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Linear Algebra - help wanted with finding geometric multiplicity and eigenspace of... watch

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    Hi there

    Please could somebody confirm if this is right:

    I need to find the geometric multiplicity and corresponding eigenspace for each eigenvalue in the following matrix:

    \begin{pmatrix} 2 & 0 \\5 & 2 \end{pmatrix}

    Det(A-\lambdaI) = 0

    Det\begin{pmatrix} 2 – \lambda & 0 \\5 & 2 – \lambda \end{pmatrix}

    Det =  (2 – \lambda)( 2 – \lambda) – 0 = 0

    Det =  (2 – \lambda) ^{2}
    So  \lambda = 2 repeated

    So, lets find the geometric multiplicity for  \lambda = 2
    So lets substitute  \lambda = 2 into A - \lambda I
    \begin{pmatrix} 2-2 & 0 \\5 & 2-2 \end{pmatrix} = 0

    \begin{pmatrix} 0 & 0 \\5 & 0 \end{pmatrix} = 0

    \begin{pmatrix} 0 & 0 \\5 & 0 \end{pmatrix} = 0 \begin{pmatrix} V_1 \\ V_2\end{pmatrix} =  \begin{pmatrix} 0 \\ 0\end{pmatrix}
    So 5V_1 = 0

    Does this mean the geometric multiplicity = 0?
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    Your third line does not follow from your first. Could you clarify that the original matrix is correctly written?
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    (Original post by Smaug123)
    Your third line does not follow from your first. Could you clarify that the original matrix is correctly written?
    Sorry, typo, I've corrected it now
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    (Original post by jackie11)
    Sorry, typo, I've corrected it now
    Geometric multiplicity of an eigenvalue is by definition never 0. Indeed, if the multiplicity were 0, it wouldn't be an eigenvalue, because that would mean the eigenspace had dimension 0, so contained only the point 0: that is, there would be no eigenvector for that eigenvalue.

    Can you give me without any thought an eigenvector for that matrix, and hence show that the geometric multiplicity is >= 1? (Hint if necessary: it's already a lower-diagonal matrix.) ETA: by "lower-diagonal" I mean "lower-triangular".

    The task is then either to find another eigenvector which is not a multiple of that one, or to show that no such eigenvector exists. There is a system of simultaneous equations to solve to find the eigenvectors. Can you tell me what that system is?
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    (Original post by Smaug123)
    Geometric multiplicity of an eigenvalue is by definition never 0. Indeed, if the multiplicity were 0, it wouldn't be an eigenvalue, because that would mean the eigenspace had dimension 0, so contained only the point 0: that is, there would be no eigenvector for that eigenvalue.

    Can you give me without any thought an eigenvector for that matrix, and hence show that the geometric multiplicity is >= 1? (Hint if necessary: it's already a lower-diagonal matrix.) ETA: by "lower-diagonal" I mean "lower-triangular".

    The task is then either to find another eigenvector which is not a multiple of that one, or to show that no such eigenvector exists. There is a system of simultaneous equations to solve to find the eigenvectors. Can you tell me what that system is?
    Thank you for your reply

    Ok, I now understand that all eigenvalues must have at least one eigenvector, but i still can't see what the eigenvector would be for the eigenvalue of 2.

    \begin{pmatrix} 2  & 0 \\5 & 2  \end{pmatrix} \begin{pmatrix} V_1  \\ V_2 \end{pmatrix} = 2\begin{pmatrix} V_1  \\ V_2 \end{pmatrix}

    So  2V_1 = 2V_1  which is not very helpful

     5V_1 + 2V_2 = 2V_2
     5V_1  = 0
    Therefore  V_1  = 0

    So how do I get anything but 0?
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    (Original post by jackie11)
    Thank you for your reply

    Ok, I now understand that all eigenvalues must have at least one eigenvector, but i still can't see what the eigenvector would be for the eigenvalue of 2.

    \begin{pmatrix} 2  & 0 \\5 & 2  \end{pmatrix} \begin{pmatrix} V_1  \\ V_2 \end{pmatrix} = 2\begin{pmatrix} V_1  \\ V_2 \end{pmatrix}

    So  2V_1 = 2V_1  which is not very helpful

     5V_1 + 2V_2 = 2V_2
     5V_1  = 0
    Therefore  V_1  = 0

    So how do I get anything but 0?
    You have determined that for any eigenvector of eigenvalue 2 (and hence for any eigenvector of any eigenvalue, since the only eigenvalue is 2), the first entry of that eigenvector must be 0. That is, all eigenvectors must lie on the span of the vector {0, 1}. Does that help?
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    (Original post by Smaug123)
    You have determined that for any eigenvector of eigenvalue 2 (and hence for any eigenvector of any eigenvalue, since the only eigenvalue is 2), the first entry of that eigenvector must be 0. That is, all eigenvectors must lie on the span of the vector {0, 1}. Does that help?
    Oh ok, yes I get it now thank you!!!
 
 
 
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