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RolyPoly
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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.
 one year ago
 one year ago
RolyPoly Group Title
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.
 one year ago
 one year ago

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marsss Group TitleBest ResponseYou've already chosen the best response.0
it 's not linearly.
 one year ago

marsss Group TitleBest ResponseYou've already chosen the best response.0
if it was linearly, that should be wv=vu. ok??
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I don't understand it.. Can you explain?
 one year ago

marsss Group TitleBest ResponseYou've already chosen the best response.0
wv=((00),(01),(12))=(0,1,1) vu=((01),(12),(23))=(1,1,1) ok??
 one year ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Hm, that is not actually right.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Please, write what \(n\) in \(\Re ^n\) ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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 \(\Re ^n\) ?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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 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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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 \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\).
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPoly Group TitleBest 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?!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Not quite. [0, 0, 0] is too trivial...Hmmm... :\
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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 \(\vec x=r\vec{u} + s\vec{v} + t\vec{w}\).
 one year ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I'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?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@klimenkov
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
:O Thanks!!! As you may observe, I'm not quite clear with the concepts, right?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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