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Given the vector space $\mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.

Is it correct that these three vectors are linearly independent if $\mathbb{C}^3$ is defined over the field $\mathbb{R}$, while they are linearly dependent if the field is $\mathbb{C}$?

I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):

A finite set of vectors $\{x_i\}$ is linearly dependent if there exists a corresponding set $\{\alpha_i\}$ of scalars, not all zero, such that

$$ \tag{1}\label{eqn_li}\sum_i \alpha_i x_i = 0, $$

and the reason I'm asking is that $\eqref{eqn_li}$ is satisfied for the scalars $\alpha_1 = i$ , $\alpha_2 = 0$, $\alpha_3 = 0$, hence there is a set of scalars $\{\alpha_i\}$, not all zeros, such that $\eqref{eqn_li}$ holds.

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  • $\begingroup$ "...for the scalars..." . What scalars?? $\endgroup$ – DonAntonio Jan 12 at 17:44
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    $\begingroup$ The given set of scalars doesn't satisfy (1). $\endgroup$ – Thomas Shelby Jan 12 at 17:45
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    $\begingroup$ Hint: $\sum\alpha_jx_j=(\alpha_1,\alpha_2,\alpha_3)$. $\endgroup$ – David C. Ullrich Jan 12 at 18:03
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Those three vectors are linearly independent both over $\mathbb C$ and over $\mathbb R$.

However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $\mathbb C$ and linearly independent over $\mathbb R$.

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Your assumption that $ix_1+0x_2+0x_3=0$ is wrong. Your three vectors are linearly independent, no matter if we view $\Bbb C^3$ as three-dimensional space over $\Bbb C$, or six-dimensional space over $\Bbb R$ (or e.g., infinite-dimensional space over $\Bbb Q$) in the apparent way.

However, $(1,0,0)$ and $(i,0,0)$ are $\Bbb R$-linearly independant and $\Bbb C$-linearly dependant.

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