- #1

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thanks.

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- Thread starter mind0nmath
- Start date

- #1

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thanks.

- #2

mathwonk

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sounds like scalar multiples of the identity no?

- #3

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they are permutation matrices?

- #4

morphism

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Nope -- a permutation matrix can have more than one eigenvalue. mathwonk got it.

- #5

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[tex]

\left[ {\begin{array}{*{20}c}

1 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0 \\

\end{array}} \right],\left[ {\begin{array}{*{20}c}

0 & 0 & 1 \\

1 & 0 & 0 \\

0 & 1 & 0 \\

\end{array}} \right]

[/tex]

- #6

- 350

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[tex](1-\lambda)(\lambda^2-1) = 0 \Rightarrow \lambda = \pm 1[/tex]

That permits two unique eigenvalues, not one.

Besides if you simply choose the coordinate system aligned with the eigenvectors, in that coordinate system the matrix (call it A) will be proportional to the identity matrix. Then the matrix [tex]A-\lambda I[/tex] will vanish in that coordinate system for some complex number [tex]\lambda[/tex], but then it would have to vanish in all coordinate systems and therefore the matrix is proportional to the identity matrix.

- #7

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[tex](1-\lambda)(\lambda^2-1) = 0 \Rightarrow \lambda = \pm 1[/tex]

That permits two unique eigenvalues, not one.

Besides if you simply choose the coordinate system aligned with the eigenvectors, in that coordinate system the matrix (call it A) will be proportional to the identity matrix. Then the matrix [tex]A-\lambda I[/tex] will vanish in that coordinate system for some complex number [tex]\lambda[/tex], but then it would have to vanish in all coordinate systems and therefore the matrix is proportional to the identity matrix.

Yep, I don't really know what was I thinking, because I use a lot of row\column permutations recently, suddenly I thought that you can do anything with them... Sorry for that. Probably I meant matrices similar to identity matrix....

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