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Is the set of all 2x2 invertible matrices a subspace of all 2x2 matrices?
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.
Why you say matercis is vector?
In math, a "vector field" is an abstract term for a set, and a number of operations, that have specific properties. Matrices of the same size, for example, all 3 x 2 matrices, combined with matrix addition and multiplication by a scalar, happens to have all those properties. You may want to read an introductory Linear Algebra book for more details.
What is difference between vector spaces and field?
A vector space is a set of all points that can be generated by a linear combination of some integer number of vectors. A field is an abstract mathematical construct that is basically a set elements that form an abelian group under two binary operations, with the distributive property. Examples: Euclidean space(x,y,z) is a vector space. The rational and real numbers form a field with regular addition and multiplication. Also, every set of congruence classes formed under a prime integer (mod algebra) is a field.
Why null vector is called vector?
The same sort of reasoning that zero is a number. It ensures that the set of all vectors is closed under addition and that, in turn, allows the generalization of many operations on vectors.Also, the way we got around the concept of having something with zero magnitude also have a direction is pretty cool. We made it up! In abstract algebra it's perfectly OK to constrain a specific algebraic structure with rules (called axioms) that the structure must follow.In your example, the algebraic structure that vectors are in is called a "vector space." One of the axioms that define a vector space is:"An element, 0, called the null vector, exists in a vector space, v, such that v + 0 = vfor all of the vectors in the vector space."Ta Da!! Aren't we clever?
How to find orthogonal vector?
Given one vector a, any vector that satisfies a.b=0 is orthogonal to it. That is a set of vectors defining a plane orthogonal to the original vector.The set of vectors defines a plane to which the original vector a is the 'normal'.
