Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know this sounds like a studpid question in here, but I just want to organize and clear what I studied.

For an $n\times n$ matirx $A$, it has independent columns when nullspace only has zero vector. And independent columns mean $A$ has rank $n$, therefore by the rank theorem, nullspace has zero dimension. That is, zero vector is zero dimension, is that right?

AND one more thing. I want to show that $\lbrace Av_1,...,Av_n \rbrace$ span $R^n$ when $\lbrace v_1,...,v_n \rbrace$ form a basis. Dimension theorem is used in here? If so, how can I show that $\lbrace Av_1,...,Av_n \rbrace$ span $R^n$?

share|cite|improve this question
Vectors don't have a dimension. – Qiaochu Yuan Dec 3 '12 at 8:23
up vote 4 down vote accepted

Yes but here's a minor nit pick: A vector doesn't have a dimension, you want to say that the subspace spanned by the zero vector has dimension zero.

For the second question you may use the rank nullity theorem. You have $\mathrm{dim} \mathrm{ker} A = 0$ and hence $\mathrm{dim} \mathrm{im} A = \mathrm{dim}R^n - \mathrm{dim} \mathrm{ker} A = \mathrm{dim}R^n$ hence the $v_i$ span $R^n$.

share|cite|improve this answer
Yes, but the problem is that I have no idea how to apply the rank-nullity theorem in here. Please state details then I can follow it. – email Dec 3 '12 at 8:16
@email I added some detail. Hope this helps. – Rudy the Reindeer Dec 3 '12 at 8:21

The concept of dimension is applied to sets of vectors, in particular subsets of vector spaces that are also subspaces. Thus, it more appropriate to say that the subspace consisting of the zero vector has dimension zero.

If you are assuming that $A$ is a square matrix and has independent columns, it has maximal rank. A matrix with maximal rank is invertible and the linear map induced by left multiplication by $A$ is an isomorphism. Isomorphisms always map a basis to another basis.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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