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Math Help - Eigenvalue proof.

  1. #1
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    Eigenvalue proof.

    Prove that every singular matrix A has 0 as an eigenvalue.

    I don't really have a clue how to prove this.

    I can see that when A is a singular matrix, the eigenvalues are found using the equation A lamda^2 + B lamda = 0 so one value of lamda is always 0.

    I don't know how to prove it though..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chr91 View Post
    Prove that every singular matrix A has 0 as an eigenvalue.

    I don't really have a clue how to prove this.

    I can see that when A is a singular matrix, the eigenvalues are found using the equation A lamda^2 + B lamda = 0 so one value of lamda is always 0.

    I don't know how to prove it though..
    There are multiple results which finish this problem immediately, so it depends what you know. Probably the easiest method is to note that if \sigma\left(A\right) denotes the multiset of eigenvalues (i.e. all the eigevalues, including multiples) then \displaystyle \det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda. Thus, if A is singular then ]\displaystyle 0=\det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda from where it follows that \lambda is in whatever the underlying field you're working with (if that last part didn't make sense replace it with "since the \lambdas are in \mathbb{C}) it follows that at least one element of \sigma\left(A\right) must be zero.

    Anothe direction is that A is singular if and only if Ax=\mathbf{0} for some x\ne \mathbf{0}, but I doubt you know what since this is, basically the question.
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  3. #3
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    \displaystyle det(A)=(a_{11}-\lambda)A_{11}+\sum_{i=2}^{n}a_{i1}A_{i1}

    \displaystyle (a_{11}-\lambda)A_{11}=(a_{11}-\lambda)(a_{22}-\lambda)...(a_{nn}-\lambda)

    \displaystyle =(-1)^n\lambda^n+...+(-1)^{n-1}\lambda^{n-1}

    \displaystyle p(0)=det(A)=\lambda_1\lambda_2...\lambda_n

    \displaystyle (-1)^{n-1}=tr(A)=\sum_{i=1}^{n}\lambda_i

    \displaystyle p(\lambda)=0 has exactly n solutions \displaystyle \lambda_1,...,\lambda_n

    \displaystyle p(\lambda)=(\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)

    \displaystyle p(0)=(\lambda_1)(\lambda_2)...(\lambda_n)=det(A)
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by chr91 View Post
    Prove that every singular matrix A has 0 as an eigenvalue.
    An alternative:

    A\;\textrm{singular}\;\Rightarrow \det A=0 \Rightarrow \det (A-0I)=0\Rightarrow \lambda=0\textrm{\;is\;eingenvalue\;of\;}A

    Regards.

    Fernando Revilla
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