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$mathbfL succeq mathbfQ$ true? For, $mathbfL$ diagonal with all entries $leq P$ and $mathbfQ$ Hermitian, with its trace limited by $P$
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I have the Hermitian and positive semidefinite matrix $mathbfQ$, with $texttrmathbfQ leq P$ and the diagonal matrix
$$
mathbfD = textdiag(d_1, ..., d_N), ,
$$
with
$$
0 leq d_1 leq ... leq d_K < P leq d_K+1 leq ... , leq d_N , , quad kin0, ..., N.
$$
The relation $mathbfD succeq mathbfQ$ is true.
Now, I construct a matrix $mathbfL$, which is identical to $mathbfD$ except, that all entries, which are greater than $P$, are set to $P$. Therefore, $mathbfL$ can be expressed as
$$
mathbfL = textdiag(d_1, ..., d_K, P, ..., P) ,.
$$
What I want to find out is, if the relation $mathbfL succeq mathbfQ$ is true?
For me, intuitively, this makes sense, but trying to prove it, I always get stuck. I checked multiple eigenvalue inequalities from different books, tried an approach where I assume that $mathbfL succeq mathbfQ$ is false and then try to induce a contradiction. Also, I try to find a counterexample to prove, that it's false.
For the cases where
- $K=0,$, i.e. all entries of $mathbfL$ are equal to $P$ and
- $K=N,$, i.e. $mathbfL = mathbfD$
it is more or less obviously true.
linear-algebra matrices eigenvalues-eigenvectors trace positive-semidefinite
asked Jul 24 at 13:20
BBC
11
add a comment |Â
up vote
0
down vote
favorite
I have the Hermitian and positive semidefinite matrix $mathbfQ$, with $texttrmathbfQ leq P$ and the diagonal matrix
$$
mathbfD = textdiag(d_1, ..., d_N), ,
$$
with
$$
0 leq d_1 leq ... leq d_K < P leq d_K+1 leq ... , leq d_N , , quad kin0, ..., N.
$$
The relation $mathbfD succeq mathbfQ$ is true.
Now, I construct a matrix $mathbfL$, which is identical to $mathbfD$ except, that all entries, which are greater than $P$, are set to $P$. Therefore, $mathbfL$ can be expressed as
$$
mathbfL = textdiag(d_1, ..., d_K, P, ..., P) ,.
$$
What I want to find out is, if the relation $mathbfL succeq mathbfQ$ is true?
For me, intuitively, this makes sense, but trying to prove it, I always get stuck. I checked multiple eigenvalue inequalities from different books, tried an approach where I assume that $mathbfL succeq mathbfQ$ is false and then try to induce a contradiction. Also, I try to find a counterexample to prove, that it's false.
For the cases where
- $K=0,$, i.e. all entries of $mathbfL$ are equal to $P$ and
- $K=N,$, i.e. $mathbfL = mathbfD$
it is more or less obviously true.
linear-algebra matrices eigenvalues-eigenvectors trace positive-semidefinite
asked Jul 24 at 13:20
BBC
11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the Hermitian and positive semidefinite matrix $mathbfQ$, with $texttrmathbfQ leq P$ and the diagonal matrix
$$
mathbfD = textdiag(d_1, ..., d_N), ,
$$
with
$$
0 leq d_1 leq ... leq d_K < P leq d_K+1 leq ... , leq d_N , , quad kin0, ..., N.
$$
The relation $mathbfD succeq mathbfQ$ is true.
Now, I construct a matrix $mathbfL$, which is identical to $mathbfD$ except, that all entries, which are greater than $P$, are set to $P$. Therefore, $mathbfL$ can be expressed as
$$
mathbfL = textdiag(d_1, ..., d_K, P, ..., P) ,.
$$
What I want to find out is, if the relation $mathbfL succeq mathbfQ$ is true?
For me, intuitively, this makes sense, but trying to prove it, I always get stuck. I checked multiple eigenvalue inequalities from different books, tried an approach where I assume that $mathbfL succeq mathbfQ$ is false and then try to induce a contradiction. Also, I try to find a counterexample to prove, that it's false.
For the cases where
- $K=0,$, i.e. all entries of $mathbfL$ are equal to $P$ and
- $K=N,$, i.e. $mathbfL = mathbfD$
it is more or less obviously true.
linear-algebra matrices eigenvalues-eigenvectors trace positive-semidefinite
asked Jul 24 at 13:20
BBC
11
I have the Hermitian and positive semidefinite matrix $mathbfQ$, with $texttrmathbfQ leq P$ and the diagonal matrix
$$
mathbfD = textdiag(d_1, ..., d_N), ,
$$
with
$$
0 leq d_1 leq ... leq d_K < P leq d_K+1 leq ... , leq d_N , , quad kin0, ..., N.
$$
The relation $mathbfD succeq mathbfQ$ is true.
Now, I construct a matrix $mathbfL$, which is identical to $mathbfD$ except, that all entries, which are greater than $P$, are set to $P$. Therefore, $mathbfL$ can be expressed as
$$
mathbfL = textdiag(d_1, ..., d_K, P, ..., P) ,.
$$
What I want to find out is, if the relation $mathbfL succeq mathbfQ$ is true?
For me, intuitively, this makes sense, but trying to prove it, I always get stuck. I checked multiple eigenvalue inequalities from different books, tried an approach where I assume that $mathbfL succeq mathbfQ$ is false and then try to induce a contradiction. Also, I try to find a counterexample to prove, that it's false.
For the cases where
- $K=0,$, i.e. all entries of $mathbfL$ are equal to $P$ and
- $K=N,$, i.e. $mathbfL = mathbfD$
it is more or less obviously true.
linear-algebra matrices eigenvalues-eigenvectors trace positive-semidefinite
asked Jul 24 at 13:20
BBC
11
asked Jul 24 at 13:20
BBC
11
asked Jul 24 at 13:20
BBC
11
asked Jul 24 at 13:20
BBC
11
11
add a comment |Â
add a comment |Â
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