# Why is the set of all nonsingular 3x3 matrices not a vector space over the reals?

1. Oct 27, 2009

### neergmas

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

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

2. Relevant equations

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

A2 Existence of 0
A3 Existence of negative

3. 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?

2. Oct 27, 2009

### D H

Staff Emeritus
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.

3. Oct 27, 2009

### neergmas

Thank you for your response, D H.

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?

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.

4. Oct 27, 2009

### D H

Staff Emeritus
Yes.

Try that again. Particularly the middle two.

5. Oct 27, 2009

### neergmas

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.

6. Oct 27, 2009

### D H

Staff Emeritus
You're problem is here:

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.

7. Oct 27, 2009

### neergmas

Lol. Of course. Your reply also clears up the problem with A3.

Thanks!

8. Oct 28, 2009

### willem2

Multiplication with a scalar is another problem. What is 0*A ?