Understanding Commutativity and Eigenvalues in the Product of Hermitian Matrices

Click For Summary
SUMMARY

The product of two Hermitian matrices, denoted as matrices A and B, is Hermitian only if they commute, expressed mathematically as [A, B] = 0. If they do not commute, the resulting matrix C = AB may have complex eigenvalues. The discussion emphasizes the importance of the Operator 2 norm, which is defined as the maximum singular value of a matrix, and clarifies that the Frobenius norm is not suitable for proving eigenvalue properties in this context. Participants are encouraged to explore the relationship between singular values and the Operator 2 norm to resolve their queries.

PREREQUISITES
  • Understanding of Hermitian matrices and their properties
  • Familiarity with matrix norms, specifically the Operator 2 norm and Frobenius norm
  • Knowledge of singular value decomposition (SVD)
  • Basic linear algebra concepts, including eigenvalues and eigenvectors
NEXT STEPS
  • Study the definition and properties of the Operator 2 norm in depth
  • Learn about singular value decomposition and its applications in matrix analysis
  • Explore the relationship between eigenvalues and matrix norms
  • Review resources on Hermitian matrices and their implications in quantum mechanics
USEFUL FOR

Mathematicians, physicists, and students studying linear algebra, particularly those interested in matrix theory and its applications in quantum mechanics and dynamic systems.

LagrangeEuler
Messages
711
Reaction score
22
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?
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Delta2
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
882
  • · Replies 31 ·
2
Replies
31
Views
4K