Are all 2x2 Matrices with det(A) = 0 a Subspace of M2x2?

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Homework Statement


Determine whether all 2x2 matrices with det(A) = 0 are a subspace of M2x2, the set of all 2x2 matrices with the standard operations of addition and scalar multiplication.

Homework Equations


Must pass in order to be a subspace
Closure property of addition - If w and v are objects in A, then w+v are contained within A
Closure property of scalar multiplication - If K is any real number scalar and v is any object in A, then kv is also in A

The Attempt at a Solution


I wasn't sure where to really start with this one so I picked a matrix with a determinant of 0
B2x2 = [[w1,w2][w1,w2] and added it to another det=0 matrix C2x2 = [[v1,v1][v2,v2]]

Added together they make D2x2 = [[w1+v1,w2+v1][w1+v2,w2+v2]] which wouldn't have a det of 0 so this wouldn't be a subspace right?
 
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You have the right idea, but to make this a proof you need to either (1) give specific values for v_1, v_2, w_1, w_2 which show that the result matrix is invertible, or (2) compute the determinant of your result matrix in terms of v_1, v_2, w_1, w_2 and prove that this expression takes nonzero values.
 
Ok then an example that would disprove this as a subspace would be

[[3,13][3,13]]+[[5,5][7,7]] = [[8,18][10,20]]

det of [[8,18][10,20]] = -20 so it fails
 
Yes, this is correct.

There is also a simpler example (which doesn't match the pattern you gave): \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) + \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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