# Product of Hermitian matrices

## 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?

Delta2

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StoneTemplePython
Gold Member
2019 Award
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?

StoneTemplePython
Gold Member
2019 Award
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.

Delta2
StoneTemplePython
Gold Member
2019 Award
Thanks, but I still do not understand. Lets 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?
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##.

StoneTemplePython
Gold Member
2019 Award
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.

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.