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Thread: Linear combinations (from a 5x7 matrix)

  1. #1
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    Linear combinations (from a 5x7 matrix)

    Sorry this might be a bit messy! I apologize!

    The subset S = {v1, v2, v3, v4, v5, v6, v7} of R^5 is given by:
    v1 = (-2, -2, -1, 1, 2), v2 = (3, 3, 2, -1, -3), v3 = (5, 5, 4, -1, -5), v4 = (3, 2, 2, -2, -4), v5 = (4, 3, 3, -2, -5), v6 = (-2, -2, -1, 2, 3), v7 = (-3, -4, -1, 3, 4)


    The reduced row echelon form of the matrix A is the matrix B where:


    A = $\displaystyle \begin{bmatrix} -2 & 3 & 5 & 3 & 4 & -2 & -3 \\ -2 & 3 & 5 & 2 & 3 & -2 & -4 \\ 1 & 2 & 4 & 2 & 3 & -1 & -1 \\ 1 & -1 & -1 & -2 & -2 & 2 & 3 \\ 2 & -3 & -5 & -4 & -5 & 3 & 4 \end{bmatrix}$

    AND

    B = $\displaystyle \begin{bmatrix} 1 & 0 & 2 & 0 & 1 & 0 & 1 \\ 0 & 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$


    (A) Write down a four element subset S1 of S so that every vector which is a linear combination of elements of S is already a linear combination of elements of S1.

    I have:
    Columns associated with v1, v2, v4 and v6 have leading entries, thus s1 = v1=a, v2 = b, v4 = c, v6 = d
    (where a, b, c, d are real values)



    (B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.


    I currently have:
    v3 = 2v1 + 3v2 (this is correct)
    v5 = v1 + v2 + v3 (this is not correct)
    v7 = v1 + v3 + 2v4 (this is not correct)


    That took a while to write up!

    Help, maybe from scratch?


    Thanks!
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  2. #2
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    Quote Originally Posted by iExcavate View Post
    Sorry this might be a bit messy! I apologize!

    The subset S = {v1, v2, v3, v4, v5, v6, v7} of R^5 is given by:
    v1 = (-2, -2, -1, 1, 2), v2 = (3, 3, 2, -1, -3), v3 = (5, 5, 4, -1, -5), v4 = (3, 2, 2, -2, -4), v5 = (4, 3, 3, -2, -5), v6 = (-2, -2, -1, 2, 3), v7 = (-3, -4, -1, 3, 4)


    The reduced row echelon form of the matrix A is the matrix B where:


    A = $\displaystyle \begin{bmatrix} -2 & 3 & 5 & 3 & 4 & -2 & -3 \\ -2 & 3 & 5 & 2 & 3 & -2 & -4 \\ 1 & 2 & 4 & 2 & 3 & -1 & -1 \\ 1 & -1 & -1 & -2 & -2 & 2 & 3 \\ 2 & -3 & -5 & -4 & -5 & 3 & 4 \end{bmatrix}$

    AND

    B = $\displaystyle \begin{bmatrix} 1 & 0 & 2 & 0 & 1 & 0 & 1 \\ 0 & 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$


    (A) Write down a four element subset S1 of S so that every vector which is a linear combination of elements of S is already a linear combination of elements of S1.

    I have:
    Columns associated with v1, v2, v4 and v6 have leading entries, thus s1 = v1=a, v2 = b, v4 = c, v6 = d
    (where a, b, c, d are real values)



    (B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.


    I currently have:
    v3 = 2v1 + 3v2 (this is correct)
    v5 = v1 + v2 + v3 (this is not correct)
    v7 = v1 + v3 + 2v4 (this is not correct)


    That took a while to write up!

    Help, maybe from scratch?


    Thanks!
    It looks like you have a mistake in the first column of A.
    Maybe fix it and carry out the reduction again.
    Last edited by PiperAlpha167; Apr 16th 2010 at 06:07 AM. Reason: Fix wording.
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  3. #3
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    Your picking of 4 linearly independent vectors went well, so now you can form any other vector within $\displaystyle \mathbb{R}^4$

    $\displaystyle S_1 = v_1, v_2, v_4, v_6 = \begin{bmatrix} -2 \\ -2 \\ 1 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix} 3 \\ 3 \\ 2 \\ -1 \\ -3\end{bmatrix},\begin{bmatrix} 3 \\ 2 \\ 2 \\ -2 \\ -4\end{bmatrix},\begin{bmatrix} -2 \\ -2 \\ -1 \\ 2 \\ 3\end{bmatrix}$

    You now want to make a linear combination of the basis elements of $\displaystyle S_1$ to generate the elements $\displaystyle v_3,v_5,v_7$

    So for example:

    $\displaystyle v_3 = \begin{bmatrix} 5 \\ 5 \\ 4 \\ -1 \\ -5\end{bmatrix} = a\cdot v_1 + b\cdot v_2 + c\cdot v_4 + d\cdot v_6 $

    You now get a system of equations you should be able to solve.
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  4. #4
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    Sorry, for the previous part, I had written the question itself out incorrectly! In the matrix, the '1' in column A should be '-1', like the brackets! The row-reduction is still correct!

    Thanks, also, how do I find 'a, b, c, d' for the linear combinations? I'm a little confused sorry brouwer
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  5. #5
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    I'll give you a simple example: I want to express $\displaystyle \begin{bmatrix}1 \\ 0\end{bmatrix}$ as a linear combination of $\displaystyle \begin{bmatrix}2 \\ 3\end{bmatrix},\begin{bmatrix}1 \\ 1\end{bmatrix}$. You'll get the following system:

    $\displaystyle a\cdot \begin{bmatrix}2 \\ 3\end{bmatrix} + b \cdot \begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 0 \end{bmatrix}$

    This is the same as the equations:

    $\displaystyle 2a+b = 1$
    $\displaystyle 3a+b = 0$

    Now we have to get $\displaystyle a$ in terms of $\displaystyle b$ and substitute in the other equation so:

    $\displaystyle a = \frac{-b}{3}$

    from the 2nd equation and plug this in the 1st equation to get

    $\displaystyle \frac{-2b}{3} + b = 1$

    And we solve for $\displaystyle b$ to get $\displaystyle b = 3$. Filling in this $\displaystyle b$ in either equation will give us our desired $\displaystyle a$. With 4 variables it's the same, but you have to substitute a few times. Good luck.
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  6. #6
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    Quote Originally Posted by iExcavate View Post

    (B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.
    Can't you just read the linear combination off the row reduced matrix? I learnt this method, but cannot understand it in terms of this 5 x 7 matrix. I didn't think there should be a need to use simultaneous equations?



    v3 = 2v1 + 3v2

    How do I read v5 and v7 off the row reduced matrix? Perhaps, that is an easier way?
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  7. #7
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    Yes it's also possible to read the linear combination off your reduced matrix. And it's really a lot easier actually, sorry.

    $\displaystyle v_3 = 2v_1+3v_2$
    $\displaystyle v_5 = v_1+v_2+v_4$

    I don't know why you put $\displaystyle v_3$ here again, as it's a linear combination of the elements of $\displaystyle S_1$. Because these are 4 linearly independent vectors and span whole $\displaystyle \mathbb{R}^4$, which basically means that you can form any other vector that looks like $\displaystyle \begin{bmatrix}a & b & c & d & 0\end{bmatrix}^T$ out of these. $\displaystyle v_7$ can be read off like the other vectors but just in terms of $\displaystyle av_1+bv_2+cv_4+dv_6$.
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