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P-Jay1
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Prove that the space of 2×2 real matrices forms a vector space of dimension 4 over
R. [12]
Im unsure, anyone any idea?
R. [12]
Im unsure, anyone any idea?
ker T is the set of all x such that Tx=0. Finding that isomorphism is one way to do it, but not the only one. Edit: Wait, you said "from T[a,b;c,d]". I don't know what you mean by that. I meant that one way to solve the problem is to find an isomorphism from the set of 2×2 matrices to ℝ4.P-Jay1 said:So what does ker(T) stand for? I am still really clueless. So do I find an isomorphism from T[a,b;c,d] to (a,b,c,d)?
Fredrik said:ker T is the set of all x such that Tx=0. Finding that isomorphism is one way to do it, but not the only one. Edit: Wait, you said "from T[a,b;c,d]". I don't know what you mean by that. I meant that one way to solve the problem is to find an isomorphism from the set of 2×2 matrices to ℝ4.
A vector space dimension refers to the number of linearly independent vectors needed to span a given vector space.
The dimension of a vector space can be found by determining the number of vectors in a basis for the vector space. A basis is a set of linearly independent vectors that span the vector space.
The dimension of a vector space is important because it helps to describe the size and complexity of the vector space. It also plays a crucial role in determining the number of solutions to a system of linear equations.
No, the dimension of a vector space is a fundamental property of the space and cannot change. However, the dimension of a subspace of a vector space may be smaller than the dimension of the original vector space.
Yes, there are two types of vector space dimensions: finite and infinite. A finite vector space has a finite number of vectors in its basis, while an infinite vector space has an infinite number of vectors in its basis.