Can the Set of 2x2 Matrices with Determinant Zero be a Subspace?

AI Thread Summary
The discussion revolves around proving that the set of all 2x2 matrices with a determinant of zero is not a subspace of M2x2. Participants emphasize that for a set to be a subspace, it must be closed under addition and scalar multiplication. A counterexample is needed to show that the sum of two singular matrices can yield a non-singular matrix, violating the closure property. The conclusion is that finding just one such counterexample is sufficient to demonstrate that the set is not a subspace. Ultimately, the original poster successfully understands the concept through the discussion.
andytran
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hi

i got this question for homework and after a few hours of head scratching, i still haven't figure it out...
the question is, prove that the set of all 2x2 matrices with determinant zero, is not a subspace of M2x2 (<-- M2x2 means just a notation for 2x2 matrices).

someone please give me some hints...

thanks!
 
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The space of singular 2x2 matrices, call it S, must be closed under addition and multiplication, i.e. for any scalar "c" in your field, and any two matrices "M" and "N" in S, the following are satisfied.

M + N \in S
and
cM \in S

so it suffices to find one counterexample such that the above conditions do not hold. You won't have much luck with the second equation, but try adding two singular matrices to make a non-singular matrix.

--J
 
Last edited:
andytran said:
hi

i got this question for homework and after a few hours of head scratching, i still haven't figure it out...
the question is, prove that the set of all 2x2 matrices with determinant zero, is not a subspace of M2x2 (<-- M2x2 means just a notation for 2x2 matrices).

someone please give me some hints...

thanks!

For M2x2 to be a subspace, for every two matrices A and B in M2x2, mA+nB
must also be in M2x2 (where m and n are any two real numbers)... in other words mA+nB must also be a 2x2 matrix with determinant zero.

If you can find two matrices A and B (that are both in M2x2) and two constants m and n, such that mA+nB does not have determinant zero... then you've proven M2x2 is not a subspace.
 
but that only be true to whatever counterexample i pick wouldn't it?
 
The closure conditions must be true for all elements in the space in order for it to be a subspace. If it's not true for one or more, it's not a subspace. Hence, you only need one counterexample.

--J
 
Justin Lazear said:
The closure conditions must be true for all elements in the space in order for it to be a subspace. If it's not true for one or more, it's not a subspace. Hence, you only need one counterexample.

--J

oh finally get it thanks

figured it out while trying to disprove you hehehe!

thx every1 for helping..
 

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