k**********g
发帖数: 33 | 1 Soppose that we are given a symmetric matrix B[x]，which is linear in the
decision variable x=[x_1,...,x_n],
how to give two linear terms a[x]=a^T*x+a_0 and b[x]=b^T*x+b_0 , such that
a[x] I <= B[x] <= b[x] I
where I is the identity matrix, A<=B means that B-A is Positive-definite matrix.
For example:
B[x]=[x1+x2+1 x2-1
x2-1 x1+2x2+2]; | t*******e
发帖数: 172 | 2 Suppose we can, take trace of this inequality.
you get
linear function <=linear function <=linear function
which can not be hold for all the x unless these three function are parallel
to each other.
However, if we make some restriction on the range of x, this problem is
trivial by simple le a_{0} negative enough and b_{0} big enough.
【在 k**********g 的大作中提到】
: Soppose that we are given a symmetric matrix B[x]，which is linear in the
: decision variable x=[x_1,...,x_n],
: how to give two linear terms a[x]=a^T*x+a_0 and b[x]=b^T*x+b_0 , such that
: a[x] I <= B[x] <= b[x] I
: where I is the identity matrix, A<=B means that B-A is Positive-definite matrix.
: For example:
: B[x]=[x1+x2+1 x2-1
: x2-1 x1+2x2+2];
| k**********g
发帖数: 33 | 3 Thanks for your reply.
I am sorry that I made a mistake.
A<=B should mean that B-A is Positive-definite matrix.
How about considering the maximum eig values? It's possible?
parallel
【在 t*******e 的大作中提到】
: Suppose we can, take trace of this inequality.
: you get
: linear function <=linear function <=linear function
: which can not be hold for all the x unless these three function are parallel
: to each other.
: However, if we make some restriction on the range of x, this problem is
: trivial by simple le a_{0} negative enough and b_{0} big enough.
| t*******e
发帖数: 172 | 4 That is what I understood before.
You take trace. if the matrix is positive then the trace should be big than
0.
【在 k**********g 的大作中提到】
: Thanks for your reply.
: I am sorry that I made a mistake.
: A<=B should mean that B-A is Positive-definite matrix.
: How about considering the maximum eig values? It's possible?
:
: parallel
| s***e
发帖数: 267 | 5 trace condition for matrix inequality is only necessary but not sufficient.
Even if trace inequality is satisfied, the matrix inequality may still fail
to hold.
E.g., max trace element <= max eigenvalue, but not vice versa. If the x's
are not bounded, then the diagonal linear equations should be parallel with
a & b. Still need more conditions to figure out the rest. | z****g
发帖数: 1978 | 6 B(X) = B_1*x_1 + B_2*x_2 + ...+ B_n*x_n
Since B(X) is symmetric in terms of polynomial(1st degree in your case), all
{B_i}s should be symmetric.
For a simple case, if limit {B_i} in real field. Every {B_i} can be
diagonalized, since they are all real symmetric. Say B_i = T_i*E_i*T_i^{t}
decompose f(x)*I into sum_i{a_i*x_i*T_i*T_i^{t}}. You can see the result
directly. |
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