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    let M be a 3x3 matrix and let s be a eigen vector

    Ms = lambda * s
    (M-identity * lambda)s = 0

    Why is determinant of (M - identity * lambda) equal to zero? I don't know understand that at all, can someone please explain to me in a simple and easy to understand way please. I've seen some youtube videos but they start talking about eigen space and stuff like which i've never heard of
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    (Original post by Bealzibub)
    let M be a 3x3 matrix and let s be a eigen vector

    Ms = lambda * s
    (M-identity * lambda)s = 0

    Why is determinant of (M - identity * lambda) equal to zero? I don't know understand that at all, can someone please explain to me in a simple and easy to understand way please. I've seen some youtube videos but they start talking about eigen space and stuff like which i've never heard of
    EDIT: btw when I wrote (AI-lambda), I actually meant (A-lambda*I) where I is the identity matrix. (AI-lambda) actually makes no sense because it is a matrix minus a scalar.

    You want to find all non-zero vectors that get sent to a multiple of the original vector. (You don't want the zero vector because that automatically is an eigenvector for any eigenvalue so it is not very interesting)

    That is Au=lambda*u where u is the vector. Such a vector is called an eigenvector and the value of lambda is called the eigenvalue corresponding to the eigenvector u.

    So you want to find for what values (that is what eigenvalues) you can find eigenvectors for.

    So you want solutions to Au=lambda*u which is the same as Au-lambda*u=0 which is the same as (AI-lambda)*u=0.

    Now it turns out that the only way you can hope to find non-zero values for u (that is find an eigenvector) is if determinant of (AI-lambda)=0 this is because if the determinant of this is not zero then you can invert it and times both sides by this inverse to get

    (AI-lambda)^(-1)(AI-lambda)*u=(AI-lambda)^(-1)*0

    Which is the same as u=0 that is the zero eigenvector precisely what we don't want and so we have to look for the determinant to be zero.

    Eigenspace is just the set of all eigenvectors corresponding to a given eigenvalue.

    So say we know 5 is an eigenvalue then the 5 eigenspace is just the vectors such that Au=5u.
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    (Original post by poorform)
    You want to find all non-zero vectors that get sent to a multiple of the original vector. (You don't want the zero vector because that automatically is an eigenvector for any eigenvalue so it is not very interesting)

    That is Au=lambda*u where u is the vector. Such a vector is called an eigenvector and the value of lambda is called the eigenvalue corresponding to the eigenvector u.

    So you want to find for what values (that is what eigenvalues) you can find eigenvectors for.

    So you want solutions to Au=lambda*u which is the same as Au-lambda*u=0 which is the same as (AI-lambda)*u=0.

    Now it turns out that the only way you can hope to find non-zero values for u (that is find an eigenvector) is if determinant of (AI-lambda)=0 this is because if the determinant of this is not zero then you can invert it and times both sides by this inverse to get

    (AI-lambda)^(-1)(AI-lambda)*u=(AI-lambda)^(-1)*0

    Which is the same as u=0 that is the zero eigenvector precisely what we don't want and so we have to look for the determinant to be zero.

    Eigenspace is just the set of all eigenvectors corresponding to a given eigenvalue.

    So say we know 5 is an eigenvalue then the 5 eigenspace is just the vectors such that Au=5u.
    Thank you, understood.
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    (Original post by Bealzibub)
    Thank you, understood.
    btw when I wrote (AI-lambda), I actually meant (A-lambda*I) where I is the identity matrix. (AI-lambda) actually makes no sense because it is a matrix minus a scalar.
 
 
 
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