Simple quadratic optimization problem

In summary, The problem discusses a quadratic expression and its minimization over a vector K. The expression involves various matrices such as P^ + , P^ - , I, Q, and K, with Q being positive definite and P^ - being positive semidefinite. It is then extended to include a diagonal matrix A and the minimum over K is considered. However, there may be a typo in the original problem as the matrix multiplication KQK^T is not possible if K is a column vector.
  • #1
nfreris2
1
0
Let [tex]P^ + ,P^ - ,I,Q \in R^{n\times n}, K\in R^{n\times 1}, M\in R^{1 \times n}[/tex], and assume that [tex]Q[/tex] is positive definite, [tex]P^ -[/tex] is positive semidefinite whence [tex](MP^ - M^T + Q)^{ - 1}[/tex] exists (where [tex]T[/tex] denotes transpose).

In what sense does [tex]K = P^ - M^T(MP^ - M^T + Q)^{ - 1}[/tex] minimize the quadratic expression [tex]P^ + : = (I - KM)P^ - (I - KM)^T + KQK^T[/tex], over [tex]K[/tex]?
Is this minimization of [tex]P^ +[/tex] over all vectors [tex]K[/tex] with respect to the usual ordering for positive semidefinite matrices [tex]A\le B[/tex] iff [tex]B - A [/tex]is positive semidefinite?

Next consider the extension [tex]P^ + : = (I - KAM)P^ - (I - KAM)^T + KAQA^TK^T[/tex], where [tex]A\in R^{n\times 1}, K\in R^{n\times n}[/tex] and [tex]K[/tex] diagonal, where all other matrices are as above.
What is the minimum over [tex]K[/tex] (with respect to the previous ordering or something )??

Any help will be deeply appreciated.
 
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  • #2
I think maybe there is a typo, because if K is a column vector (in Rnx1) and Q a square matrix then KQK* is meaningless, you can't do that matrix multiplication. However, if you meant that K is a row vector, then that implies KQK* is a scalar, so it cannot possibly add to anything to equal the nxn matrix P+.
 
  • #3
That is correct [tex]Q > 0[/tex], a scalar.

Please help with this.
 
  • #4
What are your thoughts on the problem so far?
 

1. What is a simple quadratic optimization problem?

A simple quadratic optimization problem is a type of mathematical problem in which the goal is to find the maximum or minimum value of a quadratic function, subject to certain constraints. It involves finding the optimal values for the variables in the quadratic function in order to achieve the desired outcome.

2. How is a simple quadratic optimization problem solved?

A simple quadratic optimization problem can be solved using various methods, such as the graphical method, substitution method, or the quadratic formula. The most common method is the quadratic formula, which involves finding the roots of the quadratic equation and determining the optimal value from there.

3. What are the key components of a simple quadratic optimization problem?

The key components of a simple quadratic optimization problem include a quadratic function, variables, and constraints. The quadratic function is a mathematical expression in the form of ax^2 + bx + c, where a, b, and c are coefficients. The variables represent the unknown values that need to be optimized, and the constraints define the limitations or conditions that the variables must satisfy.

4. What are some real-life applications of simple quadratic optimization problems?

Simple quadratic optimization problems have various real-life applications, including in business, economics, and engineering. For example, they can be used to optimize production costs, maximize profits, or minimize resource usage in manufacturing processes. They can also be used in financial planning to determine the optimal investment portfolio.

5. Are there any limitations to using simple quadratic optimization problems?

While simple quadratic optimization problems are useful in many situations, they do have limitations. One limitation is that they can only be used for problems with quadratic functions, which may not accurately model all real-life scenarios. Additionally, the optimal solutions obtained from these problems may not always be practical or feasible in practical applications.

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