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$$A = \begin{bmatrix}0&0&0&0&0&1\\0&0&0&0&1&0\\0&0&0&1&0&0\\1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\end{bmatrix}$$

Find the eigenvalues of $A$. What are the algebraic and geometric multiplicities of the eigenvalues of $A$? Further, find also the characteristic polynomial of $A$.

My attempt: I was reading wiki article on permutation matrix. There it was given that, To calculate the eigenvalues of a permutation matrix $P_σ$, write $σ$ as a product of cycles say, $C_1, C_2,...,C_t$ Let the corresponding lengths of these cycles be $l_1, l_2,..., l_t$ and let $R_i(1≤i≤t)$ be the set of complex solutions of $x^{l_i}=1$. Then union of all $R_i$s is the set of eigenvalues of the corresponding permutation matrix.

Now, here,

$$σ=(16)(25)(34)(41)(52)(63)$$ $$=(13)(46)(2)(5)$$

So that their length cycles respectively are, $l_1=2, l_2=2, l_3=1,l_4=1$

Hence, by above discussion $R_1=\{z∈C: x^2=1\}$ hence

$R_1=\{1,-1,-i\}$ and similarly

$R_2=\{1,-1,-i\}$,

$R_3=R_4=\{1\}$

Hence we saw, eigenvalues values are, $1,-1,-i$ but in key it was given that, $i$ is also an eigenvalue, How? Where I went wrong?

Furtther, please tell me is there is any short way to find algebraic and geometric multiplicity of Permutation matrix? Further, how could one proceed to find Characteristic polynomial of above $6×6$ permutation matrix? Is algebraic multiplicity is equals to geometric multiplicity of Permutation matrix?

Note: time given for such problems in our exam is about ten minutes. So how could one proceed, In such short time? as, by usual way to find geometric multiplicity requires, to find $dim(ker(A-λI)$ which takes time.

Please Help me.

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    $\begingroup$ Possible duplicate of Find the eigenvalues of the block matrix $\endgroup$ Jun 2, 2018 at 11:53
  • $\begingroup$ @RodrigodeAzevedo sir, none of answers, and in fact Question to does not match "Exactly" what I am asking... $\endgroup$ Jun 2, 2018 at 11:54
  • $\begingroup$ @RodrigodeAzevedo, no sir I want to discuss method for finding eigenvalues given on "Wikipedia". There they doesn't discussed that. Further, I also need to find geometric multiplicity, and Algebric multiplicity, there they just find out, eigenvalues. $\endgroup$ Jun 2, 2018 at 11:57
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    $\begingroup$ To begin with, your expression of $\sigma$ as product of cycles is wrong: $2$ and $5$ are clearly not fixed points of the permutation, as your final expression would imply. $\endgroup$ Jun 2, 2018 at 12:21
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    $\begingroup$ And the set of roots complex roots of $x^2-1$ does not have three elements (hint: the degree of that polynomial is $2$). $\endgroup$ Jun 2, 2018 at 12:23

1 Answer 1

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First off, a permutation matrix always has finite multiplicative order (some nonzero power of it equals the identity), and is therefore diagonalisable over the complex numbers; all eigenvalues are roots of unity (their order divides that of of the permutation) and their geometric multiplicities are the same as their algebraic ones.

Then, the reference you found is correct: for each cycle of length$~l$ in the permutation, the characteristic polynomial has a factor $X^l-1$, whose roots are precisely the complex $l$-th roots of unity (all of them simple roots). This is easy to see because you can conjugate the permutation to one in which the cycles each act act on a consecutive range of integers, cycling the range in the forward direction, and then the permutation matrix become block diagonal with each diagonal block being the companion matrix of such a factor $X^l-1$ (with $l$ the size of the block / length of the cycle). In conclusion, the complex eigenvalues of the permutation are those primitive $m$-th roots of unity where $m$ divides the length of some cycle of the permutation, and its multiplicity is the number of cycles whose length is divisible by$~m$.

You do need to compute the cycle lengths of the permutation correctly though (here $2$ and $4$). This is easy, but what you wrote (originally?) in the question is not even close to being a correct method.

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