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Finding values for a linearly independent subset

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    There is a vector space with real entries of all 2x2 matrices. You have to find what values of [itex]\alpha[/itex][itex]\in[/itex]ℝ make the set Z = [itex]\{
    \begin{pmatrix}
    1 & 2\\
    1 & 0
    \end{pmatrix},
    \begin{pmatrix}
    3 & 7\\
    0 & 0
    \end{pmatrix},
    \begin{pmatrix}
    2 & 6\\
    \alpha & 0
    \end{pmatrix}
    \}
    [/itex]


    2. Relevant equations
    To find for linear independence, I wrote the equation:
    [tex]a\begin{pmatrix}
    1 & 2\\
    1 & 0
    \end{pmatrix}+b
    \begin{pmatrix}
    3 & 7\\
    0 & 0
    \end{pmatrix}+c
    \begin{pmatrix}
    2 & 6\\
    \alpha & 0
    \end{pmatrix}=
    \begin{pmatrix}
    0 & 0\\
    0 & 0
    \end{pmatrix}
    [/tex]


    3. The attempt at a solution
    This is where I'm a bit confused. Is that equation correct and then you would simply write it as a matrix and then use row reductions to make the matrix into reduced row echelon form?
    Or would you not test it for the set of matrices but just for the last matrix and let it equal to the zero vector to find linear independence - and find the value of alpha that way?
     
  2. jcsd
  3. Mar 24, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That is a good approach.
    I don't understand what you mean. The third matrix will never be identical to 0, independent of the value of α. If there would be such an α, the system would be linear dependent (as it would have a zero vector), of course.
     
  4. Mar 25, 2013 #3
    Okay, thanks. So I set up the matrix:
    [tex]
    \begin{pmatrix}
    1 & 2 & 3 & 7 & 2 & 6 & | & 0 & 0\\
    1 & 0 & 0 & 0 & \alpha & 0 & | & 0 & 0
    \end{pmatrix}
    [/tex]

    I then did row operations and made the matrix into reduced row echelon form, obtaining:
    [tex]
    \begin{pmatrix}
    1 & 0 & 0 & 0 & \alpha & 0 & | & 0 & 0\\
    0 & 1 & 3/2 & 7/2 & -\alpha/2+1 & 3 & | & 0 & 0
    \end{pmatrix}
    [/tex]

    So, α is now in both rows. How am I able to solve to find the value of α? I did try through back substitution, with row 1 giving v1=-αv3. Although, there are still many values in row 2 so I can't solve it further.
     
  5. Mar 25, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You need a separate row for all 4 components of your matrices.
     
  6. Mar 25, 2013 #5
    Sorry, I don't really understand by what you mean for having a separate row for each component? And how would you create it?
     
  7. Mar 25, 2013 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You have a 4-dimensional vector space, as your matrices have 4 independent components.

    Your equation for linear independence can be written as
    [tex]a\begin{pmatrix}
    1 \\
    2 \\
    1 \\
    0\end{pmatrix}+b
    \begin{pmatrix}
    3 \\ 7\\
    0 \\ 0
    \end{pmatrix}+c
    \begin{pmatrix}
    2 \\ 6\\
    \alpha \\ 0
    \end{pmatrix}=
    \begin{pmatrix}
    0 \\ 0\\
    0 \\ 0
    \end{pmatrix}
    [/tex]
    You can convert this into a matrix in the usual way.
     
  8. Mar 25, 2013 #7
    So, you would convert that into a matrix to then solve for reduced row echelon form and as it is linearly independent, you should only have one value in each row, with the rest as zero.
    I have tried solving the matrix and obtained:
    [tex]
    \begin{pmatrix}
    1 & 2 & 0 & | & 0\\
    0 & 1 & 2 & | & 0\\
    0 & 0 & \alpha+4 & | & 0\\
    0 & 0 & 0 & | & 0
    \end{pmatrix}
    [/tex]
    However, I don't know how to solve it further from there.
    Yet, since it's given that the system is linearly independent, then where α+4 is should equal 1, moreover with the "2's" being 0 as well.
    Did I make some error then since I still have some "2's" in my matrix, since it is not yet linearly independent?
     
  9. Mar 25, 2013 #8
    As you have written, let

    [tex]a\begin{pmatrix}
    1 & 2\\
    1 & 0
    \end{pmatrix}+b
    \begin{pmatrix}
    3 & 7\\
    0 & 0
    \end{pmatrix}+c
    \begin{pmatrix}
    2 & 6\\
    \alpha & 0
    \end{pmatrix}=
    \begin{pmatrix}
    0 & 0\\
    0 & 0
    \end{pmatrix}
    [/tex]

    Then,

    [tex]\begin{pmatrix}
    a+3b+2c & 2a+7b+6c\\
    a+c\alpha & 0
    \end{pmatrix}
    =
    \begin{pmatrix}
    0 & 0\\
    0 & 0
    \end{pmatrix}[/tex]

    This can be written as a system of linear equations (which is another representation of what mfb has written above):

    [tex]
    a+3b+2c=0\\
    2a+7b+6c=0\\
    a+c\alpha=0\\
    [/tex]

    If we reduce the system to its echelon form we get

    [tex]
    a+3b+2c=0\\
    b+2c=0\\
    c(\alpha+4)=0\\
    [/tex]

    For the the set of 3 matrices to be linearly independent, [tex]\alpha\neq-4[/tex].
     
    Last edited: Mar 25, 2013
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