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RolyPoly Group Title

Concept-checking questions Consider vectors \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\), if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\), does that mean that i) \(\vec{w}\) is linearly dependent on \(\vec{u}\) and \(\vec{v}\) ? ii) sp(\(\vec{u}\), \(\vec{v}\), \(\vec{w}\)) = sp(\(\vec{u}\), \(\vec{v}\)) ?

  • one year ago
  • one year ago

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  1. oldrin.bataku Group Title
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    (i) follows straight from the definition of linear dependence :-) For (ii), recognize that the linear span of a set of vectors is the set of all its finite linear combinations; if \(\vec{u},\vec{v},\vec{w}\) are linearly dependent, i.e. \(\vec{w}\) is a linear combination of \(\vec{u},\vec{v}\), then it should be immediately apparent that the spanning sets are identical.

    • one year ago
  2. RolyPoly Group Title
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    If I can show \(\vec{u}=(u_1, u_2, u_3)\), \(\vec{v}=(v_1, v_2, v_3)\), and \(\vec{w}=(w_1, w_2, w_3)\) span \(\Re^3\), can I immediately draw a conclusion that the vectors are linearly independent?

    • one year ago
  3. oldrin.bataku Group Title
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    I believe so, yes...

    • one year ago
  4. UnkleRhaukus Group Title
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    yes yes yes.

    • one year ago
  5. RolyPoly Group Title
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    And so, if they do not span \(\Re^3\), they are linearly dependent, right?

    • one year ago
  6. UnkleRhaukus Group Title
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    yeah three vectors that aren't all linearly independent cannot span \[\mathbb R^3\]

    • one year ago
  7. RolyPoly Group Title
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    Could "yeah three vectors that aren't all linearly independent cannot span \(\Re^3\)" be the justification of the answer?

    • one year ago
  8. UnkleRhaukus Group Title
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    i wouldn't word it like that

    • one year ago
  9. RolyPoly Group Title
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    Since it cannot span \(\Re^3\), that means one of the vectors is a linear combination of the other vectors, therefore, the vectors are linearly dependent. ^ Could that be one?

    • one year ago
  10. UnkleRhaukus Group Title
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    thats is better yes

    • one year ago
  11. RolyPoly Group Title
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    That's pretty nice :) Thanks! May I ask one more question about linear dependence here?

    • one year ago
  12. UnkleRhaukus Group Title
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    sure

    • one year ago
  13. RolyPoly Group Title
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    Checking the linear dependence of vectors, say, \(\vec{u} =(u_1,u_2,u_3), \vec{v} =(v_1,v_2,v_3),\) and \(\vec{w} =(w_1,w_2,w_3)\) Method I) \[\left[\begin{matrix}u_1 & v_1 & | & w_1 \\ u_2 & v_2 & | & w_2 \\u_3 & v_3 & | &w_3\end{matrix}\right]\] Solve it, if it is a) consistent (that is it has solution), then it is linearly dependent b) inconsistent (that is it has no solution), the it is linearly independent Method II) Find the determinant of \[\left[\begin{matrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\u_3 & v_3 & w_3\end{matrix}\right]\] If the determinant is a) equal to 0, it is linearly dependent b) not equal to 0, it is linearly independent Is that correct? Is there any other way to check the linear dependence?

    • one year ago
  14. UnkleRhaukus Group Title
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    for the first method if (0,0,0) is the only solution the the vectors are independent the second method is right, and im sure there are other methods but i can think of them

    • one year ago
  15. RolyPoly Group Title
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    I think for the first method, it's actually testing if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\) Please leave a comment here if you can think of any of them. That would be really helpful! Thank you for your help!! :)

    • one year ago
  16. oldrin.bataku Group Title
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    If \(\vec{w}\) can be written as a linear combination of \(\vec{u}, \vec{w}\) then they are dependent.

    • one year ago
  17. oldrin.bataku Group Title
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    \(\vec{v}\) **

    • one year ago
  18. RolyPoly Group Title
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    And that's why "if it is a) consistent (that is it has solution), then it is linearly dependent b) inconsistent (that is it has no solution), the it is linearly independent" :|

    • one year ago
  19. UnkleRhaukus Group Title
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    if you find the angle between vectors is π/2 the vectors will be independent

    • one year ago
  20. RolyPoly Group Title
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    How to find the angle between three vectors??

    • one year ago
  21. UnkleRhaukus Group Title
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    \[\theta=\arccos\left(\frac{\vec u \cdot \vec v}{||\vec u||~~||\vec v||}\right)\]

    • one year ago
  22. UnkleRhaukus Group Title
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    For multiple vectors you would have to find the angle between each possible pair of vectors

    • one year ago
  23. RolyPoly Group Title
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    For the case of two vectors, I still can do it. But not for three or more vectors..

    • one year ago
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