Understanding Commutativity and Eigenvalues in the Product of Hermitian Matrices

[LagrangeEuler]

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
$$A=\sum_k \lambda_k|k \rangle \langle k|$$
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?

 

[StoneTemplePython]

hint: the Operator 2 Norm is submultiplicative -- find a way to use this.

 

[LagrangeEuler]

Thanks. So
$$||AB||\leq ||A|| \cdot ||B||$$
this is submultiplicativity? But which norm?

 

[StoneTemplePython]

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 $\big \Vert C \big\Vert_2$

 

[LagrangeEuler]

Sorry, but I am not sure what is Operator 2 norm.

 

[StoneTemplePython]

Sorry, but I am not sure what is Operator 2 norm.

https://math.stackexchange.com/questions/2996827/frobenius-and-operator-2-norm
is a nice discussion of the operator 2 norm and the Frobenius norm
simplistically, it is the maximum singular value of a matrix

it is also known as the induced 2 norm. As usual MIT has a lot of nice free resources, e.g.

https://ocw.mit.edu/courses/electri...-spring-2011/readings/MIT6_241JS11_chap04.pdf

 

[LagrangeEuler]

Thanks, but I still do not understand. Let's take matrix from the example
$$I_2=\begin{bmatrix} 1 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 1 \\[0.3em] \end{bmatrix}$$
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

 

[LagrangeEuler]

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 $|\lambda|>1$.

 

[StoneTemplePython]

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.

 

[LagrangeEuler]

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.