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I am sure that A,B and E are wrong, but I do not know which is right C or D, and why, could anyone help me please?

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    $\begingroup$ Every basis is a maximal linearly independent subset, so no linearly independent subset contains a basis as its proper subset. $\endgroup$
    – Hanul Jeon
    Jul 29, 2017 at 10:55

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Take nontrivial scalar multiples of elements of $B $. This gives a basis $B'$ which is disjoint from $B $. I.e. if $B=\{b_1, b_2,\dots,b_n\}$, let $B'=\{cb_1, cb_2,\dots, cb_n\}$, where $c\neq1$.Then $ B'$ and $B $ are disjoint. The answer is D.

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$C$ is also wrong. Basis is a maximal linearly independent set. Thus, $B$ can't be a proper subset of a linearly independent set.

Thus, the correct answer is $D$.

For example, let $V=R^2$ and $B=\{(0,1),(1,0)\}$. Then $\{(1,2),(3,4)\}$ is also a basis of $V$ disjoint from $B$.

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I think $D$ is true: take a suitable invertible map on $V$ such that no member of $B$ is mapped to any other, and take the image of $B$. E.g. Multiply all basis elements by $2$.

$C$ fails as a basis is a maximal linearly independent subset of $V$.

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