I'm writing a linear algebra exam soon and I'd like to know a fast way to solve a task like that:
$$u=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}, v= \begin{pmatrix} 1\\
0\\ 1 \end{pmatrix}, w = \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}$$Show that $(u,v,w)$ is a basis of the vector space $\mathbb{R}^{3}.$
I would start by checking if these vectors are linearly independent. I do this by checking determinant $\neq 0$:
Using Sarrus rule, we know that determinant is $1$ and thus these vectors are linearly independent.
Now let $x,y,z \in \mathbb{R}$
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\lambda_{1}\begin{pmatrix}
0\\
0\\
1
\end{pmatrix}+ \lambda_{2}\begin{pmatrix}
1\\
0\\
1
\end{pmatrix}+ \lambda_{3}\begin{pmatrix}
0\\
1\\
1
\end{pmatrix}$$
$$x = \lambda_{2}$$
$$y = \lambda_{3}$$
$$z = \lambda_{1}+\lambda_{2}+\lambda_{3}$$
The solution is unique too and thus $(u,v,w)$ is a basis of the vector space $\mathbb{R}^{3}$
Did I solve the task correctly and can you tell me better ways if there are some?
Best Answer
Once you have proved that the $3$ vectors are linearly independent, you automatically have that they are a basis for $\mathbb{R}^3$, since they generate a subspace with dimension $3$ of a space of dimension $3$ - so they must generate the entire space! As for proving linear independence, the determinant approach proposed in the question is general and works well.
In this particular case, a simpler approach is to see that $v-u=(1 0 0)^T$, $w-u=(010)^T$, $u=(001)^T$ form what is called the canonical basis of $\mathbb{R}^3$, so $u,v,w$ must also form a basis of $\mathbb{R}^3$.