# Does zero vector have zero dimension?

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

For an $$n\times n$$ matrix $$A$$, 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 $$R^n$$?

• Vectors don't have a dimension. – Qiaochu Yuan Dec 3 '12 at 8:23

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 $R^n$.
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.