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May 25th

May 25th

# MATH2605 Homework 6

## Problem 1

Problem 2 of Homework 5 had an error. Did you notice that? I have a fix so that you have a symmetric Hessian matrix.
Let’s do the problem again.

Compute a quadratic function f (x,y) that passes the point (2,3). At the point (2,3), the function f (x,y) has the
gradient (1,1), the maximum curvature 3 along the direction (2,1), and the minimum curvature 2 along the direction
(1,−2).

Now, let’s build a Hessian from curvature (not necessarily min. or max.) requirements.
Compute a quadratic function f (x,y) that passes the point (2,3). At the point (2,3), the function f (x,y) has the
gradient (1,1), the curvature 3 along the direction (2,1), the curvature 2 along the direction (1,1), and the curvature 1
along the direction (3,2). Notice that these directions are not eigenvectors.
Hint: Let , and solve for a ,b, and c.

## Problem 2

We will compute . What is a simple function f (x) whose solution is ? The function f (x) should not have .
Set up the Newton ite ration x = g(x) for f (x). Submit f (x) and x = g(x).
Using x0 = 3 as the initial guess, perform the iteration. Submit .
Using x0 = −0.1 as the initial guess, perform the iteration. Submit .

## Problem 3

We may start learning methods to orthonormalize a matrix. In this problem, we will learn an iterative method.

Suppose that a matrix is given, and we want to compute an orthogonal matrix that is close to A. In fact,
the matrix closest to A in a matrix norm can be computed from the singular value decomposition . If the SVD is
, the orthogonal matrix closest to A is . However, singular value decomposition is expensive to
compute and complex to implement . So, let’s use the Newton’s iteration that is simple and fast.

The idea is using A as the initial guess for the fol lowing Newton iteration, which hopefully converge to a nearby
orthogonal matrix. Consider . Its Newton iteration is

where .

Using Matlab, create a random matrix A by using the command A = (rand(3,3) - 0.5)*10;. Compute the orthogonal
matrix from the singular value decomposition of A. Submit A and , which is the orthogonal matrix closest
to A in a matrix norm.

If we apply the above iteration by A = A - 1/2*inv(A’)*(A’*A - eye(3));, we will see that A quickly converges to an
orthogonal matrix. The value of A will be fixed after a few iterations. Report per each iteration.
After A is converged, compare with ˜A. Are they same?
After A is converged, add a small random noise by using the command A = A + (rand(3,3)-0.5)*0.1;. Then, A will
no longer be orthogonal, but will be quite close to an orthogonal matrix since only small noise is added, and hence,
and . Therefore, we simplify the iteration (1) to obtain

which may converge if . Notice that this iteration does not require a matrix inverse, and therefore, much
cheaper and simpler.

Using the A (with the small noise) as the initial guess, try the iteration (2). Does the iteration converge to an orthogonal
matrix?

Add much larger noise by the command A = A + (rand(3,3)-0.5)*10; Using (2), does the iteration converge to an
orthogonal matrix?

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