Quadratic Functions. Consider the strictly convex quadratic function f(x)=1/2 x:t Px-q^t x+r, where P in R^{n times n} is a positive definite matrix, q in R^n and r in R.

Gerardo Aguilar

Gerardo Aguilar

Answered question

2022-10-28

Quadratic Functions
Consider the strictly convex quadratic function f ( x ) = 1 2 x t P x q t x + r ,, where P R n × n is a positive definite matrix, q R n and r R . Let H := { H : H  is a  k dimensional subspace in  R n } . Clearly, the restriction of f to any H H is again a strictly convex function. For any H H , we will use x H to denote the unique optimal point of the following problem
min. x H f ( x ) .
Now consider the map, ψ ( H ) = x H .
Prove / Disprove: The map ψ is bijective.
Remark: It is assumed that P is invertible and q 0 .

Answer & Explanation

cdtortosadn

cdtortosadn

Beginner2022-10-29Added 19 answers

Step 1
Whether ψ is surjective depends on the definition of its codomain. For example, if n = 2 and k = 1, the image of ψ is a curve on R 2 . So, if the codomain of ψ is R 2 , ψ is certainly not surjective.
Step 2
At any rate, ψ may not be injective even if the origin is not a global minimum. Consider a = ( 1 , 0 , 0 ) T and f ( x ) = x a 2 . Let { e 1 , e 2 , e 3 } be the canonical basis of R 3 . Then x H = a when H = span { e 1 , e 2 } or span { e 1 , e 3 }.

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