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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 :)

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    $\begingroup$ Yes, you are correct. $\endgroup$
    – Chee Han
    Mar 2, 2017 at 18:50
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    $\begingroup$ It is perhaps worth point out that the question in the title contains an error: a subspace cannot be a basis of a vector space, because a subspace always contains the zero vector, and a set containing the zero vector cannot be linearly independent. Presumably you meant "subset" instead of "subspace". $\endgroup$
    – mweiss
    Mar 2, 2017 at 19:16
  • $\begingroup$ Note that if you only have two vectors, for linear independence you just need to check that they’re not scalar multiples of each other. $\endgroup$
    – amd
    Mar 3, 2017 at 3:58

3 Answers 3

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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!

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  • $\begingroup$ @egregg: Thanks for the edit, I must have missed that extra $\alpha$. + I have never seen the use of 'array' to denote a matrix. Thank you very much. $\endgroup$
    – Student
    Mar 2, 2017 at 21:42
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Yes, that is a very nice way to complete the problem. All the reasoning and notation look good to me.

Without the determinant trick, one could still "brute force" by considering a linear combination of the two which is equal to zero, and you will still come to the same conclusion.

In this special case of dimension two, you can also just show that one of the vectors is not a scalar multiple of the other (for that is what it means for two vectors to be linearly dependent.) This doesn't generalize to more vectors as easily, though.

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While I don't know your instructor's preferred notation, I believe that your reasoning is correct. Only in the case where $\alpha = \frac{1}{2}$ do you get two vectors that are parallel.

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