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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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  1. marsss Group Title
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    it 's not linearly.

    • one year ago
  2. RolyPoly Group Title
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    Why?

    • one year ago
  3. marsss Group Title
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    if it was linearly, that should be w-v=v-u. ok??

    • one year ago
  4. RolyPoly Group Title
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    I don't understand it.. Can you explain?

    • one year ago
  5. marsss Group Title
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    w-v=((0-0),(0-1),(1-2))=(0,-1,-1) v-u=((0-1),(1-2),(2-3))=(-1,-1,-1) ok??

    • one year ago
  6. klimenkov Group Title
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    Can you calculate this? \(\left|\begin{matrix} 1&2&3\\ 0&1&2\\ 0&0&1 \end{matrix}\right|\)

    • one year ago
  7. RolyPoly Group Title
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    1

    • one year ago
  8. RolyPoly Group Title
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    But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)

    • one year ago
  9. klimenkov Group Title
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    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
  10. RolyPoly Group Title
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    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
  11. klimenkov Group Title
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    Hm, that is not actually right.

    • one year ago
  12. klimenkov Group Title
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    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
  13. RolyPoly Group Title
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    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
  14. klimenkov Group Title
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    Please, write what \(n\) in \(\Re ^n\) ?

    • one year ago
  15. klimenkov Group Title
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    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
  16. RolyPoly Group Title
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    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
  17. RolyPoly Group Title
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    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.

    • one year ago
  18. klimenkov Group Title
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    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
  19. klimenkov Group Title
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    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
  20. RolyPoly Group Title
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    "\(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) 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
  21. RolyPoly Group Title
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    Not quite. [0, 0, 0] is too trivial...Hmmm... :\

    • one year ago
  22. klimenkov Group Title
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    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
  23. klimenkov Group Title
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    But don't take a concrete vector \(\vec x\), like (1,2,3). Solve it in general.

    • one year ago
  24. RolyPoly Group Title
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    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?

    • one year ago
  25. RolyPoly Group Title
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    @klimenkov

    • one year ago
  26. klimenkov Group Title
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    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
  27. RolyPoly Group Title
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    :O Thanks!!! As you may observe, I'm not quite clear with the concepts, right?

    • one year ago
  28. klimenkov Group Title
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    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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