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Linear combination of vectors

  1. May 9, 2009 #1
    HI everyone,

    v1=[1 4 2 8]^t
    v2=[2 5 3 9]^t
    v3=[11 14 12 18]^t
    v4=[4 3 2 1]^t

    I have to express vector v=[7 9 6 8]^t in two ways as a linear combination v=c1v1+c2v2+c3v3+c4v4 of {v1,v2,v3,v4}

    Please reply as soon as possible.

    Thank You in advance.
  2. jcsd
  3. May 9, 2009 #2


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    Reply how? Tell you the answer? I won't do that! Have you tried anything yourself?

    Have you for example, replace v, v1, v,2, v3, and v4 in the equation
    v=c1v1+c2v2+c3v3+c4v4 by [7 9 6 8]^t, [1 4 2 8]^t, etc. to get
    [7 9 6 8]^t= c1[1 4 2 8]^t+ c2[2 5 3 9]^t+ c3[11 14 12 18]^t+ c4[4 3 2 1]^t.

    Go ahead and do the vector calculation on the right and set each component on the left equal to the corresponding component on the right. That will give you four equations to solve for the four numbers c1, c2, c3, and c4.
  4. May 9, 2009 #3
    Hi HallsOfIvy

    Ya i did tried it by myself.
    Bu am not able to find values for "c1,c2,c3,c4".

    According to me it gives no solution.

    So how can I express "v" as linear combination of "v1,v2,v3,v4"?
  5. May 9, 2009 #4


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    As long as you refuse to show what you have done, we cannot see or explain what you have done wrong! I see no problem with solving the equations
    c1+ 2c2+ 11c3+ 5c4= 7
    4c1+ 5c2+ 14c3+ 3c4= 9
    2c1+ 3c2+ 12c3+ 2c4= 6
    8c1+ 9c2+ 18c3+ c4= 8
  6. May 9, 2009 #5

    D H

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    I do. Two problems. That 5 in the first row should be 4, and more importantly, that you had a typo in that particular element doesn't matter.

    krocks, are you sure you have the numbers correct?
  7. May 10, 2009 #6

    Ya the question i mentioned is absolutely correct.
    The equation is:

    9=4c1+ 5c2+ 14c3+ 3c4

    This equation is showing no answer becuse the matrix formed by using coefficients of "c1,c2,c3,c4" is singular.

    So is there any way to represent "v" as linear combination of "v=c1v1+c2v2+c3v3+c4v4" ??
  8. May 10, 2009 #7

    D H

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    The matrix is indeed singular. What is its null space?
  9. May 10, 2009 #8
    Please reply friends.

    Just tell me the answer only. i am in need of it.
  10. May 11, 2009 #9


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    Looks to me like there is an error in the problem.

    After correcting my previous mis-copy, I get the augmented matrix
    [tex]\begin{bmatrix}1 & 2 & 11 & 4 & 7 \\ 4 & 5 & 14 & 3 & 9 \\ 2 & 3 & 12 & 2 & 6 \\ 8 & 9 & 18 & 1 & 8\end{bmatrix}[/tex]

    But after row-reducing, I get
    [tex]\begin{bmatrix}1 & 2 & 11 & 4 & 7 \\ 0 & 1 & 10 & \frac{13}{3} & \frac{19}{3} \\ 0 & 0 & 0& 1& 1\\ 0 & 0 & 0 & 0 & 1\end{bmatrix}[/tex]
    and, because of that "1" in the last row, there is no solution. Are you sure you have all of the numbers right? Those vectors are not independent so do not span all of R4 and the "target" vector, <7, 9, 6, 8>, is not in their span- it cannot be written as a linear combination of the given vectors.

    If it were in their span, then, because they are not independent, you would be able to write it as a linear combination of them in infinitely many ways.
  11. May 11, 2009 #10

    Thanks a ton for help. The question which i got from my professor is exactly the same which I mentioned. I think there's some problem in question itself. i will ask about it from my professor and will post the reply soon here.

    Thanks :)
  12. May 11, 2009 #11


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    This was schoolwork? Why wasn't it posted under the homework section?

    You should talk to Derillo- he posted exactly this question under the homework section!
  13. May 12, 2009 #12
    This row operation is not difficult. Even you got the answer, you better try it again by yourself.
  14. May 13, 2009 #13

    I apologize for posting a homework question here.
    Actually i was new to this site that's why i wasn't aware of such rules.

    But i'll keep that in mind in future
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