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Thread: 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 $\displaystyle \sigma\left(A\right)$ denotes the multiset of eigenvalues (i.e. all the eigevalues, including multiples) then $\displaystyle \displaystyle \det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda$. Thus, if $\displaystyle A$ is singular then $\displaystyle ]\displaystyle 0=\det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda$ from where it follows that $\displaystyle \lambda$ is in whatever the underlying field you're working with (if that last part didn't make sense replace it with "since the $\displaystyle \lambda$s are in $\displaystyle \mathbb{C}$) it follows that at least one element of $\displaystyle \sigma\left(A\right)$ must be zero.

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

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

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

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

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

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

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

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

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

    Regards.

    Fernando Revilla
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