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In Axler's Linear Algebra Done Right,

Beside the proof of the Replacement Theorem: that is the length of the list of linearly independent vectors is less than or equal to the list of spanning vectors, there is a comment:

"Suppose that for each positive integer m, there exists a linearly independent list of m vectors in V . Then Replacement Theorem implies that V is infinite dimensional."

How does Replacement Theorem imply this?

Thanks.

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2 Answers 2

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The dimension of the space is defined as the cardinality of a basis set. Suppose the space is finite dimensional with dimension $n$. Then a maximal linearly independent set is $n$ elements. But for any $m\in\mathbb{N}$ there exists a linearly independent set with $m$ elements, so there certainly exists one for $n+1$ which contradicts the fact that the space has dimension $n$. The space is consequently infinite dimensional.

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A corollary of the Replacement Theorem is that every linearly independent set can be extended to form a basis for the vector space. So if linearly independent sets can be arbitrarily large, then the only possibility is that a basis must have infinitely many elements.

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