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Linear Algebra

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm studying for my linear algebra midterm, one of the challenge questions from my textbook is as follows:

    Using the procedure of Example 8 of Chapter 2.3, find whether or not {(0,1,0,1),(-1,1,4,1),(-1,0,2,2)} is or is not a basis for the hyperplane 4[itex]x_{1}[/itex]-[itex]x_{2}[/itex]+[itex]x_{3}[/itex]+[itex]x_{4}[/itex]=0 in [itex]ℝ^{n}[/itex]

    Example 8:
    Show that [itex]\beta[/itex]={[1,2,-1],[1,1,1]} is a basis for the plane -3[itex]x_{1}[/itex]+2[itex]x_{2}[/itex]+[itex]x_{3}[/itex]=0

    We observe that [itex]\beta[/itex] is clearly linearly independent since neither vector is a scalar multiple of the other. Thus, we need to show that every vector in the plane can be written as a linear combination of the vectors in [itex]\beta[/itex]. To do this, observe that any vector [itex]\vec{X}[/itex] in the plane must satisfy the condition of the plane. Hence, every vector in the plane has the form
    [itex]\vec{X}[/itex] = [([itex]x_{1}[/itex]),([itex]x_{2}[/itex]),(3[itex]x_{1}[/itex]-2[itex]x_{2}[/itex])]
    Since [itex]x_{3}[/itex]=3[itex]x_{1}[/itex]-2[itex]x_{2}[/itex]
    Therefore, we now just need to show that the equation
    t1(1,2,-1)+t2(1,1,1)=[([itex]x_{1}[/itex]),([itex]x_{2}[/itex]),(3[itex]x_{1}[/itex]-2[itex]x_{2}[/itex])]
    is always consistent
    Row reducing the corresponding augmented matrix gives
    [(1,0,0), (1,1,0)|(([itex]x_{1}[/itex]),(2[itex]x_{1}[/itex]-[itex]x_{2}[/itex]),(0))]

    2. Relevant equations



    3. The attempt at a solution
    I'm not entirely sure where to start with this one. I've been working really hard in this class, but it's not sticking. Thank you
     
  2. jcsd
  3. Oct 3, 2013 #2
    The first step is to determine whether the given vectors, which I'll call ##v_1, v_2, v_3## are independent. You can stack them up into a 4 x3 matrix and triangularize it. If you wind up with a zero row, they are not independent.

    But suppose they are independent. For them to be a basis you must be able to write any point ##(x_1,x_2,x_3,x_4)## in the hyperplane as a linear combination fo the v's. That is there must be numbers a, b, c such that

    ##(x_1,x_2,x_3,x_4)## = ##av_1 + bv_2 + cv_3##.

    Now how do you know that ##(x_1,x_2,x_3,x_4)## is a vector in the hyperplane? Well, you solve the hyperplane equation for ##x_4 \text { in terms of } (x_1,x_2,x_3)## then plug in the ##x_4## you got into the 4th position. By doing this you incorporated the requirements of the hyperplane into your vector.

    Can you start now? If not, ask some specific questions.
     
  4. Oct 3, 2013 #3
    By the way, linear algebra seems to baffle almost everyone the first time around, so if it isn't sticking that is not unusual.

    One hint might be to go over the vocabulary very carefully and make sure you know what it all means. There is so much new vocabulary that it is hard to absorb it all and relate each thing to the others. But it probably would help you if you took the trouble to do it.

    A second hint is to look around at other books. Some books are completely unclear and your text might be one of those. Some other text, or even the Schaum's outline, might be easier to decipher.
     
  5. Oct 4, 2013 #4

    I found the RREF and found that the set is linearly independent. I really don't understand what to do after that though

    Am I solving for x4 in the hyperplane equation? as in x4=-x3+x2-4x1 ?
     
  6. Oct 4, 2013 #5
    Am I making a solution vector that includes the variables x1, x2, x3 and x4?
     
  7. Oct 4, 2013 #6
    ***
    Yes. Now you have a vector v with 4 components that looks like v = ##(4x_1,-x_2,x_3,-x_3+x_2-4x_1)## This is the nature of each point in the hyperplane.

    Now can you find numbers a,b,c so that a(0,1,0,1)+b(-1,1,4,1) + c(-1,0,2,2) = v? To start, you have

    ##0a-b-c =-4x_1 ##

    You have to set up 3 more equations. Then they are either solvable or not. If so, it's a basis; if not then not a basis.
     
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