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Math Help - Non-Bijective Linear Maps

  1. #1
    MHF Contributor Swlabr's Avatar
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    Non-Bijective Linear Maps

    Do all non-bijective linear maps have an eigenvalue 0?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Swlabr View Post
    Do all non-bijective linear maps have an eigenvalue 0?
    Hi Swlabr.

    For A=\begin{pmatrix}a&b\\c&d\end{pmatrix},

    \det(A-\lambda I)\ =\ 0

    \implies\ \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|\ =\ 0

    \implies\ (a-\lambda)(c-\lambda)-bc\ =\ 0

    \implies\ \lambda^2-(a+d)\lambda+ad-bc\ =\ 0

    \implies\ \lambda(\lambda-(a+d))\ =\ 0 if ad-bc=0

    \implies\ \lambda\ =\ 0\ \mbox{or}\ \lambda\ =\ a+d

    So the answer is no, they can have a nonzero eigenvalue if a+d\ne0.
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by TheAbstractionist View Post
    Hi Swlabr.

    For A=\begin{pmatrix}a&b\\c&d\end{pmatrix},
    \det(A-\lambda I)\ =\ 0
    ....
    /snip
    ....
    So the answer is no, they can have a nonzero eigenvalue if a+d\ne0.
    I think the question was whether all non-bijective maps have a 0 eigenvalue, rather than whether non-bijective maps have all eigenvalues as zeros.But of course, then the question is trivial
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Hi Swlabr.

    For A=\begin{pmatrix}a&b\\c&d\end{pmatrix},

    \det(A-\lambda I)\ =\ 0

    \implies\ \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|\ =\ 0

    \implies\ (a-\lambda)(c-\lambda)-bc\ =\ 0

    \implies\ \lambda^2-(a+d)\lambda+ad-bc\ =\ 0

    \implies\ \lambda(\lambda-(a+d))\ =\ 0 if ad-bc=0

    \implies\ \lambda\ =\ 0\ \mbox{or}\ \lambda\ =\ a+d

    So the answer is no, they can have a nonzero eigenvalue if a+d\ne0.


    ...but are they forced to have one that is 0? Alternatively, are non-bijective linear maps forced to have at least one eigenvalue?

    I'm asking as apparently for V a finite dimensional vector space, T \in End(V) such that T is nilpotent (there exists n \in \mathbb{N} such that T^n=0) then zero is the only eigenvalue. It is clear that zero is the only possible eigenvalue, but I'm unsure if it is indeed an eigenvalue. (Note that if T is nilpotent that it is not a bijection...which I hoped would be the route to the solution...)
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Isomorphism View Post
    I think the question was whether all non-bijective maps have a 0 eigenvalue, rather than whether non-bijective maps have all eigenvalues as zeros.But of course, then the question is trivial

    Yes-that is what I was asking. What do you mean 'the answer is trivial'? Are you perhaps inferring that the zero-vector is an Eigenvector?...
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by Swlabr View Post
    Yes-that is what I was asking. What do you mean 'the answer is trivial'? Are you perhaps inferring that the zero-vector is an Eigenvector?...
    Eigenvalues are defined for square matrices. In such cases injectivity implies bijectivity, or rather non-bijectivity implies non-injectivity. But non-injective linear maps are non-invertible and hence 0 is an eigenvalue.
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  7. #7
    MHF Contributor Swlabr's Avatar
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    But why do non-invertible maps have a zero eigenvalue? Can you perhaps write up a short proof?

    Thanks.
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by Swlabr View Post
    But why do non-invertible maps have a zero eigenvalue? Can you perhaps write up a short proof?

    Thanks.
    \det(T) = 0 \Leftrightarrow \det(T - 0I) = 0
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