# linearly independent set that has many elements as possible

In $\mathbb{R}^2$ specify a linearly independent set that has many elements as possible.

What do such sets have in common?

In $\mathbb{R}^3$ do the same.

(I have an answer in mind, but I would like to know if this wording is clear... and if I have really answered my own question fully and correctly. Thank you.)

I removed the 2nd question since people are responding as if it is related or as if I don't know the definition of linear independence.

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The first question is clear. It appears to be a basic question designed to make you apply the definition of linear independence to familiar objects, the vector spaces $\mathbb{R}^2$ and $\mathbb{R}^3$. (That's why the responses so far give hints rather than solutions.)

You should note that linear independence is a property of sets of vectors-- so the "sets" in the first question are implied to be sets of vectors. So both questions are really about sets of vectors.

Asaf Karagila is pointing out that the property in question 2, that "no vector is a constant multiple of another vector in the set," is a different property than linear independence. In fact, question 2 has a very different answer than question 1. If you aren't sure of the answers, explain your thinking and your uncertainty, and I'm sure you'll get a more direct hint.

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"no vector is a constant multiple of another vector in the set," is a different property than linear independence. I'm well aware of that I've moved it to another question. –  a little don Oct 20 '10 at 13:29

think about what linear independence means, and make sure you understand the definition of "the subspace generated (or spanned) by the set {x,y,...}." Once you have done that, it should be clear. If you still have trouble try looking at the vector space $\mathbb{R}$ and do the same.

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I know from this response that the first question is "clear" Thanks. –  a little don Oct 20 '10 at 6:48

Just to clarify about the three dimensional case, it is not enough to ask that it won't be a multiple of another vector in the set, you must request for something stronger:

Suppose you have $\{v_1,\ldots ,v_k\}$ which are linear independent, then $v_{k+1}$ can be added to this set while retaining the independence if $v_{k+1}$ cannot be written as the sum: $c_1 v_1 + c_2 v_2 + \cdots + c_k v_k$, where $c_i$ are scalars (in this case real numbers).

(The said sum is called "linear combination", so using this term one can restate that $v_{k+1}$ is linearly independent of $\{v_1,\ldots, v_k\}$ if it cannot be written as a linear combination of $v_1,\ldots ,v_k$.)

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Is this an answer to the 2nd set of questions or the first set? I don't know what "it is not enough to ask that it won't be a multiple of another vector in the set" is referring to? –  a little don Oct 20 '10 at 6:57
Just trying to clarify that when the dimension is greater than 2 then linear independence is not equivalent to being a scalar multiplication of another vector. –  Asaf Karagila Oct 20 '10 at 7:46
It's not equivalent in dimension 2 either! –  Jonas Kibelbek Oct 20 '10 at 7:50
@Jonas, true. However I was referring to the case where you have a non-maximal independent set. Then it is true. You are very much correct though. –  Asaf Karagila Oct 20 '10 at 8:08
@Jonas, to clarify on my previous comment - when you have a pair of vectors you want to see if it's independent you only have to verify one is not a scalar multiple of the other. –  Asaf Karagila Oct 20 '10 at 8:21