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    Consider the solution of the matrix system M x = r, where the matrix and right-hand side are given by

    M= \begin{pmatrix} 4& 4& 0\\4 & 5 & 2\\0 & 2& 4\end{pmatrix}; r = \begin{pmatrix} 2\\ 5\\ 6\end{pmatrix}



    The previous parts to this question was proving what the eigenvectors were (simple dot product) and writing down associated eigenvalues.

    The question I'm stuck on:

    We can write the vector x as a linear combination of the eigenvectors.

    Writing  x = \sum\limits_{i} \beta_i V_i; r =  \sum \limits_{i} \alpha_i V_i, derive a formula for  \beta_i in terms of  \alpha_i and the eigenvalues \lambda_i

    Edit: Is something up with latex? Wow...
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    (Original post by Jagwar Ma)
    Consider the solution of the matrix system M x = r, where the matrix and right-hand side are given by

    M= \begin{pmatrix} 4& 4& 0\\4 & 5 & 2\\0 & 2& 4\end{pmatrix}; r = \begin{pmatrix} 2\\ 5\\ 6\end{pmatrix}



    The previous parts to this question was proving what the eigenvectors were (simple dot product) and writing down associated eigenvalues.

    The question I'm stuck on:

    We can write the vector x as a linear combination of the eigenvectors.

    Writing  x = \sum\limits_{i} \beta_i V_i; r =  \sum \limits_{i} \alpha_i V_i, derive a formula for  \beta_i in terms of  \alpha_i and the eigenvalues \lambda_i

    Edit: Is something up with latex? Wow...
    Yes, something appears to be up in latex land...anyway, press on...

    How far have you got? Have you derived the relation \sum_{i} \beta_{i} \lambda_{i} V_{i} = \sum_{i} \alpha_{i} V_{i} ?

    If V_{3} is the eigenvector corresponding to the zero eigenvalue, what does that imply about \beta_{3} ?
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    (Original post by Gregorius)
    Yes, something appears to be up in latex land...anyway, press on...

    How far have you got? Have you derived the relation \sum_{i} \beta_{i} \lambda_{i} V_{i} = \sum_{i} \alpha_{i} \V_{i} ?

    Now you should notice that the V_{i} are not linearly independent (as the matrix is singular). If V_{3} is the eigenvector corresponding to the zero eigenvalue, what does that imply about \beta_{3} ?
    v3 is not the eigenvector that corresponds to the zero eigenvalue though?

    I did manage to solve it, but now I'm trying to use the result to determine the solution vector x
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    (Original post by Jagwar Ma)
    v3 is not the eigenvector that corresponds to the zero eigenvalue though?
    I gave them an arbitrary ordering for convenience: v3 = (2,-2,1)
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    (Original post by Gregorius)
    I gave them an arbitrary ordering for convenience: v3 = (2,-2,1)
    Ah, yeah that's my V1

    I'm trying to solve to find the solution vector x but don't really understand my lecturers working (to a similar problem).
    mx = r
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    (Original post by Jagwar Ma)
    Ah, yeah that's my V1

    I'm trying to solve to find the solution vector x but don't really understand my lecturers working (to a similar problem).
    mx = r
    OK, so I take v1 = (4,5,2); v2 = (-1,0,2) and v3 = (2,-2,1). Then by inspection, r = v1 + 2 v2. So you now have to solve

    \displaystyle \beta_{1} \lambda_{1} V_{1} + \beta_{2} \lambda_{2} V_{2}= V_{1} + 2 V_{2}

    remembering the comment about  \beta_{3}.
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    (Original post by Gregorius)
    OK, so I take v1 = (4,5,2); v2 = (-1,0,2) and v3 = (2,-2,1). Then by inspection, r = v1 + 2 v2. So you now have to solve

    \displaystyle \beta_{1} \lambda_{1} V_{1} + \beta_{2} \lambda_{2} V_{2}= V_{1} + 2 V_{2}

    remembering the comment about  \beta_{3}.
    r = (2, 5, 6) I've already completed that stage.

    Trying to find the x in Mx = r
 
 
 
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