# Dimension of a set of vectors

Is dimension of vector space spanned by a set of vectors the rank of the matrix ( vectors as columns) or its nullity.

Consider the vectors $$\alpha_1=(1,1,0,1,0,0),\\ \alpha_2=(1,1,0,0,1,0),\\ \alpha_3=(1,1,0,0,0,1),\\ \alpha_4=(1,0,1,1,0,0),\\ \alpha_5=(1,0,1,0,1,0),\\ \alpha_6=(1,0,1,0,0,1).$$ The rank of the matrix obtained is 4. And hence it is said the dimension is 4. But I had always thought that the dimension is (no. of columns - rank).

Reducing to echelon form $$\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1\\ 0& 0 & 0 & 1 & 1 & 1\\ 0& 0 & 0& 0 & 0& 0\\ 0& 0 & 0 & 0 & 0 & 0\end{bmatrix}$$ Evidently there are two free variables say $$u,v$$, therefore on writing the span, I got $$span=u(-1,0,-1,-1,0,1),v(-1,-1,0,-1,1,0)$$.

The span obtained is of dimension 2 and not 4. This is what confused me.

• Note that the third component of the vectors tells us no linear combination of $\alpha_1,\,\alpha_2,\,\alpha_3$ can be $\alpha_4$.
– J.G.
Jun 15, 2019 at 7:17
• Yes I get it. There are four linearly independent vectors and hence dimension of basis is 4. But what about the span I ve written above. Jun 15, 2019 at 8:14