# Homework Help: Inner Product Spaces of 2x2 Matrix

1. Sep 17, 2009

### abbasb

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

Show that <U,V> = u1.v1 + u2.v3 + u2.v3 + u4.v4 is NOT an inner product on M2x2

2. Relevant equations
U: row 1 = [u1 u2] row 2 = [u3 u4] V: row 1 = [v1 v2] row 2 = [v3 v4]

3. The attempt at a solution

As I went through each of the axioms, I found that they were all correct (at least the way I did it), I've been stuck for quite some time now. I found that the symmetry axiom, additivity axiom and homogeneity axiom all remained true. So I'm assuming the positivity axiom doesn't hold true, I'm just confused as how to show this.

2. Sep 17, 2009

### Staff: Mentor

Your formula looks like it might have a typo, with u2.v3 appearing twice. Is that how the problem is worded?

3. Sep 17, 2009

### abbasb

Sorry, its supposed to be <U,V> = u1.v1 + u2.v3 + u3.v2 + u4.v4

and U and V are supposed to be matrices...couldn't find a better way to show them as so.

4. Sep 17, 2009

### Staff: Mentor

I'm leaning toward the positivity thing, too. Can you come up with a 2x2 matrix U for which <U, U> = u12 + u2u3 + u3u2 + u42 < 0? Try playing around with a mix of positive and negative numbers for which the sume of u2u3 and u3u2 is more negative than the two squared terms are positive.

Last edited: Sep 17, 2009
5. Sep 17, 2009

### abbasb

But shouldn't I be looking for a 2x2 matrix such that <U,U> = u1^2 + 2(u2.u3) + u4^2 < 0? And the only way I could find such a matrix is through trial and error?

6. Sep 17, 2009

### Staff: Mentor

First question: yes. u12 and u42 are always going to be nonnegative, so can you fiddle with values of u2 and u3 so that 2u2u3 is more negative than the two squared terms are positive?
Second question: Yes, but using judicious trial and error. You're only dealing with 2x2 matrices, and you can start with 1s in the main diagonal positions. That leaves you only two other positions to fill in.

BTW, I edited my previous post to fix a subscript.