Decide if the subset $ \left\{v_{1},v_{2}\right\}=\left\{
\begin{pmatrix} 1\\ 2 \end{pmatrix}, \begin{pmatrix} \alpha\\ 1
\end{pmatrix}\right \}$ is a basis of the vector space $V =
\mathbb{R}^{2}$ for an arbitrary fixed $\alpha \in \mathbb{R}, \alpha
\neq \frac{1}{2}$
I first check if both vectors are linearly independent. I use the determinant trick for it.
$$\begin{vmatrix}
1 & \alpha\\
2 & 1
\end{vmatrix}$$
$$\text{Determinant}= 1 \cdot 1 – 2\alpha$$
This can never equal zero as $\alpha \neq \frac{1}{2}$ and thus both vectors are linearly independent.
Moreover the dimension of the vector space is $2$ since $V=\mathbb{R}^{2}$ and we have the same number of linearly independent vectors, $2$.
Thus the subset $\left\{v_{1},v_{2}\right\}$ is indeed a basis of the vector space $V$.
Can you please tell me if I did it correctly? It's very important for me to know as I would do it like that in the exam. And also tell me if the notation is fine 🙂
Best Answer
Yes that is correct. There are several ways to do this, but computing the determinant is probably the 'fastest' way (keeping in mind your question about the use of a calculator to compute the row reduction of a matrix).
Note that you could also have used Gaussian Elimination: this would give you the following: $$\begin{pmatrix} 1 & \alpha\\ 2 & 1 \end{pmatrix}$$ if we now replace row 2 by (row 2 - 2 times row 1), we find $$\begin{pmatrix} 1 & \alpha\\ 0 & 1 - 2\alpha \end{pmatrix}$$ This results in a zero row if and only if $1 - 2 \alpha = 0$ and hence if and only if $\alpha = \frac{1}{2}$.
Another way would be to consider the following linear combination: $$\lambda_1v_1 + \lambda_2v_2 = 0$$ for $\lambda_1, \lambda_2 \in \mathbb{R}$ and to show that both $\lambda_1 = \lambda_2 = 0$. However, considering this coordinate wise, this results in the matrix $$\left(\begin{array}{cc|c} \!\!1 & \alpha & 0\!\!\\ \!\!2 & 1 & 0\!\! \end{array}\right),$$ (I used bars to stress that this is the augmented matrix) which is already solved.
All of this to tell that there are numerous ways to solve this question (and to show you that using the determinant is the fastest way!)
P.S. best of luck with your linear algebra exam!