Inner Product Spaces of 2x2 Matrix

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Homework Help Overview

The discussion revolves around the properties of an inner product defined for 2x2 matrices, specifically examining the expression = u1.v1 + u2.v3 + u3.v2 + u4.v4. Participants are tasked with demonstrating that this expression does not satisfy the criteria for an inner product.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the axioms of inner products, noting that symmetry, additivity, and homogeneity appear valid, while questioning the positivity axiom. There is discussion about potential typos in the expression and attempts to construct specific matrices to test the positivity condition.

Discussion Status

Participants are actively engaging with the problem, suggesting various approaches to demonstrate the failure of the positivity axiom. Some have proposed testing specific 2x2 matrices with both positive and negative entries to explore the conditions under which the inner product could be negative.

Contextual Notes

There is a noted confusion regarding the correct formulation of the inner product expression, as well as the challenge of finding a suitable matrix that meets the criteria for demonstrating the failure of the positivity axiom. Participants are encouraged to use trial and error within the constraints of 2x2 matrices.

abbasb
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Homework Statement



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

Homework Equations


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

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.
 
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abbasb said:

Homework Statement



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

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

Homework Equations


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

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.
 
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.
 
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:
Mark44 said:
I'm leaning toward the positivity thing, too. Can you come up with a 2x2 matrix U for which <U, U> = u12 + u2u3 + u3u2 + u22 < 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.

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?
 
abbasb said:
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?
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.
 

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