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I know that linear independence means each vector is not the linear combination of the others. But, I don't know why when we check whether a set of vectors are linearly independent, we only check for the equation $a_1v_1 + a_2v_2 +... + a_nv_n = 0$ has only trivial solution $x_1 = x_2=...=x_p=0$.

Why not check for other vectors, like $(1,0,...,0)$? Is it enough to just check for the $0$ vector and say that it is linearly independent?

My thought:

The argument of linear dependence is more easy for me:

If a system of vector $v_1,v_2,...,v_n$ is called linearly dependent if $0$ can be represented as nontrivial linear combination.

I am think does the above line implies that "if a system of vector $v_1,v_2,...,v_n$ is called linearly independence if $0$ can only be represented as trivial linear combination." which is the statement of linearly independence.

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    $\begingroup$ The idea is that if a bunch of things add to $0$, for example, $a + b + c = 0$, then you can just move one of those to the other side, eg $c= -a -b$ and suddenly the value of $c$ is entirely determined by $a$ and $b$, so it's not really a free variable anymore. It's only when we can't do this (the criterion of linear independence) that we know that all the variables (vectors, etc) are actually independent of one another. $\endgroup$ Aug 24, 2019 at 5:43

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This is the definition of linear independence. It is equivalent to the statement that none of the vectors $v_1,\dots,v_n$ can be written as a linear combination of the remaining vectors.

Here's why $0$ is special: You can always write it as a linear combination of any vectors, by taking $0v_1+\dots+0v_n$. Linear independence is the criterion for that to be the only solution. If you picked $(1,0,\dots,0)$, you wouldn't know, first of all, that it is a linear combination of $v_1,\dots,v_n$ at all, and, even if it were, you wouldn't know what linear combination had to work.

Here's an equivalent formulation of linear independence: Any vector $v$ that is in the span of $v_1,\dots,v_n$ (i.e., that can be written as a linear combination of them) must be a unique linear combination of them. There cannot be two different ways of writing it as a linear combination.

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  • $\begingroup$ This comment is merely to provide another way of phrasing the last paragraph (which I found helpful when learning). A set of vectors $\beta = \{v_1, \dots, v_n\}$ in a vector space $V$ is linearly independent if the "linear-combination mapping" $L_{\beta}: \Bbb{R}^n \to V$ defined by $L_{\beta}(a_1, \dots, a_n) = \sum_{i=1}^n a_iv_i$ is an injective mapping (injectivity captures the idea of uniqueness). $\endgroup$
    – peek-a-boo
    Aug 24, 2019 at 5:12
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It's all the same in the end. If you decided to show that $\sum a_i\textrm{v}_i=\sum b_i\textrm{v}_i=\textrm{u}$, then $\sum (a_i-b_i)\textrm{v}_i=\textrm{u}-\textrm{u}=\textrm{0}$ is a non-trivial combination of the zero vector. Conversely, if you knew $\sum z_i\textrm{v}_i=\textrm{0}$ was a non-trivial combination of the zero vector and you knew $\sum a_i\textrm{v}_i=\textrm{u}$, then $\sum (a_i+z_i)\textrm{v}_i=\textrm{u}+\textrm{0}=\textrm{u}$ would be a different linear combination.

The zero vector is probably the one people traditionally lean toward because it is specifically privileged in the space, and is at least nearly always the easiest vector to solve before. Also, it gives us some information about the nullspace.

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You could just as easily say that you are checking if any one of $v_1,\dots, v_n$ is a linear combination of the others: For say $a_1v_1+\dots+a_nv_n=0$ nontrivially. Then one of the $a_i\neq0$, and so, solving for $v_i$ we get $v_i=\dfrac {-a_1}{a_i}v_1+\dots +\hat{v_i}+\dots+\dfrac {-a_n}{a_i}v_n$ (where $\hat v$ means $v$ is omitted).

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