Thread: linear independent/vectors

1. linear independent/vectors

Show wether the vectors are linear independent and if they generate $\displaystyle \mathbb{R}^3$.
a)$\displaystyle (2,-2,0),(1,-2,1),(-1,4,-3)$
My attempt : I don't understand anything about my teacher's notes. So I've read on wikipedia that in order to determine if some vectors are l.i. I could calculate the determinant of a matrix formed by them. That's what I did here and I found it to be different from $\displaystyle 0$, so the 3 vectors are l.i.
Now how can I show that they generate $\displaystyle \mathbb{R}^3$ or not? I guess I must find a basis (but I don't know how to procede and what it is) and check if it generates $\displaystyle \mathbb{R}^3$.
Thanks in advance.

2. Originally Posted by arbolis
Show wether the vectors are linear independent and if they generate $\displaystyle \mathbb{R}^3$.
a)$\displaystyle (2,-2,0),(1,-2,1),(-1,4,-3)$
My attempt : I don't understand anything about my teacher's notes. So I've read on wikipedia that in order to determine if some vectors are l.i. I could calculate the determinant of a matrix formed by them. That's what I did here and I found it to be different from $\displaystyle 0$, so the 3 vectors are l.i.
Now how can I show that they generate $\displaystyle \mathbb{R}^3$ or not? I guess I must find a basis (but I don't know how to procede and what it is) and check if it generates $\displaystyle \mathbb{R}^3$.
Thanks in advance.
If you can show these vectors are linearly independent then they would generate $\displaystyle \mathbb{R}^3$ - why?
Thus, we just need to check if they are linearly indepedent.

Say $\displaystyle a\begin{bmatrix} 2\\-2\\0 \end{bmatrix} + b\begin{bmatrix}1\\-2\\1 \end{bmatrix} + c \begin{bmatrix}-1\\4\\-3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

This gives us,
$\displaystyle 2a + b - c = 0$
$\displaystyle -2a - 2b + 4c = 0$
$\displaystyle 0a + b - 3c = 0$

We need to show this only have a trivial solution i.e. $\displaystyle a=b=c=0$.

This is equivalent to showing,
$\displaystyle \det \begin{bmatrix}2& 1 & -1 \\ -2&-2&4 \\ 0 & 1 & - 3 \end{bmatrix} \not = 0$

3. <2, -2, 0> - 3<1, -2, 1> = <2, -2, 0> + <-3, 6, -3> = <-1, 4, -3>

The vectors are not linearly independent.

4. This is equivalent to showing,
This is what I've done.
$\displaystyle \det A=2(6+4)-(6+0)-(-2+0)=22-6\neq 0$ thus they are linear independent. Am I calculating it badly?
And about
If you can show these vectors are linearly independent then they would generate - why?
I asked it myself : is that possible that they generate for example $\displaystyle \mathbb{R}^2$? And I thought not, because it would mean that 2 of them are linear dependent and the remaining vector is linear independent from the 2 firsts, but I don't know how to show this.
I probably did an error since icemanfan claims that they are linear dependent...

5. The determinant is (2)(-2)(-3) + (1)(4)(0) + (-1)(-2)(1) - (-1)(-2)(0) - (2)(4)(1) - (1)(-2)(-3) = 12 + 0 + 2 - 0 - 8 - 6 = 0.

6. Sorry, I made an error of sign. It was $\displaystyle \det A= 2(6-4)+...$

7. Sorry not to start it in a new thread, but the entire question states "If they are not l.i., characterize implicitly the subspace generated and give a basis of this subspace".
I've really no idea about how to do that. Maybe they generate $\displaystyle \mathbb{R}^2$, but I don't know how to show it...nor how to pick off a basis.

8. you can do both parts of your problem at the same time without even using determinant: first let $\displaystyle A=\begin{bmatrix}2 & 1 & -1 \\ -2 & -2 & 4 \\ 0 & 1 & -3 \end{bmatrix}, \ \bold{b}=\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix}.$ then $\displaystyle \bold{b}$ is in the subspace spanned by the columns of $\displaystyle A,$

if and only if the equation $\displaystyle A \bold{x}=\bold{b}$ has a solution. so you need to consider the augmented matrix $\displaystyle [A \mid \bold{b}]=\begin{bmatrix} 2 & 1 & -1 & b_1 \\ -2 & -2 & 4 & b_2 \\ 0 & 1 & -3 & b_3 \end{bmatrix} \sim \begin{bmatrix}2 & 1 & -1 & b_1 \\ 0 & -1 & 3 & b_1+b_2 \\ 0 & 0 & 0 & b_1+b_2+b_3 \end{bmatrix}.$

now it's clear that the columns of $\displaystyle A$ are linearly dependent because the third row of the echelon form of $\displaystyle A$ is all 0. this answers the first part of your question. also it's clear that $\displaystyle A \bold{x}=\bold{b}$ is

consistent iff $\displaystyle b_1+b_2+b_3=0,$ which gives us $\displaystyle b_3=-b_1-b_2.$ hence $\displaystyle \bold{b}=\begin{bmatrix}b_1 \\ b_2 \\ -b_1-b_2 \end{bmatrix}=b_1 \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} + b_2 \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix}.$ so the subspace generated by the columns of $\displaystyle A$ is a plane which goes through

the origin and is spanned by the vectors $\displaystyle v_1=\begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \ v_2=\begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix}.$ clearly $\displaystyle \{v_1, v_2 \}$ is a basis for the subspace. this answers the second part of your question.

9. Thank you very much NonCommAlg, I could follow you entirely despite the fact that all this stuff is new to me.