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Math Help - Eigenvalue Question

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    Eigenvalue Question

    Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

    ??
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ruprotein View Post
    Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

    ??
    i'm a bit rusty in linear algebra (even though i took it just last semester) so here's your answer in a nutshell, you have to transform this into a rigorous proof.

    Recall that eigen values and eigen vectors are "unchanged" under linear transformations. so the matrix resulting from applying a linear transformation on a finite demensional matrix will have the same eigenvalues and eigenvectors as the first matrix.

    now, if we have an eigenvalue = 0, then det(I*lambda - A) = 0, this means that detA = 0, which means that the matrix is not invertible, since for a matrix A to be invertible, we must have detA not= 0

    so that should get you started. sorry i can't help more, i was lost for most of that class
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  3. #3
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    Quote Originally Posted by ruprotein View Post
    Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

    ??
    I prove it one way.

    Let the standard matrix for T be A, that is [T]=A.

    We say that A is invertible by hypothesis, meaning
    det(A) not = 0.

    The eigenvalues k are solutions to the polynomial,

    det(k*I-A)=0

    Now, if k is zero, then,

    det(k*I-A)=det(0*I-A)=det(-A)=(-1)^n * det(A) !=0.
    Because det(A)!=0.

    Hence k=0 is not and eigenvalue.
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