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Exercises 29 and 30 show that every basis for Rn must contain exactly n vectors.Let $S=\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\}$ be a set of $k$ vectors in $\mathbb{R}^{n},$ with $k<n$ .Use a theorem from Section 1.4 to explain why $S$ cannot be a basis for $\mathbb{R}^{n}$ .

the columns of the matrix do not span $\mathbb{R}^{n}$ .

Calculus 3

Chapter 4

Vector Spaces

Section 3

Linearly Independent Sets; Bases

Vectors

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in this problem, we're given a setup vectors as and we're told that this set of K vectors ah isn't an intermission space. And and we're also told that Kay is greater than ain't so this system point by this vectors would be like a and a system metrics, would we would have a number of roles and K number of homes. So since chase greater than a, then it means that Rosa A would be less than cold. So this implies that, uh, there wouldn't be enough. Give it elements for each home. So they cannot the revenge or each hole. And using your abate being section 1.7 or your textbook. This system for the columns can not be linearly independent, so that cannot forma basis.

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