For which value $t$ we get a basis in $\mathbb{R}^{3}$ by these $3$ vectors?

Question

For which $t \in \mathbb{R}$ we have a basis in $\mathbb{R}^{3}$ if the vectors are

$$\begin{pmatrix} 1\\ 2\\ 2t \end{pmatrix}, \begin{pmatrix} 1\\ 2t\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ t\\ 1+t \end{pmatrix}$$

This is no homework, just another task that could be asked in my linear algebra exam.

I have no idea how to solve such a task. I have tried Gauss but it didn't work (I could get $3$ zeroes in bottom left corner of the matrix but getting $3$ more zeroes in top right seemed impossible).

Then I thought try to insert some value for $t$ such that the determinant $\neq 0$.

So I just took $t = 0$ because the determinant would be $-2$ and thus the vectors would be linearly independent which indicates they are a basis.

But then there would be many solutions... Did I do it correctly anyway? Is the notation correct too ?

Answer 1

I think the easiest way is to use the determinant. Then You can put the polynomial equal to zero.

You should put your vectors in columns of a $3 \times 3$ matrix. If the determinant of a matrix is $0$, then it means that the columns are linearly dependent and they cannot make a basis for $\mathbb{R}^3$. Therefore, you calculate the determinant of the matrix, which is a polynomial with respect to $t$, and put it equal to zero.

What you achieve at the end is

$-2=0$

It means that for no value of $t$ the determinant can be $0$ and the vectors are always linearly independent.

Answer 2

The first two vectors are always linearly independent as one cannot be a scalar multiple of another: $2 = 2t, 2t = 0$ is impossible.

Thus assume by contradiction that the third vector is a linear combination of the other two, i.e that:

$$a \begin{pmatrix} 1\\ 2\\ 2t \end{pmatrix} + b \begin{pmatrix} 1\\ 2t\\ 0 \end{pmatrix} = \begin{pmatrix} 1\\ t\\ 1+t \end{pmatrix}$$

Comparing components, $a + b = 1, 2a + 2bt = t, 2at = 1 + t$. Repeated substitution gives:

$$2a + (2 - 2a)t = t, \ 2at = 1 + t$$ $$\implies 2a + 2t - (1 + t) = t \implies 2a - 1 = 0, \ a = \frac{1}{2}$$

but then from the second equation, $1 + t \ne t$ for all $t$.

Hence these three vectors form a basis in $\mathbb R^3$ for all $t \in \mathbb R$.