# Math subject GRE exam 9768 Q.30 (condition that a basis of a real vector space must satisfy)

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?

• Every basis is a maximal linearly independent subset, so no linearly independent subset contains a basis as its proper subset. Jul 29, 2017 at 10:55

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.
$C$ 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 $B$.
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 $V$.