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The solutions to homework 5 of the amsc/cmsc 460 course, which covers the reduction of a hessenberg matrix to triangular form using gaussian elimination with partial pivoting, and the properties of positive definite matrices. It includes the implementation of a matlab function hesssolve(b, p, mult, u) to solve the system hx=b, and the proof of various facts related to positive definite matrices.
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Answer:
n = size(H,1); for i = 1:n- if p(i) temp = b(i); b(i) = b(i+1); b(i+1) = temp; end b(i+1) = b(i+1) - mult(i)*b(i); end
for j=n:-1: x(j) = b(j)/L(j,j); b(1:j-1) = b(1:j-1) - L(1:j-1,j)*x(j); end
Answer:
a. If A is singular then there is a nonzero vector x such that Ax = 0. But then 0 = xT(Ax) = xTAx > 0 , a contradiction.
b. Let x 6 = 0. Then y = A−^1 x 6 = 0. Since x = Ay, we have
xTA−^1 x = yTATA−^1 Ay = yTAA−^1 Ay = yTAy > 0.
c. Let x = (x 1 x 2 · · · xk) T 6 = 0. Construct a vector as follows
y = zeros(n); for j=1:k y(i(j)) = x(j); end
Then it is easily verified that yTAy = xTBx. But since y 6 = 0, we have yTAy > 0.
d. Assume, to the contrary that aij (i 6 = j) is an element of largest magnitude and consider the matrix
B =
aii aij aij ajj
which is positive definite. Now if aij ≥ 0 , then
aii aij aij ajj
= aii + ajj − 2 aij ≤ 0.
On the other hand if aij ≤ 0 , then
aii aij aij ajj
= aii + ajj + 2aij ≤ 0.
In either case we find that B is not positive definite — a contradiction.
e. First suppose that B has linearly independent columns. Then for any nonzero x, y = Bx 6 = 0. Hence xT(BTB)x = (Bx)T(Bx) = yTy = ‖y‖^22 > 0
so that A = BTB is positive definite.
On the other hand if A = BTB is positive definite, then for any nonzero x,
0 < xT(BTB)x = ‖Bx‖^22.
Thus if x 6 = 0, then Bx 6 = 0, and hence the columns of B are independent.