# Find an upper bound

1. Jan 23, 2010

### tom08

1. The problem statement, all variables and given/known data

assume that x and y are vectors, and A is a matrix.

can anyone kindly help me to find an upper bound C w.r.t $$\| A \|$$ s.t.

$$\| x-Ay \| \leq C \cdot \| x-y\|$$

2. Jan 23, 2010

### CompuChip

Some quick-and-dirty trial gives me
C = sup( ||(A - I) v|| )
where I is the identity matrix and the supremum is taken over all vectors v.

I wonder if you can do any better, without more information on A.

3. Jan 23, 2010

### tom08

Thank you for ur kind help. if all the entries of A is between 0 and 1, can we get a nicer upper bound ?

4. Jan 23, 2010

### tom08

Could u show me any hints about ur estimate for C.

I only figure out that $$\|x-Ay\|=\|(x-Ax)+(Ax-Ay)\|\leq\|I-A\| \|x\|+\|A\| \|x-y\|$$

I don't know how to continue... could anyone kindly give me more hints ?

Last edited: Jan 24, 2010
5. Jan 24, 2010

### CompuChip

I did
|| x - A y|| = || (x - y) + (y - Ay) ||

But when all entries of A are between 0 and 1, then you can define ||A|| by
|| A || = max(i, j)( |Aij| )
and use that to get a better estimate.

6. Jan 24, 2010

### tom08

if || x - A y|| = || (x - y) + (y - Ay) ||,
then || x - A y|| <= || (x - y) || + || (y - Ay) ||,

but how could u find that C = sup |x-Ax| for all x ?

notice that my esitmate is C*||x-y||

7. Jan 25, 2010

### tom08

can someone give me a hand?