LA: Span R^3 - Find Values of m & Solutions for Vector w

  • Thread starter daniel_i_l
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In summary: The only real difference would be in the interpretation of the reduced matrix and its rows. For instance, the first row would be 1 2 -3 -m. This means that 1 * v1 + 2 * v2 - 3 * v3 - m * v4 = 0. That's the same thing as saying that -v2 - v4 + 1/3 * v3 = 1/3 * v1 (or -v2 - v4 + 1/3 * v3 = (1/3, 0, 0)). Now, when you reduced the matrix to:1 2 -3 -m a0 -
  • #1

daniel_i_l

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Homework Statement


you have 4 vectors:
v1=(2,1,-1), v2=(-m,-1,3), v3=(-3,2,m+1), v4=(1,2,1)
a) for which values of m do the vectors span R^3
b) the vector w=(m+1, m-1, 1). for which m is there a solution to:
x1v1 + x2v2 + x3v2 + x4v4 = w ?

Homework Equations


the definition of spanning - every vector in R^3 can be made of those 4 vectors.


The Attempt at a Solution


a) i made the following matrix where the first few coloums are v4,v1,v3,v4 and the last one is some vector (a,b,c):
Code:
[1  2   -3 -m  a]
[2  1    2 -1  b]
[1 -1 (m+1) 3  c] 
and reduced it to:
[1  2   -3    -m   a]
[0 -3    8  (2m-1) b-2a]
[0  0 (m-4) (4-m) c-b+a]
now the only way that there isn't a solution is if there's a row with the form:
[0 0 0 0 alpha] where alpha is any scalar. since the only row that can possibly look like this is the third one where m=4 then the answer is that they span R^3 for every m except 4.
Is that right?
b) here's where i got confused, if m doesn't equal 4 then there's a solution because according to a) the vectors v1,v2,v3 and v4 span R^3 which includes w. and if m=4 then there's no solution because for w:
c-b+a = 3 (not 0)
so the answer for b should be the same as a). is that true? it doesn't seem likely that there isn't some trick or something.
Thanks.
 
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  • #2
Woah. If you have 3 independent vectors they automatically span R^3. I can give you four vectors (1,0,0), (2,0,0), (3,0,0) and (4,0,0) but they do not span R^3.

Your solution is right. If m is not 4 then you can reduce the matrix to find all solutions (pick such an m if you want to see). Note how these 4 vectors 'overspan' R^3 so that you would get infinitely many solutions.

We know that for m not equal to four there is a solution to w (as then the 4 vectors span R^3). What happens when you try to solve the system when m=4?
 
  • #3
When m=4 then there's no solution because you can reduce the matrix to a matrix that has the row:
[0 0 0 0 3]
is that right? if so why would they give two questions with the same answer?
 
  • #4
I think you are reducing the transpose of the matrix you want to reduce. Reduce this one (and I would say forget about the a,b,c). You are missing one of the values of m that makes this group of vectors fail to span.

[ 1 2 1 ]
[ - m - 1 3 ]
[ - 3 2 m + 1 ]
[ 1 2 1 ]
 
  • #5
i made the matrix that i posted above to check when an arbitrary vector is a linear combination of the four given vectors, to do that you'd put in the vectors as coloums right?
 
  • #6
If you are going to do row reduction I think you should put in the vectors as rows. When the reduction is finished you can tell by inspection how many of the vectors are linearly independent. If there are three of them, then they span R^3.
 
  • #7
Yes, you can have four vectors, as long as one is linearly dependent. The range has to be 3 dimensional.
 
  • #8
If he row reduces, linear dependence will show up as zero vectors in the list.
 
  • #9
yes, you get the same answer regardless of whether you put the vectors in as rows and coloums as long as you know how to interpert the final matrix. my question is if my answer for b is correct.
thanks
 
  • #10
daniel_i_l said:
yes, you get the same answer regardless of whether you put the vectors in as rows and coloums as long as you know how to interpert the final matrix. my question is if my answer for b is correct.
thanks

I can't argue with that. Today I think it's correct. Yesterday, I was getting a quadratic for m. I think I was wrong yesterday.
 
  • #11
If you know how to add ordered triples, then I don't see why the issue about putting the vectors into the matrix as rows or columns would arise.
 

1. How do I determine the values of m in LA: Span R^3?

To determine the values of m in LA: Span R^3, you will need to solve for m using the given information and equations. This may involve using techniques such as Gaussian elimination or row reduction to eliminate variables and isolate m. Once you have solved for m, you can use the value to determine the solutions for vector w.

2. What is the significance of LA: Span R^3 in mathematical concepts?

LA: Span R^3 is significant in mathematical concepts because it represents the set of all possible linear combinations of three-dimensional vectors in a vector space. This concept is essential in understanding vector spaces, linear transformations, and systems of linear equations.

3. How do I find solutions for vector w in LA: Span R^3?

To find solutions for vector w in LA: Span R^3, you will need to use the values of m that you have determined. Substitute these values into the equations and solve for the components of vector w. This will give you the solutions for vector w that satisfy the given conditions and equations.

4. What are some real-life applications of LA: Span R^3?

LA: Span R^3 has numerous real-life applications, including computer graphics, physics, engineering, and economics. It is used to model and solve problems involving three-dimensional vectors and their linear combinations. For example, it can be used to calculate forces and motion in physics or to create 3D graphics in computer animation.

5. What are some common mistakes when working with LA: Span R^3?

Some common mistakes when working with LA: Span R^3 include not setting up the equations correctly, not using the correct techniques to solve for m, and making errors in calculations. It is also essential to check for consistency and to ensure that the solutions for vector w satisfy all the given conditions and equations. It is always helpful to double-check your work and seek clarification if needed.

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