Proving Square Matrix of Order n Has LU-Factorization

[borjstalker]

Homework Statement

Hi, I have a question: Prove that a square matrix of order n has an LU-factorization if and only if the origin is not contained in the numerical range.

Homework Equations

The numerical range is found using the Euclidean inner scalar product <Ax,x> where <x,x>=1 and. The numerical range is A(W)=(<Ax,x> such that <x,x> =1 for all x reals)

The Attempt at a Solution

I tried considering the 2x2 matrix case but got stuck.:) Thanks in advance!

 

[mighty2000]

Thank you for your question. I am happy to help you prove this statement. Let's start by defining what an LU-factorization is. An LU-factorization of a square matrix A is a decomposition of A into the product of a lower triangular matrix L and an upper triangular matrix U, such that A = LU. Now, let's also define the numerical range of a matrix A. The numerical range of A, denoted as A(W), is the set of all possible values that can be obtained by taking the inner scalar product <Ax,x>, where x is a real vector and <x,x> = 1. In other words, the numerical range is the collection of all possible eigenvalues of A.

Now, to prove that a square matrix A of order n has an LU-factorization if and only if the origin is not contained in the numerical range, we need to prove two statements: (1) If A has an LU-factorization, then the origin is not contained in the numerical range. (2) If the origin is not contained in the numerical range, then A has an LU-factorization.

Let's start with statement (1). If A has an LU-factorization, then A = LU, where L is lower triangular and U is upper triangular. Now, let's assume that the origin is contained in the numerical range of A. This means that there exists a vector x such that <Ax,x> = 0 and <x,x> = 1. Substituting A = LU, we get <LUx,x> = 0. Since L and U are both triangular matrices, we can write this as <Ux,Lx> = 0. But this means that Ux and Lx are orthogonal, which implies that Ux = 0 or Lx = 0. But since U is upper triangular, Ux = 0 can only happen if x = 0, which contradicts our assumption that <x,x> = 1. Similarly, Lx = 0 can only happen if x = 0, which again contradicts our assumption. Therefore, our assumption that the origin is contained in the numerical range must be false, and hence, the origin is not contained in the numerical range.

Now, let's prove statement (2). If the origin is not contained in the numerical range of A, then there exists a vector x such that <Ax,x> = λ and