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

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Homework Help Overview

The discussion revolves around determining whether all 2x2 matrices with a determinant of zero form a subspace of the set of all 2x2 matrices, M2x2, under standard operations of addition and scalar multiplication.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the closure properties required for a set to be a subspace, specifically examining the addition of matrices with determinant zero and questioning whether the result maintains a determinant of zero.

Discussion Status

Some participants have provided examples to illustrate their reasoning, while others suggest that specific values or determinant calculations are necessary to substantiate claims about the subspace properties. There is an acknowledgment of correct reasoning in the examples presented.

Contextual Notes

Participants are working within the constraints of proving subspace properties, specifically focusing on closure under addition and scalar multiplication, and are encouraged to provide concrete examples or calculations to support their claims.

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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).
 

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