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I understood basis as a set of vector $v_{1},v_{2},...,v_{n}$ as the set whose linear combination will span the entire vector space say $ \mathbb R^{n}$ which makes perfect sense in intuitive terms. There are $n$ independent vectors, linear combination spans the entire space and the dimensions equals the number of linearly independent vector.

But I came across a very interesting question, which asked, is this a vector space? If yes, then find its dimension and basis. and asks this about,

  • all skew symmetric matrices of $2 \times 2$ dimension

Its interesting to note, it is associative,commutative under addition operator and scaling of the same is a subset of the same space. (do correct me if my choice of words is right here)

Which implies its a vector space, with a basis of $\begin{bmatrix} 0 & 1\\ -1 &0 \end{bmatrix}$ That implies dimension is 1.

So my question is, if basis indeed can span a linear system (represented by the matrix)? If my interpretation is right, then can anyone give me an intuitive "feel" of the basis, dimension and vector space.

Help much appreciated

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I don't understand the question. What do you mean by "linear system"? –  Qiaochu Yuan May 27 '12 at 5:28
    
A matrix represents a linear system isnt it? So a vector space can even represent a space of matrices which follow a certain property as shown above. –  Soham May 27 '12 at 5:32
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A matrix represents a lot of things. A vector space can be any set which is equipped with the appropriate operations, and that includes sets of matrices. –  Qiaochu Yuan May 27 '12 at 5:33
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Possibly related : (Although remotely) math.stackexchange.com/q/116717/22386 –  Inquest May 27 '12 at 6:22
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A basis has to have two properties: (i) the span must equal the entire space; and (ii) the set must be linearly independent. Also, most vector spaces have more than one basis, so it makes no sense to talk about "finding the basis" for a vector space. –  Arturo Magidin May 28 '12 at 21:01

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Qiaochu Yuan answered the question.

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