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.
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$\begingroup$Appeal to intuition: What if the vectors were all zero? What ought the dimension be? Is that the rank or the nullity of the corresponding matrix?
$\endgroup$ $\begingroup$The rank of a matrix is defined as the dimension of the vector space spanned by its columns. Its not a consequence but a definition.
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