Matrix norm

1. Jan 9, 2009

saltine

1. The problem statement, all variables and given/known data
How do I analytically solve for k to satisfy:
$$||kA-I||<1$$?
Here, k is a real number scalar, A is a known matrix.

2. Relevant equations

3. The attempt at a solution
I am confused because if A was a number, I could break it into two cases where kA-I is positive and negative. But now A is a matrix so I can't do so. How should I look at the problem and what other equations are relevant?

Suppose the norm of A is 5, e.g. A = [5 0;0 5]. Then I know that the upper limit of k is 2/5, so that kA-I can be at most [1 0;0 1]. I also know that the lower limit of k has to be 0, because if k is ever negative, the norm would be greater than one, since the norm of I is already 1.

But how do I solve for this kind of result given an arbitrary matrix A?

- Thanks

2. Jan 10, 2009

HallsofIvy

Staff Emeritus
There are several different definitions of "norm" of a matrix. Which are you using here?

3. Jan 10, 2009

saltine

The largest gain in magnitude to a vector:

For y = Ax, the norm of A is the largest 'a' satisfying |y| = a|x| from all possible choices of x, where |.| is the 2-norm of the vector. Does this definition make sense?

4. Jan 10, 2009

Dick

The vector norm of Ax squared is Ax.Ax (dot product). So you can find the operator norm of A by finding the square root of the largest eigenvalue of A*A^(T).

5. Jan 19, 2009

saltine

So in my original equation where I am trying to find the unknown k such that $$||kB-I||<1$$, my $$A$$ is $$kB-I$$. To find the eigenvalues of $$AA^T$$, I do $$0 = \lambda I - AA^T = \lambda I - (kB-I)(kB-I)^T$$

Is there a way to pull out k so that I could solve for k in terms of the old eigenvalues or the old norm?

- Thanks

6. Jan 19, 2009

Dick

I don't think you can really 'pull out the k' in any useful way. I think you just have to put the matrix B in and crank it out. It could get pretty complicated unless the matrix is small.