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Is the set of all 2x2 invertible matrices a subspace of all 2x2 matrices?
Wiki User
∙ 11y ago
I assume since you're asking if 2x2 invertible matrices are a "subspace" that you are considering the set of all 2x2 matrices as a vector space (which it certainly is).
In order for the set of 2x2 invertible matrices to be a subspace of the set of all 2x2 matrices, it must be closed under addition and scalar multiplication.
A 2x2 matrix is invertible if and only if its determinant is nonzero. When multiplied by a scalar (let's call it c), the determinant of a 2x2 matrix will be multiplied by c^2 since the determinant is linear in each row (two rows -> two factors of c). If the determinant was nonzero to begin with c^2 times the determinant will be nonzero, so an invertible matrix multiplied by a scalar will remain invertible. Therefore the set of all 2x2 invertible matrices is closed under scalar multiplication.
However, this set is not closed under addition. Consider the matrices {[1 0], [0 1]} and {[-1 0], [0 -1]}. Both are invertible (in this case, they are both their own inverses). However, their sum is {[0 0], [0 0]}, which is not invertible because its determinant is 0.
In conclusion, the set of invertible 2x2 matrices is not a subspace of the set of all 2x2 matrices because it is not closed under addition.
Wiki User
∙ 11y ago
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Is a to power of 4 multiplied by b to power of 5 invertible?
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