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Thread: I need a counter example

  1. #1
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    I need a counter example

    Hello!

    Any kind soul can provide me with a counter example for this:

    False statement: Every square matrix has an eigenvalue.

    Thanks
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  2. #2
    Super Member Deadstar's Avatar
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    Do you mean have no REAL eigenvalues? Or none at all?

    Here is an example of a matrix with no real eigenvalues.

    $\displaystyle \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by noob mathematician View Post
    Hello!

    Any kind soul can provide me with a counter example for this:

    False statement: Every square matrix has an eigenvalue.

    Thanks
    $\displaystyle {\bf{0}}_{2 \times 2}$

    CB
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  4. #4
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    Sorry for posting under calculus

    Hm.. I think that $\displaystyle \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$ has an eigenvalue of 0? Am i wrong?

    Ya I guess the question is asking about real and non-real eigenvalues.

    Thanks
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  5. #5
    Super Member Showcase_22's Avatar
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    $\displaystyle Det|M-\lambda I|=0$

    $\displaystyle \begin{vmatrix}
    0-\lambda&0\\
    0&0-\lambda
    \end{vmatrix}=0$

    $\displaystyle \lambda^2=0 \Rightarrow \ \lambda=0$

    But when $\displaystyle \lambda=0$, we get an eigenvalue of $\displaystyle \begin{pmatrix}
    0\\
    0 \end{pmatrix}$.

    The problem is that this 0 vector has no direction which contradicts the definition of an eigenvalue.

    (ie. eigenvectors have the same direction when they undergo a transformation. They're only scaled by the eigenvalue. A 0 vector does not have a direction so cannot be an eigenvector).
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by noob mathematician View Post
    Sorry for posting under calculus

    Hm.. I think that $\displaystyle \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$ has an eigenvalue of 0? Am i wrong?

    Ya I guess the question is asking about real and non-real eigenvalues.

    Thanks
    An eigen vector $\displaystyle x$ of $\displaystyle A$ is a non-zero vector such that:

    $\displaystyle Ax= \lambda x$

    for some scalar $\displaystyle \lambda$ (there is usually no requirement that $\displaystyle \lambda$ be non-zero here). So every non-zero vector is an eigen vector of the zero matrix, and $\displaystyle 0$ is the corresponding eigenvalue.

    So yes, I was wrong the zero matrix does have an eigen value (I had what had to be non-zero confused).

    CB
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