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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it was linearly, that should be wv=vu. ok??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand it.. Can you explain?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wv=((00),(01),(12))=(0,1,1) vu=((01),(12),(23))=(1,1,1) ok??

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Can you calculate this? \(\left\begin{matrix} 1&2&3\\ 0&1&2\\ 0&0&1 \end{matrix}\right\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Because 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
 3 years ago
Best ResponseYou've already chosen the best response.0Wow! 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
 3 years ago
Best ResponseYou've already chosen the best response.1Hm, that is not actually right.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1You 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh!!!! 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
 3 years ago
Best ResponseYou've already chosen the best response.1Please, write what \(n\) in \(\Re ^n\) ?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
 3 years ago
Best ResponseYou've already chosen the best response.01) 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
 3 years ago
Best ResponseYou've already chosen the best response.01) 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 nonzero, nonparallel 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
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
 3 years ago
Best ResponseYou've already chosen the best response.1Now 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
 3 years ago
Best ResponseYou've already chosen the best response.0"\(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) 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
 3 years ago
Best ResponseYou've already chosen the best response.0Not quite. [0, 0, 0] is too trivial...Hmmm... :\

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Take 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
 3 years ago
Best ResponseYou've already chosen the best response.1But don't take a concrete vector \(\vec x\), like (1,2,3). Solve it in general.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry for late reply! I got \(r = x_1 , \ s = 2x_1 + x_2, \ t=x_12x_2+x_3\) These are all scalars, I supposed. Aren't they?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0:O Thanks!!! As you may observe, I'm not quite clear with the concepts, right?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. 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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