What does singularity have to do with whether or not it is a Vector Space?

Click For Summary
The set of all 2x2 singular matrices is not a vector space because it fails to meet the closure property under addition. A counterexample demonstrates this: adding two singular matrices results in a non-singular matrix. This violation indicates that at least one of the vector space axioms is not satisfied. Specifically, while individual matrices may be singular, their sum can be non-singular, disqualifying the set from being a vector space. Therefore, the conclusion is that the set of singular matrices does not form a vector space.
dietcookie
Messages
15
Reaction score
0

Homework Statement



Determine whether the set, together with the indicated operations, is a vector space. If it is not, identify at least one of the ten vector space axioms that fails.

"The set of all 2 x 2 singular matrices with the standard operations."



Homework Equations





The Attempt at a Solution



This is from the solution guide,

This set is not a vector space because it is not closed under addition. A counterexample is,
1 0 + 0 0 = 1 0
0 0 0 1 0 1
Each matrix on the left is singular, while the sum is non-singular.

I must be drawing a blank, because I don't see how singularity (in this case singular, meaning it is not invertible and the determinant equals 0) has to do with the 10 axioms and the counterexample proves that at least one of the axioms has failed. Thanks guys.

I'm having trouble using latex for a matrix, we have singular matrix A and B A + B produces a non-singular matrix.
A=
[1 0]
[0 0]

B=
[0 0]
[0 1]

C=
[1 0]
[0 1]
 
Last edited:
Physics news on Phys.org
Hint: Look at the definition of a vector space, V. If A and B are elements of V, then what about A + B? What about scalar multiplication with A or B?

You are asked to say whether the set of all 2 x 2 singular matrices with standard operations is a vector space. Look at a couple examples of 2 x 2 singular matrices.
 
You aren't drawing a blank. You are exactly correct that that set of matrices isn't closed under addition. End of discussion. Not a vector space.

[Edit] Woops. I didn't see that was from the solution guide. Oh well, maybe you see now.
 
I'm having a hard time grasping this, am I going about this the right way? In the counterexample, it seems to me that they verified axiom 1 failed, A and B are in the set (of singular matrices) but A+B is not in the same set (because it is singular?). Is that all it is? Because A+B is non-singular, it is not an element in the 2x2 singular matrix set.
 
dietcookie said:
I'm having a hard time grasping this, am I going about this the right way? In the counterexample, it seems to me that they verified axiom 1 failed, A and B are in the set (of singular matrices) but A+B is not in the same set (because it is singular?). Is that all it is? Because A+B is non-singular, it is not an element in the 2x2 singular matrix set.

Yes, that's all there is to it. When you add two elements of a vector space you must get another element of the vector space, and in this case, you don't always. So it isn't a vector space.
 
Note that to prove something is a vector space, you must verify that all of the axioms are true for all vectors in the set. To prove that it is not a subspace you only have to provide a counter example to one axiom.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K