1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I did not check the computations, but your logic is flawless. $\endgroup$ – José Carlos Santos Sep 12 '18 at 15:58
  • $\begingroup$ 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 $\mathbb{R}^3$ $\endgroup$ – nikola Sep 12 '18 at 15:58
  • $\begingroup$ How do you know the three vectors are linearly independent quickly? If you giuve me a solution I'll give credit @nikola $\endgroup$ – Jwan622 Sep 12 '18 at 16:03
  • $\begingroup$ 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. $\endgroup$ – dmtri Sep 12 '18 at 16:07
1
$\begingroup$

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 $\Bbb R^n$.

$\endgroup$
  • $\begingroup$ Did you add a1 and a3 to get -a3 = 0? and so a3 must be 0? $\endgroup$ – Jwan622 Sep 12 '18 at 16:19
  • $\begingroup$ @Jwan622 Exactly what method you use to solve that final set of equations is entirely up to you. Adding the two equations, left side to left side and right side to right side is one possible way to get there, yes. $\endgroup$ – Arthur Sep 12 '18 at 16:52
0
$\begingroup$

Take a look at the zero-components. The second vector can’t be a linear combination of the other vectors. Since these other vectors aren’t parallel, the three vectors are linearly independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.