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United States Linear Algebra - Vector Equations

  1. Dec 30, 2011 #1
    See Attachment for Question
    See Attachment for Answer from back of book
    I do not see how part a and part b are asking me two different things.

    I interpret the first part of part a
    "Is b in {a_1,a_2,a_3}?"
    as
    Is b a solution of the system represented by matrix A?
    [1,0,-4,4
    0,3,-2,1
    -2,6,3,-4]
    I got up to here
    [1,0,-4,4
    0,1,-2/3,1/3
    0,0,1,-2]
    and saw that the system was consistent and stopped and put yes for the first part.
    for the second part of part a
    "How many vectors are in {a_1,a_2,a_3}?"
    I said infinitely many because in the first part of part a I could easily get the matrix in reduced echelon form if I continued and so the fourth column of the matrix could be anything.

    For part b.
    Is b in W? How many vectors are in W?
    I don't understand how this any different than part a because
    b=[4;1;-4] and W=Span{a_1,a_2,a_3}
    replacing W and B in the question with this information I get
    "Is b=[4;1;-4] in Span{a_1,a_2,a_3}? How many vectors are in Span{a_1,a_2,a_3}?"
    which looks just like part a to me
    "Is b in {a_1,a_2,a_3}? How many vectors are in {a_1,a_2,a_3}?"

    I don't understand how part a and part b are different and I guess what exactly I'm being asked even sense the questions are different some how.

    I have no idea what I'm even being asked by
    "Show that a_1 is in W. [Hint: Row operations are unnecessary.]"

    Thanks for any help. This a question from my home work from my Linear Algebra class, my first class in linear algebra.
     

    Attached Files:

  2. jcsd
  3. Dec 30, 2011 #2

    Mark44

    Staff: Mentor

    No. This is really a very simple question. The set here contains just the vectors shown in it. Is b one of the vectors in that set?
    This too is a very simple question that has nothing to do with matrices. Just count the vectors in it.

    W, or Span{a1, a2, a3}, is different. This notation represents all of the vectors that are linear combinations (i.e., sums of scalar multiples of) the vectors in this set.


     
  4. Dec 30, 2011 #3

    Mark44

    Staff: Mentor

    BTW - What have you started putting "United States ..." in the titles of your threads?
     
  5. Dec 30, 2011 #4
    I don't know I thought that it would be a good idea to put the course title in the title of the thread so that way people could tell right off from reading the title weather or not they can help with the question. I also thought it might be a good idea to include the country in which I'm taking the course because sometimes the curriculum are drastically different depending on were you go to school, at least this is what I've heard from other people.
     
  6. Dec 30, 2011 #5

    Mark44

    Staff: Mentor

    I don't believe that including the name of the country provides any useful information.
     
  7. Dec 30, 2011 #6
    Ok so for the first part of question a
    "Is b in {a_1,a_2,a_3}"
    it's just asking me if a is in the set not the span of a_1,a_2,a_3 and sense b=/=a_1=/=a_2=/=a_3 the answer is no
    and there are exactly three vectors in the set.
    This answer makes sense to me now.

    for the second question
    "Is b in w? How many vectors are in W?"
    so b has to be a scalar multiple of one of the vectors in W, W is defined as the span of(a_1,a_2,a_3), so b has to be a scalar multiple of either a_1,a_2, or a_3.
    I don't see how you can
    c*a_1 or c*a_2 or c*a_3 to get b
    were is the scalar multiple
    hm I don't see why the answer is yes

    "How many vectors are in W?"
    Ok it makes sense to me now why the answer is infinitely many.

    "Show that a_1 is in W"
    the answer they provide now makes sense to me.

    I guess I just don't understand why "Is b in w? How many vectors are in W?" this is yes now.
     
  8. Dec 30, 2011 #7
    Ya your probably right lol.
     
  9. Dec 30, 2011 #8

    Mark44

    Staff: Mentor

    Right, right, and good. I also made a correction above.
    Not necessarily. b could be a scalar multiple of one of the vectors in the set, or it could be a linear combination of any two of them or of all three of them.
    No. b is a linear combination (look it up) of the three vectors. That does NOT mean that it is a scalar multiple of any one of them. That's not what linear combination means.
     
  10. Dec 30, 2011 #9
    Thanks
    so I understand what a linear combination is now but I can't seem to come up with one that equals b. Is there any way to conclude that the answer yes without coming up with one?
    this gets it close
    -(a_3 + a_2 /3) = [4;1;-7]
    note the negative sign
     
    Last edited: Dec 30, 2011
  11. Dec 30, 2011 #10

    Mark44

    Staff: Mentor

    Based on where I think you are in your linear algebra knowledge, the only way to show that b is in W is to show the specific linear combination of the vectors in W that produces b.

    What you're doing is solving the equation
    [tex]c_1\begin{bmatrix}1 \\ 0 \\ -2\end{bmatrix} + c_2\begin{bmatrix}0 \\ 3 \\ 6\end{bmatrix} + c_3\begin{bmatrix}-4 \\ -2 \\ 3\end{bmatrix} = \begin{bmatrix}4 \\ 1 \\ -4\end{bmatrix}[/tex]

    Write an augmented matrix for this system and row reduce it to get the constants.
     
  12. Dec 30, 2011 #11
    Ah thank you that makes sense.

    Question
    my book lists for reduced echelon form one of the conditions to be
    The leading entry in each nonzero row is 1.

    Is a case like such
    [1 8
    0 #]
    #=/=0
    considered to be in reduced echelon form? This system is inconsistent I know but is the matrix considered to be in reduced or do I have to multiply by row 2 by 1/# because it's considered a leading entry?
    [1 8
    0 1]
    the problem would then be all the values in the column above it would also have to be zero like so
    [1 0
    0 1]
    which is weird...

    Such a question came up in my mind while taking a quiz and I wasn't sure what to do so I just left it like
    [1 8
    0 1]

    and originally had like

    [1 8
    0 12]

    and I wasn't sure if I left it like that if I would loose a point or not. I was asked to reduce a matrix to reduced echelon form.
     
    Last edited: Dec 30, 2011
  13. Dec 30, 2011 #12

    Mark44

    Staff: Mentor

    Yes, to be reduced (i.e., in reduced row-echelon form), the leading entry in row 2 needs to be 1 and each entry above or below it needs to be 0. To just be in echelon form, the leading entry doesn't need to be 1 and the entries above or below can be whatever. I think that's correct.
     
  14. Dec 30, 2011 #13
    Interesting... hmm... Here's what my book says (see attachment) when I read it I assumed that the conditions for it to be in reduce echelon form met that it also had to meet the conditions to be in echelon form because my book lists the conditions in ascending order so I thought this is what it implied but now that I look through my book it doesn't directly state this any where I'm not really sure
     

    Attached Files:

  15. Dec 30, 2011 #14

    Mark44

    Staff: Mentor

    A matrix can meet the first three conditions (i.e., be in echelon form) without meeting the last two conditions. A matrix in reduced echelon form has to satisfy all five conditions.

    "If a matrix in echelon form satisfies the following additional conditions..."
     
  16. Dec 30, 2011 #15
    ah yes thanks your right
     
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