Determinant of nearly companion matrix

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For the matrix
$$
A=beginpmatrix0 &0 &0 &0 &0 & 0&1&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&0&0&0&1&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

I would like to compute the determinant (resp. characteristic polynomial).

The matrix looks nearly the same as the companion matrix (only the first and third row differ)

$$
B=beginpmatrix0 &1 &0 &0 &0 & 0&0&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&1&0&0&0&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

for which it is known that the characteristic polynomial is given by
$$
t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3-ct^2-bt -a
$$

My question is if I can deduce the characteristic polynomial of matrix $A$ from that of $B$.

(of course i could simply compute the determinant of A by, e.g. gauss. but i am interested to know whether it suffices to know the determinant of B)

matrices determinant

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asked Jul 20 at 20:06

Rhjg

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up vote
2
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For the matrix
$$
A=beginpmatrix0 &0 &0 &0 &0 & 0&1&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&0&0&0&1&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

I would like to compute the determinant (resp. characteristic polynomial).

The matrix looks nearly the same as the companion matrix (only the first and third row differ)

$$
B=beginpmatrix0 &1 &0 &0 &0 & 0&0&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&1&0&0&0&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

for which it is known that the characteristic polynomial is given by
$$
t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3-ct^2-bt -a
$$

My question is if I can deduce the characteristic polynomial of matrix $A$ from that of $B$.

(of course i could simply compute the determinant of A by, e.g. gauss. but i am interested to know whether it suffices to know the determinant of B)

matrices determinant

share|cite|improve this question

asked Jul 20 at 20:06

Rhjg

258214

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

For the matrix
$$
A=beginpmatrix0 &0 &0 &0 &0 & 0&1&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&0&0&0&1&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

I would like to compute the determinant (resp. characteristic polynomial).

The matrix looks nearly the same as the companion matrix (only the first and third row differ)

$$
B=beginpmatrix0 &1 &0 &0 &0 & 0&0&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&1&0&0&0&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

for which it is known that the characteristic polynomial is given by
$$
t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3-ct^2-bt -a
$$

My question is if I can deduce the characteristic polynomial of matrix $A$ from that of $B$.

(of course i could simply compute the determinant of A by, e.g. gauss. but i am interested to know whether it suffices to know the determinant of B)

matrices determinant

share|cite|improve this question

asked Jul 20 at 20:06

Rhjg

258214

For the matrix
$$
A=beginpmatrix0 &0 &0 &0 &0 & 0&1&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&0&0&0&1&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

I would like to compute the determinant (resp. characteristic polynomial).

The matrix looks nearly the same as the companion matrix (only the first and third row differ)

$$
B=beginpmatrix0 &1 &0 &0 &0 & 0&0&0&0&0&0&0&0\
0&0&1&0&0&0&0&0&0&0&0&0&0\
0&0&0&1&0&0&0&0&0&0&0&0&0\
0&0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0&0\
0&0&0&0&0&0&0&0&1&0&0&0&0\
0&0&0&0&0&0&0&0&0&1&0&0&0\
0&0&0&0&0&0&0&0&0&0&1&0&0\
0&0&0&0&0&0&0&0&0&0&0&1&0\
0&0&0&0&0&0&0&0&0&0&0&0&1\
a&b&c&d&e&f&g&h&i&j&k&l&mendpmatrix
$$

for which it is known that the characteristic polynomial is given by
$$
t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3-ct^2-bt -a
$$

My question is if I can deduce the characteristic polynomial of matrix $A$ from that of $B$.

(of course i could simply compute the determinant of A by, e.g. gauss. but i am interested to know whether it suffices to know the determinant of B)

matrices determinant

share|cite|improve this question

asked Jul 20 at 20:06

Rhjg

258214

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 20:06

Rhjg

258214

asked Jul 20 at 20:06

Rhjg

258214

asked Jul 20 at 20:06

Rhjg

258214

258214

add a comment | 

add a comment | 

1 Answer
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They are similar and different at the same time.

I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.

Anyway, if you note the common part:

$G(t)=t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$

Then $begincasesP_A(t)=G(t)-(a+c)t^5-bt^4\ P_B(t)=G(t)-ct^2-bt-aendcases$

Eventually you could rewrite:

$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$

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answered Jul 20 at 21:07

zwim

11k627

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

They are similar and different at the same time.

I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.

Anyway, if you note the common part:

$G(t)=t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$

Then $begincasesP_A(t)=G(t)-(a+c)t^5-bt^4\ P_B(t)=G(t)-ct^2-bt-aendcases$

Eventually you could rewrite:

$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$

share|cite|improve this answer

answered Jul 20 at 21:07

zwim

11k627

add a comment | 

up vote
1
down vote



accepted

They are similar and different at the same time.

I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.

Anyway, if you note the common part:

$G(t)=t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$

Then $begincasesP_A(t)=G(t)-(a+c)t^5-bt^4\ P_B(t)=G(t)-ct^2-bt-aendcases$

Eventually you could rewrite:

$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$

share|cite|improve this answer

answered Jul 20 at 21:07

zwim

11k627

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

They are similar and different at the same time.

I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.

Anyway, if you note the common part:

$G(t)=t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$

Then $begincasesP_A(t)=G(t)-(a+c)t^5-bt^4\ P_B(t)=G(t)-ct^2-bt-aendcases$

Eventually you could rewrite:

$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$

share|cite|improve this answer

answered Jul 20 at 21:07

zwim

11k627

They are similar and different at the same time.

I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.

Anyway, if you note the common part:

$G(t)=t^13-mt^12-lt^11-kt^10-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$

Then $begincasesP_A(t)=G(t)-(a+c)t^5-bt^4\ P_B(t)=G(t)-ct^2-bt-aendcases$

Eventually you could rewrite:

$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$

share|cite|improve this answer

answered Jul 20 at 21:07

zwim

11k627

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 21:07

zwim

11k627

answered Jul 20 at 21:07

zwim

11k627

answered Jul 20 at 21:07

zwim

11k627

11k627

add a comment | 

add a comment | 

 

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