Determine all 2 by 2 matrices in with coefficients in $mathbb{Z}/19mathbb{Z}$ of order 5 (up to similarity)
$begingroup$
Note that $x^5 - 1 = (x - 1)(x^2 - 4x + 1)(x^2 + 5x + 1)$ in $mathbb{Z}/19mathbb{Z}$.
Suppose $A$ is a 2 by 2 matrix with coefficients in $mathbb{Z}/19mathbb{Z}$ satisfying $A^5 - I = 0$.
Then the minimal polynomial of $A$ divides $x^5 - 1$.
As the characteristic polynomial of a 2 by 2 matrix has degree 2, the minimal polynomial (which divides the characteristic polynomial) has degree 1 or 2.
So the minimal polynomial (and characteristic polynomial) of $A$ must be $x^2 - 4x + 1$ or $x^2 + 5x + 1$ (if the minimal polynomial is $x - 1$ then it is just the identity, with order 1).
If we let $A = begin{pmatrix} a & b \ c & d end{pmatrix}$, we can manually compute coefficients such that its characteristic polynomial $x^2 - (a + d)x + ad - bc$ equals one of the two polynomials above. Is there a way to show that matrices with such characteristic polynomials exist without doing this?
matrices
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
$endgroup$
add a comment |
$begingroup$
Note that $x^5 - 1 = (x - 1)(x^2 - 4x + 1)(x^2 + 5x + 1)$ in $mathbb{Z}/19mathbb{Z}$.
Suppose $A$ is a 2 by 2 matrix with coefficients in $mathbb{Z}/19mathbb{Z}$ satisfying $A^5 - I = 0$.
Then the minimal polynomial of $A$ divides $x^5 - 1$.
As the characteristic polynomial of a 2 by 2 matrix has degree 2, the minimal polynomial (which divides the characteristic polynomial) has degree 1 or 2.
So the minimal polynomial (and characteristic polynomial) of $A$ must be $x^2 - 4x + 1$ or $x^2 + 5x + 1$ (if the minimal polynomial is $x - 1$ then it is just the identity, with order 1).
If we let $A = begin{pmatrix} a & b \ c & d end{pmatrix}$, we can manually compute coefficients such that its characteristic polynomial $x^2 - (a + d)x + ad - bc$ equals one of the two polynomials above. Is there a way to show that matrices with such characteristic polynomials exist without doing this?
matrices
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
$endgroup$
$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18
add a comment |
$begingroup$
Note that $x^5 - 1 = (x - 1)(x^2 - 4x + 1)(x^2 + 5x + 1)$ in $mathbb{Z}/19mathbb{Z}$.
Suppose $A$ is a 2 by 2 matrix with coefficients in $mathbb{Z}/19mathbb{Z}$ satisfying $A^5 - I = 0$.
Then the minimal polynomial of $A$ divides $x^5 - 1$.
As the characteristic polynomial of a 2 by 2 matrix has degree 2, the minimal polynomial (which divides the characteristic polynomial) has degree 1 or 2.
So the minimal polynomial (and characteristic polynomial) of $A$ must be $x^2 - 4x + 1$ or $x^2 + 5x + 1$ (if the minimal polynomial is $x - 1$ then it is just the identity, with order 1).
If we let $A = begin{pmatrix} a & b \ c & d end{pmatrix}$, we can manually compute coefficients such that its characteristic polynomial $x^2 - (a + d)x + ad - bc$ equals one of the two polynomials above. Is there a way to show that matrices with such characteristic polynomials exist without doing this?
matrices
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
$endgroup$
Note that $x^5 - 1 = (x - 1)(x^2 - 4x + 1)(x^2 + 5x + 1)$ in $mathbb{Z}/19mathbb{Z}$.
Suppose $A$ is a 2 by 2 matrix with coefficients in $mathbb{Z}/19mathbb{Z}$ satisfying $A^5 - I = 0$.
Then the minimal polynomial of $A$ divides $x^5 - 1$.
As the characteristic polynomial of a 2 by 2 matrix has degree 2, the minimal polynomial (which divides the characteristic polynomial) has degree 1 or 2.
So the minimal polynomial (and characteristic polynomial) of $A$ must be $x^2 - 4x + 1$ or $x^2 + 5x + 1$ (if the minimal polynomial is $x - 1$ then it is just the identity, with order 1).
If we let $A = begin{pmatrix} a & b \ c & d end{pmatrix}$, we can manually compute coefficients such that its characteristic polynomial $x^2 - (a + d)x + ad - bc$ equals one of the two polynomials above. Is there a way to show that matrices with such characteristic polynomials exist without doing this?
matrices
matrices
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
asked Dec 6 '18 at 20:55
RaekyeRaekye
24539
24539
$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18
add a comment |
$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18
$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18
$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18
add a comment |
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$begingroup$
See: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Micah
Dec 6 '18 at 21:18