Given set of vectors S $\epsilon$ $R^{n}$ and other 2 set of vectors X, Y $\epsilon$ $R^{n}$ such that
$X\subset S\subset Y$
Option A – False
Counter Example: S = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} \alpha \\ \beta\end{bmatrix}&\end{Bmatrix}$ – Linearly Dependent
X = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}&\end{Bmatrix}$ – Linearly independent
Option B – True
Let S = {$v_{1}$,$v_{2}$, $v_{3}$ }
X = {$v_{1}$,$v_{2}$}
Given that vectors of set S are linearly independent i.e, $c_{1}v_{1} + c_{2}v_{2} + c_{3}v_{3}=0$, where
$c_{1}=c_{2}=c_{3}=0$
Now we need to show prove X = {$v_{1}$,$v_{2}$} is linearly independent
i.e, $c_{1}v_{1}+c_{2}v_{2}=0$ has only trivial solution
Assume $c_{1}v_{1}+c_{2}v_{2}=0$
Add 0 on both sides
$c_{1}v_{1}+c_{2}v_{2}+0*v_{3}=0$
As it is given that vectors S = {$v_{1}$,$v_{2}$, $v_{3}$ } are linearly independent; $c_{1}=c_{2}=0$ is the
only possible solution.
(Subset of linearly independent set of vectors are always linearly independent)
Option C – True
Superset of linearly dependent set of vectors are always linearly dependent.
This can be done by making newly added vector redundant
Example: S= $\begin{Bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 4 \end{bmatrix}&\end{Bmatrix}$ – Linearly Dependent,
X = $\begin{Bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 4 \end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}&\end{Bmatrix}$ – still linearly dependent
$\begin{bmatrix} 2\\ 4 \end{bmatrix}=2 *$ $\begin{bmatrix} 1\\ 2 \end{bmatrix}$$+ 0* \begin{bmatrix} 0\\ 1 \end{bmatrix}$
Option D – False
Counter Example:
S = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}&\end{Bmatrix}$ – Linearly independent
Y = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} \alpha \\ \beta\end{bmatrix}&\end{Bmatrix}$ – Linearly dependent