= Linear independence depends on the scalar field
{title2=$\dim_{\mathbb R}\mathbb C=2,\quad\dim_{\mathbb C}\mathbb C=1$}
Consider the <vectors> $1,i$ in the <complex numbers>, where $i$ is the <imaginary unit>. With <scalar> <field> $\mathbb R$, a <linear relation> $a\cdot1+b\cdot i=0$ has $a,b\in\mathbb R$. Taking its <real part> and <imaginary part> gives $a=b=0$, so the pair is <linearly independent>. Every <complex number> $x+iy$ also has <coefficients> $(x,y)$ in the <real numbers>, so $1,i$ form a <basis> and the <dimension of a vector space> is two.
With <scalar> <field> $\mathbb C$, the same pair is <linearly dependent>, since
$$
i\cdot1+(-1)\cdot i=0
$$
uses the nonzero <coefficients> $(i,-1)$ from the <complex numbers>. The single <vector> $1$ is now a <basis>, because $z=z\cdot1$ for every <complex number> $z$, giving <dimension of a vector space> one. For example $3+i=3\cdot1+1\cdot i=(3+i)\cdot1+0\cdot i$: the second representation uses a <coefficient> unavailable over the <real numbers>. Specifying the <field> is part of the definition of <linear independence>.
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