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RolyPoly

  • 2 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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  1. marsss
    • 2 years ago
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    it 's not linearly.

  2. RolyPoly
    • 2 years ago
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    Why?

  3. marsss
    • 2 years ago
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    if it was linearly, that should be w-v=v-u. ok??

  4. RolyPoly
    • 2 years ago
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    I don't understand it.. Can you explain?

  5. marsss
    • 2 years ago
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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??

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

  7. RolyPoly
    • 2 years ago
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    1

  8. RolyPoly
    • 2 years ago
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    But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)

  9. klimenkov
    • 2 years ago
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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.

  10. RolyPoly
    • 2 years ago
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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.

  11. klimenkov
    • 2 years ago
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    Hm, that is not actually right.

  12. klimenkov
    • 2 years ago
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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?

  13. RolyPoly
    • 2 years ago
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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.

  14. klimenkov
    • 2 years ago
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    Please, write what \(n\) in \(\Re ^n\) ?

  15. klimenkov
    • 2 years ago
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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 \(\Re ^n\) ?

  16. RolyPoly
    • 2 years ago
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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.

  17. RolyPoly
    • 2 years ago
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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.

  18. klimenkov
    • 2 years ago
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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 \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\).

  19. klimenkov
    • 2 years ago
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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.

  20. RolyPoly
    • 2 years ago
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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?!

  21. RolyPoly
    • 2 years ago
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    Not quite. [0, 0, 0] is too trivial...Hmmm... :\

  22. klimenkov
    • 2 years ago
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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 \(\vec x=r\vec{u} + s\vec{v} + t\vec{w}\).

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

  24. RolyPoly
    • 2 years ago
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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?

  25. RolyPoly
    • 2 years ago
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    @klimenkov

  26. klimenkov
    • 2 years ago
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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.

  27. RolyPoly
    • 2 years ago
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    :O Thanks!!! As you may observe, I'm not quite clear with the concepts, right?

  28. klimenkov
    • 2 years ago
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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.

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