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

  • 2
    $\begingroup$ I don't understand the question. What do you mean by "linear system"? $\endgroup$ – Qiaochu Yuan May 27 '12 at 5:28
  • $\begingroup$ 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. $\endgroup$ – Soham May 27 '12 at 5:32
  • 4
    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan May 27 '12 at 5:33
  • 1
    $\begingroup$ Possibly related : (Although remotely) math.stackexchange.com/q/116717/22386 $\endgroup$ – Inquest May 27 '12 at 6:22
  • 1
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 28 '12 at 21:01

Qiaochu Yuan answered the question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.