The set of 3-tuples numerical range of Hermitian operators is not convex
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I am trying to find a counter example for the following claim.
If $A_1, A_2, dots, A_n$ are Hermitian operators then the set of all $n$-tuples of the form $left { left( left < A_1 f, f right >, left < A_2 f, f right > , dots left < A_n f, f right > right)mid |f|=1right}$ may not be convex.
The book gives a hint about $n=3$ and the dimension of matrix is 2.
I tried some matrix but as I know, the numerical range of some $2 times 2$ Hermitian operator A is contained in the interval $left[ lambda _1 , lambda _2 right ]$, in which $lambda 1$ and $lambda_2$ are eigenvalues of $A$.
Then, the desired numerical range is always a cube and convex.
Could you please help me to give some counter example?
Thank you in advance.
linear-algebra functional-analysis operator-theory
asked Nov 14 at 8:52
Trần Linh
27219
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up vote
0
down vote
favorite
I am trying to find a counter example for the following claim.
If $A_1, A_2, dots, A_n$ are Hermitian operators then the set of all $n$-tuples of the form $left { left( left < A_1 f, f right >, left < A_2 f, f right > , dots left < A_n f, f right > right)mid |f|=1right}$ may not be convex.
The book gives a hint about $n=3$ and the dimension of matrix is 2.
I tried some matrix but as I know, the numerical range of some $2 times 2$ Hermitian operator A is contained in the interval $left[ lambda _1 , lambda _2 right ]$, in which $lambda 1$ and $lambda_2$ are eigenvalues of $A$.
Then, the desired numerical range is always a cube and convex.
Could you please help me to give some counter example?
Thank you in advance.
linear-algebra functional-analysis operator-theory
asked Nov 14 at 8:52
Trần Linh
27219
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to find a counter example for the following claim.
If $A_1, A_2, dots, A_n$ are Hermitian operators then the set of all $n$-tuples of the form $left { left( left < A_1 f, f right >, left < A_2 f, f right > , dots left < A_n f, f right > right)mid |f|=1right}$ may not be convex.
The book gives a hint about $n=3$ and the dimension of matrix is 2.
I tried some matrix but as I know, the numerical range of some $2 times 2$ Hermitian operator A is contained in the interval $left[ lambda _1 , lambda _2 right ]$, in which $lambda 1$ and $lambda_2$ are eigenvalues of $A$.
Then, the desired numerical range is always a cube and convex.
Could you please help me to give some counter example?
Thank you in advance.
linear-algebra functional-analysis operator-theory
asked Nov 14 at 8:52
Trần Linh
27219
I am trying to find a counter example for the following claim.
If $A_1, A_2, dots, A_n$ are Hermitian operators then the set of all $n$-tuples of the form $left { left( left < A_1 f, f right >, left < A_2 f, f right > , dots left < A_n f, f right > right)mid |f|=1right}$ may not be convex.
The book gives a hint about $n=3$ and the dimension of matrix is 2.
I tried some matrix but as I know, the numerical range of some $2 times 2$ Hermitian operator A is contained in the interval $left[ lambda _1 , lambda _2 right ]$, in which $lambda 1$ and $lambda_2$ are eigenvalues of $A$.
Then, the desired numerical range is always a cube and convex.
Could you please help me to give some counter example?
Thank you in advance.
linear-algebra functional-analysis operator-theory
linear-algebra functional-analysis operator-theory
asked Nov 14 at 8:52
Trần Linh
27219
asked Nov 14 at 8:52
Trần Linh
27219
asked Nov 14 at 8:52
Trần Linh
27219
asked Nov 14 at 8:52
Trần Linh
27219
asked Nov 14 at 8:52
Trần Linh
27219
27219
add a comment |
add a comment |
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