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

[andytran]

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!

 

[Justin Lazear]

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

 

[learningphysics]

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.

 

[andytran]

but that only be true to whatever counterexample i pick wouldn't it?

 

[Justin Lazear]

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

 

[andytran]

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..