Suppose we have three vectors which are linearly independent. So they would be able to span a 3 dimensional vector space. But how we are sure that number 3 would be enough to span 3-D vector space or in general N independent vectors are enough to span a N dimensional space? Actually I am looking for some intuition and a rigorous mathematical proof .
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In a nutshell - too many vectors and some are linear combinations of others and can be thrown out. Too few, and one can find some $x$ in the space that is not in the span of that set. – Sean Roberson Nov 11 '21 at 17:21
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Suppose $W$ is a subspace of $V$ such that $\dim(W)=3=\dim(V)$. Further suppose there is some $v\in V$ that doesn't belong to $W$. Can you show that ${w_1,w_2,w_3,v}$ is linearly independent, where ${w_1,w_2,w_3}$ is a basis for $W$? What does this mean? – Matthew H. Nov 11 '21 at 17:37
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It's essentially the definition of "dimension", and the standard Exchange Theorem (aka Replacement theorem) showing that the dimension is well-defined. This is covered in multiple posts in this site: here, here, here, here, and here, among many other places. – Arturo Magidin Nov 11 '21 at 17:39