Can Vector v Be Expressed as a Linear Combination of v1, v2, v3, and v4?

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Discussion Overview

The discussion revolves around the problem of expressing a vector v as a linear combination of four given vectors v1, v2, v3, and v4. Participants explore the conditions under which this is possible, including the implications of the coefficients and the properties of the corresponding matrix.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents the vectors v1, v2, v3, v4, and the target vector v, seeking to express v as a linear combination of the others.
  • Another participant suggests substituting the vectors into the equation to derive a system of equations for the coefficients c1, c2, c3, and c4.
  • A participant expresses difficulty in finding solutions for the coefficients, suggesting that there may be no solution.
  • Some participants point out potential errors in the equations and the need for clarity on the calculations performed.
  • It is noted that the matrix formed by the coefficients is singular, indicating that the vectors may not span the space needed to represent v.
  • One participant provides a row-reduced form of the augmented matrix, concluding that there is no solution due to the dependence of the vectors.
  • Another participant suggests that if the target vector were in the span of the given vectors, it could be expressed in infinitely many ways.
  • There is a discussion about the appropriateness of posting homework questions in this forum, with some participants expressing concern over the context of the question.

Areas of Agreement / Disagreement

Participants generally agree that the matrix is singular and that the target vector cannot be expressed as a linear combination of the given vectors. However, there is disagreement about the correctness of the problem statement and whether it was posed accurately.

Contextual Notes

Limitations include the potential for errors in the original problem statement and the dependence of the vectors, which affects the ability to find a solution. The discussion reflects various interpretations of the problem and the calculations involved.

Who May Find This Useful

Students and educators interested in linear algebra, particularly in understanding linear combinations, vector spaces, and the implications of singular matrices.

krocks
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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.
 
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krocks said:
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.
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.
 
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"?
 
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
 
HallsofIvy said:
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
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?
 
Hi

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

7=c1+2c2+11c3+4c4
9=4c1+ 5c2+ 14c3+ 3c4
6=2c1+3c2+12c3+2c4
8=8c1+9c2+18c3+c4

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" ??
 
The matrix is indeed singular. What is its null space?
 
Hey
Please reply friends.

Just tell me the answer only. i am in need of it.
Please
 
Looks to me like there is an error in the problem.

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

But after row-reducing, I get
\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}
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.
 
  • #10
Hey HALLSOFIVY

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 :)
 
  • #11
krocks said:
Hey HALLSOFIVY

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 :)

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!
 
  • #12
This row operation is not difficult. Even you got the answer, you better try it again by yourself.
 
  • #13
Hi

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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