Show that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for V $V=\mathbb{R}^3$
Let,
$\alpha_1=\begin{bmatrix}1\\0\\1\end{bmatrix},~\alpha_2=\begin{bmatrix}0\\1\\1\end{bmatrix},~\alpha_3=\begin{bmatrix}1\\-1\\2\end{bmatrix}$
Show that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for V and find its dual basis in $\mathbb{R}_3$.
I think we should show that elements are linear independent. (But is this enough for being basis?) I couldnt do the rest.
 A: $\text{Solution sketch:}$ Since the dimension of the vector space is $3$, if $\alpha_1,\alpha_2$ and $\alpha_3$ are linearly independent, then $\mathcal A=\{\alpha_1,\alpha_2,\alpha_3\}$ will be a basis for $V=\mathbb R^3$. Observe that
$$A=\begin{pmatrix}1&0&1\\0&1&-1\\1&1&2 \end{pmatrix}\overset{R_3-R_1}\sim \begin{pmatrix}1&0&1\\0&1&-1\\0&1&1 \end{pmatrix}\overset{R_3-R_2}\sim \begin{pmatrix}1&0&1\\0&1&-1\\0&0&2 \end{pmatrix}\implies\text{rk}(A)=3$$
This shows that $\mathbb R^3=\langle \alpha_1,\alpha_2,\alpha_3\rangle$.
If we want the dual basis for $V^*$, we have to find a set of linear maps $\mathcal A^*=\{\alpha_1^*,\alpha_2^*,\alpha_3^*\}=\{f_1,f_2,f_3\}$ such that $f_i(\alpha_j)=\begin{cases}1&\text{ if }i=j\\0&\text{ otherwise} \end{cases}$.
For the first function $f_1$ you have to impose $f_1(\alpha_1)=1$ and $f_2(\alpha_1)=f_1(\alpha_3)=0$ and solve the system. You do the same thing for the other maps.
A: One way of determining whether a set of vectors is linearly independent is to stick them in a determinant. If the determinant is not $0$ then you're done. So calculate
$$
\begin{array}{|ccc|}
\ 1 & 0 & 1 \ \\
\ 0 & 1 & 1 \ \\
\ 1 & -1 & 2 \ 
\end{array}
$$
and it shold give you a non-zero answer.
For your dual basis point, can you define it?
