# Thread: Why is outer product not backward stable?

1. ## Why is outer product not backward stable?

So I am new to the topic of numerical analysis and was trying to figure out why the outer product of 2 vectors is not backwards stable. I understand the equation would be f(a,b) = ab* and that the resulting matrix is of rank 1. My book says that it is not backwards compatible simply because ~f(a,b) is not of rank 1 which means you cannot rewrite the function ~f(a,b) = (a)(1 + e)((b)(1 + e))*. I was trying to set up the proof to ultimately disprove it but I did not even know how to do that. I assume the reason why it is not backwards compatible is that each element in the resulting MxN matrix is going to have a different error associated with it so thats why its unlikely to have a rank of 1. If anyone can clear this up for me I would greatly appreciate it!

2. ## Re: Why is outer product not backward stable?

Hey redhawk87.

Just to clarify when you mean the product do you mean a*b^T (as in transpose) where you get a number? (So basically it's an inner product?)