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If $V$ is a vector space of finite dimension over $F$, then a basis of $V$ is a maximal, linearly independent set in $V$.

Is this conjecture true? If so, how to prove it?

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  • $\begingroup$ What is your definition of basis of a vector space? (There is a number of equivalent ways of defining a basis...) $\endgroup$ – Angelo Lucia Dec 16 '14 at 17:25
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Suppose $B=\{b_1,\dots,b_n\}$ is a basis for $V$ with $\dim V=n$. Then $B$ is necessarily a linearly independent set. Let $v_{n+1}\in V$ and consider $B'=B\cup\{v_{n+1}\}$. Since $v_{n+1}\in V$, it can be written as a linear combination of $v_1,\dots,v_n$. But that means $v_{n+1}$ is linearly dependent on $B$. So $B\cup\{v_{n+1}\}$ is not a linearly independent set and hence cannot be a basis for $V$.

So yes, your statement is true.

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Yes, it is true. Suppose $B=\{v_1,\ldots,v_n\}$ is a basis for $V$. Let $u\in V-B$ be any other vector. Then, by definition of a basis, $B$ spans $V$, so $$u=a_1v_1+\cdots+ a_nv_n$$ for some $a_i\in F$ and hence $B\cup\{u\}$ is linearly dependent. So $B$ is a linearly independent set such that adding any vector to it yields a linearly dependent set. That is the definition of a maximal linearly independent set.

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A basis is a spanning set, so can express any (other) vector as a linear combination. Rephrasing: to a basis, there is no such thing as another linearly independent element.


The theorem would appear to be true of any vector space, finite-dimensional or not.

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