Why 3x3 Matrices Don't Form a Vector Space Over Reals

AI Thread Summary
The set of all nonsingular 3x3 matrices does not form a vector space over the reals because it fails to meet the necessary conditions for closure under addition. Specifically, the sum of two nonsingular matrices can result in a singular matrix, violating the requirement that for all x, y in V, x + y must also be in V. Additionally, the existence of a zero vector is problematic, as the zero matrix is singular and not part of the set. The discussion highlights the importance of understanding the definitions and properties of vector spaces, particularly regarding closure and the existence of a zero element. Overall, nonsingular 3x3 matrices do not satisfy the criteria to be considered a vector space over the real numbers.
neergmas
Messages
18
Reaction score
1

Homework Statement



The set of all nonsingular 3x3 matrices does not form a vector space over the real numbers under addition. Why?

Homework Equations



A vector space over F, under addition, is a nonempty set V such that

A1 Addition is associative
A2 Existence of 0
A3 Existence of negative
A4 Addition is commutative

The Attempt at a Solution



Is the reason because the sets of all nonsingular 3x3 matrices include those composed of complex numbers which are not reals and therefore the addition of such matrices, which all satisfy A1-A4, are not over the reals?
 
Physics news on Phys.org
neergmas said:
Is the reason because the sets of all nonsingular 3x3 matrices include those composed of complex numbers which are not reals and therefore the addition of such matrices, which all satisfy A1-A4, are not over the reals?
The question doesn't say anything about complex numbers. Why bring them in?

Your list of conditions is incomplete (you have omitted a very important condition). However, that doesn't matter. The set of all non-singular 3x3 matrices fails at least one of the conditions that you did list.
 
Thank you for your response, D H.

D H said:
The question doesn't say anything about complex numbers. Why bring them in?

Because the 3x3 matrices could be composed of complex numbers and that would automatically disqualify them from being elements of a vector space over the real numbers? Is that reasoning nonsense?

D H said:
Your list of conditions is incomplete (you have omitted a very important condition). However, that doesn't matter. The set of all non-singular 3x3 matrices fails at least one of the conditions that you did list.

Is the omission: For all x,y \in V, x+y \in V?

Which condition does it fail? Say we are given A,B,C \in R_{3x3}:

(A+B)+C = A+(B+C)? Yes
A+0 = A? Yes
Let D = -1A. A+D=0? Yes
A+B = B+A? Yes

I really want to understand the concept of "vector space". Thank you for your help.
 
neergmas said:
Is the omission: For all x,y \in V, x+y \in V?
Yes.


Which condition does it fail? Say we are given A,B,C \in R_{3x3}:

(A+B)+C = A+(B+C)? Yes
A+0 = A? Yes
Let D = -1A. A+D=0? Yes
A+B = B+A? Yes
Try that again. Particularly the middle two.
 
I don't see it. Is it the fact that we are talking about vectors and a 3x3 matrix is not a vector -- of course a 3x3 matrix can be thought of as a partitioned 3x1 column vector or a 1x3 row vector. But I still think those meet the criteria. There is something in the definition that I don't understand.

The 3x3 sets must "form a vector space" under addition. Which means that all members of 3x3 must meet the four listed criteria. And as long as the member comes from R3x3, I don't see the problem.
 
neergmas said:
I don't see it.

You're problem is here:

neergmas said:
A+0 = A? Yes
What is this "0" you use without proving its existence? It most certainly is not the matrix \bmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\endbmatrix, because that matrix is quite singular.
 
Lol. Of course. Your reply also clears up the problem with A3.

Thanks!
 
Multiplication with a scalar is another problem. What is 0*A ?
 
Back
Top