Is this set a basis for $R^3$. How to verify?

Say I have $S = {(1,0,-1), (2,1,1), (-3,0,2)}$. Is my method correct for determining if it's a basis? How do I check if I'm right?

We need to determine if $S$ spans $R^3$ and if it's linearly independent.

First, check if $S$ spans $R^3$:

Let $u_1, u_2, u_3$ be a random vector.

$$(u_1, u_2, u_3) = c_1(1,0,-1) + c_2(2,1,1) + c_3(-3,0,-2)$$ $$= (c_1 + 2c_2 - 3c_3, c_2, -c_1 + c_2 + 2c_3$$

the corresponding equations:

$$c_1 + 2c_2 - 3c_3 = u_1$$ $$c_2 = u_2$$ $$-c_1 + c_2 + 2c_3 = u_3$$

and the resulting matrix:

$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$

Determine determinant using expansion by cofactors in the second row:

determinant = $0$ * (cofactor of 2nd row, 1st column) + 1 * (cofactor of 2nd row, 2nd column) + $0$ * (cofactor of 2nd row, third column) = 5

So the fact that the determinant is non-zero means that there is a unique solution for this system of equations for any random vector in $R^3$

Is it linearly dependent?

$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 3 & -1 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

Because the last row is $[0,0,1]$ this implies that the only solution is trivial so it is linearly dependent.

So $S$ is a basis for $R^3$. Is this logic right?

• I did not check the computations, but your logic is flawless. – José Carlos Santos Sep 12 '18 at 15:58
• you could have done this way easier. Simply, you have three vectors, and if they are linearly independent, they span the three-dimensional space hence $\mathbb{R}^3$ – nikola Sep 12 '18 at 15:58
• How do you know the three vectors are linearly independent quickly? If you giuve me a solution I'll give credit @nikola – Jwan622 Sep 12 '18 at 16:03
• You may save some time, if you would like to use the fact that in a $n$ dimentional Vector Space, $n$ linearly independent vectors span it. – dmtri Sep 12 '18 at 16:07

In this case (the vectors have zeroes which line up nicely), checking linear independence is rather quick. Let $v_1,v_2,v_3$ be your three vectors from $S$, and let $a_1,a_2,a_3$ be numbers such that $$a_1v_1+a_2v_2+a_3v_3=\vec 0$$ By considering the second component of this vector equation, we immediately get $a_2=0$. That means we're left with $a_1-3a_3=0$ from the first component and $-a_1+2a_3=0$ from the third. We see that $a_1$ and $a_3$ are also $0$, which proves linear independence.
As for spanning $\Bbb R^3$, any set of $n$ linearly independent vectors from $\Bbb R^n$ spans $\Bbb R^n$.