /> 1 \\[0.3em]<br /> \end{bmatrix}
Then ##||I_2||_F=\sqrt{Tr(I_2^TI_2)}=\sqrt{Tr(2)}=\sqrt{2}## and I am not sure how they find that ##||I_2||_2=1##
which vector ##v## are they using?
I looked this link
https://math.stackexchange.com/questions/2996827/frobenius-and-operator-2-norm
 
Using Frobenius norm I know that
||A||=\sqrt{(A,A)}=\sqrt{Tr(A^*A))}
and
||AB||\leq ||A|| ||B||
|\lambda_A|\leq ||A||
|\lambda_B|\leq ||B||
|\lambda|\leq ||AB||
but I am still not sure how to prove that ##|\lambda|>1##.
 
no. you need to use the operator 2 norm. The Frobenius norm isn't the right tool for this job. Do you know what a singular value is? If not, this problem may be out of reach.
 
  • #10
I checked it on wikipedia
Something like
##AX_k=\sigma_kY_k##
##A^*Y_k=\sigma_kX_k##
but it is hard to me to understand how to find ##\sigma_k## on the concrete example.
 

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