Let $V$ be a vector space over $K$. Let $A\subseteq V$. We say that $A$ is linearly independent if every finite subset of $A$ is linearly independent.
Can you give an example?
Let $V$ be a vector space over $K$. Let $A\subseteq V$. We say that $A$ is linearly independent if...
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linear-algebra
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2$A = \{(1,0)\}$, $V =\Bbb R^2$ – 2017-02-26
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1Let $V$ be the set of fuctions from $\mathbb{R}$ to $\mathbb{R}$ as a vector space over $\mathbb{R}$. Then $A=\left\{ e^{\lambda t}:\lambda \in \mathbb{R} \right\}$ is lin. indep, right? – 2017-02-26
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0You also have the set of sequences $\Bbb R^\Bbb N$ (*or [$\ell^\infty$](https://en.wikipedia.org/wiki/Sequence_space) or however else you wish to refer to it as*) and the linearly independent set $\{(1,0,0,0,\dots),(0,1,0,0,\dots),(0,0,1,0,\dots),\dots\}$. there are of course infinitely many other correct examples. – 2017-02-26
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1@Kahler Yes. That's also an example. – 2017-02-26
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0@Kahler are you specifically interested in examples of infinite sets $A$? – 2017-02-26
2 Answers
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Here's a possibly deceptive example: consider the space $\ell^2$ of square-summable infinite sequences. Let $$ e_1 = (1,0,0,0,\dots)\\ e_2 = (0,1,0,0,\dots)\\ e_3 = (0,0,1,0,\dots) $$ and so forth. Let $x$ be the sequence $$ x = (1,1/2,1/4,1/8,\dots) $$ You may be surprised to find that, by our definition, the set $A = \{x\} \cup \{e_i : i \in \Bbb N\}$ is actually linearly independent, even though we might say that $$ x = \sum_{i=1}^\infty \frac 1{2^{i-1}} e_i $$
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If $V$ is finite-dimensional, the standard basis $\{e_i\}_{i=1}^{n}$ is linearly independent since its every finite subset, including itself, is linearly independent. If $V$ is infinite dimensional, such as $F[x]$, the vector space of polynomials with coefficients in a field $F$, has a basis $\{x^i\}_{i=0}^{\infty}$, and it is linearly independent since its arbitrary finite subset, ${x^{i_1}, ...,x^{i_k}}$ is linearly independent, by definition of linear independence.