anonymous
  • anonymous
Show that the vector \(u\) = (1, 2, 3), \(v\) = (0, 1, 2), and \(w\) = (0, 0, 1) span \(\Re ^3\). Are these vectors linearly independent? Please justify your answer.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
it 's not linearly.
anonymous
  • anonymous
Why?
anonymous
  • anonymous
if it was linearly, that should be w-v=v-u. ok??

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anonymous
  • anonymous
I don't understand it.. Can you explain?
anonymous
  • anonymous
w-v=((0-0),(0-1),(1-2))=(0,-1,-1) v-u=((0-1),(1-2),(2-3))=(-1,-1,-1) ok??
klimenkov
  • klimenkov
Can you calculate this? \(\left|\begin{matrix} 1&2&3\\ 0&1&2\\ 0&0&1 \end{matrix}\right|\)
anonymous
  • anonymous
1
anonymous
  • anonymous
But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)
klimenkov
  • klimenkov
Because nothing will change. Determinant doesn't change if you change the role of rows and columns. You've got, that this determinant is not equal to zero, that means these 3 vectors are linearly independent.
anonymous
  • anonymous
Wow! I actually did it in this way, was it correct? Consider \(r\vec{u} + s\vec{v}+t\vec{w}\) for all scalars r, s, t \[\left[\begin{matrix} 1&2&3 & | &0\\ 0&1&2 & | &0\\ 0&0&1 & | &0\end{matrix} \right]\] \[ ~ \left[\begin{matrix} 1&0&0 & | &0\\ 0&1&0 & | &0\\ 0&0&1 & | &0\end{matrix} \right]\] [r, s, t] = [0, 0, 0] Since we have only [0, 0, 0] is the (trivial) solution, so the three vectors are linearly independent.
klimenkov
  • klimenkov
Hm, that is not actually right.
klimenkov
  • klimenkov
You need: \( r\cdot\left(\begin{matrix} 1\\ 2\\ 3 \end{matrix}\right)+ s\cdot\left(\begin{matrix} 0\\ 1\\ 2 \end{matrix}\right)+ t\cdot\left(\begin{matrix} 0\\ 0\\ 1 \end{matrix}\right)= \left(\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right) \) And you system of the equations must be: \( \left\{ \begin{matrix} 1\cdot r+0\cdot s+0\cdot t=0\\ 2\cdot r+1\cdot s+0\cdot t=0\\ 3\cdot r+2\cdot s+1\cdot t=0 \end{matrix} \right. \) You've got another system. Understand?
anonymous
  • anonymous
Ahh!!!! So I... actually.... and TOTALLY .....got... that..... wrong :| Actually, I am confused. Do I get this part right? If the three vectors span \(\Re ^n\), and the determinant is not zero, they are linearly independent. If the three vectors span \(\Re ^n\), and the determinant is zero, they are linearly dependent.
klimenkov
  • klimenkov
Please, write what \(n\) in \(\Re ^n\) ?
klimenkov
  • klimenkov
I have some questions for you. Please, answer them all only by yourself. 1) What is a linear combination? 2) What is a linear independence? 3) What means \(n\) vectors span \(\Re ^n\) ?
anonymous
  • anonymous
1) Linear combination: \(\vec{v}=c_1\vec{v_1} + c_2\vec{v_2} +...+c_n\vec{v_n}\), \(c_1, c_2, ..., c_n\) 2) Linear independence: \(c_1, c_2, ..., c_n\) for not all n being zero 3) n vectors span \(\Re ^n\): the n vectors cannot be expressed in term of the others. Eg: [1, 2, 3] and [1, 4, 5] => span \(\Re ^2\) but [1, 2, 3] and [2, 4, 6] => span \(\Re ^1\) Something wrong with last one, I hope you don't mind me checking it now.
anonymous
  • anonymous
1) c1,c2,...,cn => weights (Well, I can only understand it as some scalars) 3) span: Given that \(\vec{v_1}, \vec{v_2}, ..., \vec{v_p}\) are in \(\Re ^n\) set of all linear combinations of vectors = subspace of \(\Re ^n\) = spanned by \(\vec{v_1}, \vec{v_2}, ..., \vec{v_p}\) subspace = (need to be googled) Consider vector v, and w. If they are both non-zero, non-parallel vectors in \(\Re ^n\) all possible linear combinations of v and w can be expressed in the form of rv +sw, for all scalars r and s. all possible linear combinations => filled a plane => plane spanned by v and w. Sorry, too much irrelevant things here, maybe.
klimenkov
  • klimenkov
I think it will be easy to unterstand. \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) span \(\Re ^n\) if any vector from \(\Re ^n\) can be represented as a linear combination of \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\).
klimenkov
  • klimenkov
Now back to your task. Using the system I wrote above you can get, that the vectors are independent. Now you need to show that any vector from \(\Re^3\) can be shown as their linear combination.
anonymous
  • anonymous
"\(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) span \(\Re ^n\) if any vector from Rn can be represented as a linear combination of \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\)" So, in this case, we need to find some vectors \(r\vec{u} + s\vec{v} + t\vec{w}\) from \(\Re ^3\). Set \(r\vec{u} + s\vec{v} + t\vec{w} = [0, 0, 0]\) to solve r, s, t?!
anonymous
  • anonymous
Not quite. [0, 0, 0] is too trivial...Hmmm... :\
klimenkov
  • klimenkov
Take any vector from \(\Re^3\) - \(\vec x=(x_1,x_2,x_3),\, x_i\in \Re\) and show that you can find scalars (weights) - \(r,s,t\) so \(\vec x=r\vec{u} + s\vec{v} + t\vec{w}\).
klimenkov
  • klimenkov
But don't take a concrete vector \(\vec x\), like (1,2,3). Solve it in general.
anonymous
  • anonymous
I'm sorry for late reply! I got \(r = x_1 , \ s = -2x_1 + x_2, \ t=-x_1-2x_2+x_3\) These are all scalars, I supposed. Aren't they?
anonymous
  • anonymous
klimenkov
  • klimenkov
Yes, that's right, but you've put an extra minus in the expression for \(t\). So, you've showed that these three vectors span \(\Re^3\). Congratulations.
anonymous
  • anonymous
:O Thanks!!! As you may observe, I'm not quite clear with the concepts, right?
klimenkov
  • klimenkov
Yes. Actually, it is not a big problem. All you need is to learn honestly. The biggest part in the solving math problems is to understand the statement of the question. Try to use all the possible sources to understand the problem. In this way you will learn a lot.

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