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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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  • $\begingroup$ Note that the third component of the vectors tells us no linear combination of $\alpha_1,\,\alpha_2,\,\alpha_3$ can be $\alpha_4$. $\endgroup$
    – J.G.
    Jun 15, 2019 at 7:17
  • $\begingroup$ 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. $\endgroup$
    – Harry
    Jun 15, 2019 at 8:14

2 Answers 2

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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?

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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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