Determining for which parameter values a set: spans a vector space, is linearly independent, a base for a vector space.

Question

Let $V = \mathbb{R}^{3}$

and $v_1 = \begin{bmatrix}1\\ a\\ 1\end{bmatrix}$, $v_2 = \begin{bmatrix}1\\ a+1\\ 1\end{bmatrix}$, $v_3 = \begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$ $a\in \mathbb{R}$

Let $S = \left \{ v_1, v_2, v_3 \right \}$

For which $a\in \mathbb{R}$ :

$S$ spans $V$

$S$ is linearly independent

$S$ is a base for $V$.

So here's how I look at it: I can immidiately tell that $v_1, v_2$ are linearly independent, because each one of their coordinates is the same except for the middle one, and the equation $a = a+1$ has no solution - therefore for each $a\in \mathbb{R}$ $v_1$ will no be equal to $v_2$.

I run the same procces between $v_1$ and $v_3$ and between $v_2$ and $v_3$ and I find out that they're linearly independent when $a \neq 1, \frac{1}{2}$

Since I found that these 3 vectors are linearly independent for $a \neq 1, \frac{1}{2}$ I know that $S$ is a base for $V$, because $\left | S \right | = dimV$

And therfore $S$ ofcourse spans $V$ for these values.

However, this is an easy case of "look and understand" (Assuming that I got it right)..

I would like to to get a more direct and general approach to attack this kind of question. thank you!

Answer 1

As explained in the comments, it is not sufficient to check for pairwise linear independence to conclude linear independence of the set ($S$).

and $v_1 = \begin{bmatrix}1\\ a\\ 1\end{bmatrix}$, $v_2 = \begin{bmatrix}1\\ a+1\\ 1\end{bmatrix}$, $v_3 = \begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$ $a\in \mathbb{R}$

Note that for any $a \in \mathbb{R}$, you have: $$(1+2a)\begin{bmatrix}1\\ a\\ 1\end{bmatrix} +(2-2a)\begin{bmatrix}1\\ a+1\\ 1\end{bmatrix} =\begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$$ which means that $v_1$, $v_2$ and $v_3$ are always linearly dependent.

I would like to to get a more direct and general approach to attack this kind of question. thank you!

For three vectors in $\mathbb{R^3}$, a handy equivalent property uses the determinant: the matrix formed by the three vectors has a non-zero determinant if and only if the vectors are linearly independent.

You can easily verify that for all $a \in \mathbb{R}$ (the first and last rows are identical): $$\begin{vmatrix} 1 & 1 & 3 \\ a & a+1 & a+2 \\ 1 & 1 & 3 \end{vmatrix}=0$$

Answer 2

Hint:

Let be $M\in\mathbb R^{3\times3}$ the matrix with columns $v_1$, $v_2$ and $v_3$. $S$ is linearly independent if and only if $M$ has full rank. The row rank of $M$ is obvious.