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I am going through 3b1b linear algebra video, and at the end of 2nd video in the playlist, he defines basis as The basis of vector space is a set of linearly independent vectors that span the full space.

My question is, isn't it redundant to say the part that span the full space because a set of linearly independent vector always span a subspace or a space. For example, if we have two 3x1 independent vectors then their span a 2d subspace in a 3d vector space.

So, why it is not sufficient just to define a basis as a set of linearly independent vectors?

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    $\begingroup$ Because not all sets of linearly independent vectors spans the whole space. $\endgroup$
    – lulu
    Aug 29, 2021 at 16:59
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    $\begingroup$ Those linearly independent vectors might not be enough to "grab" every vector in the space. For instance, $(0,0,1)$ and $(0,1,0)$ are independent in $\mathbb{R}^3$, but they cannot generate things with non-zero first coordinate. $\endgroup$
    – Randall
    Aug 29, 2021 at 17:01

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Let $V=\mathbb{R}^3$, the vectors $(1,0,0)$ and $(0,1,0)$ are linearly independent. As you say they certainly form a basis for some space, namely $\mathbb{R}^2\times\{0\}$. They do not however form a basis for all of $\mathbb{R}^3$.

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    $\begingroup$ So the number of vectors in a basis will ALWAYS equal to the dimension of the vector space? $\endgroup$ Aug 29, 2021 at 17:10
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    $\begingroup$ @QuaziIrfan Yes, but that's because the dimension of a vector space is defined to be the number of elements of one of its bases. All bases of the same vector space have the same number of elements! $\endgroup$ Aug 29, 2021 at 19:16
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    $\begingroup$ @QuaziIrfan We just call them linearly independent. $\endgroup$ Aug 29, 2021 at 20:16
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    $\begingroup$ "So the number of vectors in a basis will ALWAYS equal to the dimension of the vector space?". You need to be careful with this. It assumes a finite dimensional vector-space - but not all vector spaces are finite dimensional. And you can have an infinite set of linearly independent vectors that dont span the space. (Consider the space of all polynomials - this is infinite dimensional, and the set of even powers of x are an infinite set of linearly independant elements, but x is not in span({1,x^2, x^4, ...}) ) $\endgroup$ Aug 30, 2021 at 3:37
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    $\begingroup$ @MichaelAnderson While this is true, it doesn't contradict your quoted sentence, which says that the number of vectors in any basis is the same as the dimension. The converse of this is false (there are sets of linearly independent vectors the same size as the dimension without being a basis) but the forward direction is just the dimension theorem. $\endgroup$ Aug 30, 2021 at 7:07
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$\emptyset$ is a set of linearly independent vectors, for one example of a case where you have linearly independent vectors but far from spanning (though an extreme case, of course). As a less trivial example, consider $\{(1,0,0), (0,1,0)\} \subseteq \mathbb{R}^3$

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(An analogy too long for a comment)

Define $[n]$ to be the set $\{1,2,...,n\}$. We can define that a set $A$ has cardinality $n$ if and only if there is a function $f: [n] \rightarrow A$ which is one-to-one and onto. Your question is similar to asking why you can't drop the condition of being onto here. After all, if there is a one-to-one function between $[n]$ and $A$ then we can prove that $n$ is the cardinality of something (either the whole set or a subset).

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