Solving for k to Satisfy ||kA-I||<1

[saltine]

Homework Statement

How do I analytically solve for k to satisfy:
$$||kA-I||<1$$?
Here, k is a real number scalar, A is a known matrix.

Homework Equations

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

 

[HallsofIvy]

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

 

[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?

 

[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).

 

[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

 

[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.