Problem on characteristic polynomial of a matrix












2












$begingroup$


Without computing $det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 times 4$ matrix given by:



$$A = begin{bmatrix} 0 & 0 & 0 & -4 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 5 \ 0 & 0 & 1 & 0 end{bmatrix}$$










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:30










  • $begingroup$
    That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:51
















2












$begingroup$


Without computing $det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 times 4$ matrix given by:



$$A = begin{bmatrix} 0 & 0 & 0 & -4 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 5 \ 0 & 0 & 1 & 0 end{bmatrix}$$










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:30










  • $begingroup$
    That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:51














2












2








2


1



$begingroup$


Without computing $det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 times 4$ matrix given by:



$$A = begin{bmatrix} 0 & 0 & 0 & -4 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 5 \ 0 & 0 & 1 & 0 end{bmatrix}$$










share|cite|improve this question











$endgroup$




Without computing $det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 times 4$ matrix given by:



$$A = begin{bmatrix} 0 & 0 & 0 & -4 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 5 \ 0 & 0 & 1 & 0 end{bmatrix}$$







linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 6:43









Robert Lewis

48.9k23168




48.9k23168










asked Jan 17 at 6:28









Prashant ThakuriPrashant Thakuri

112




112








  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:30










  • $begingroup$
    That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:51














  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:30










  • $begingroup$
    That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:51








2




2




$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:30




$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Anyway, I think that may be a companion matrix...
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:30












$begingroup$
That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:51




$begingroup$
That is a companion matrix: en.wikipedia.org/wiki/Companion_matrix
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm





In your peculiar case however, your matrix has the form of a companion matrix:



$$
A=begin{bmatrix}0&0&dots &0&-c_{0}\1&0&dots &0&-c_{1}\0&1&dots &0&-c_{2}\vdots &vdots &ddots &vdots &vdots \0&0&dots &1&-c_{{n-1}}end{bmatrix}
$$



and by identification the characteristic polynomial is
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$





In the general case if you want to play the game of not using $det(lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:



For a $A$ a $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$



However there are a lot of computations and it is easier to directly compute $det(lambda I -A)$.



Anyway, here is the details for your example:
$$
A=left(
begin{array}{cccc}
0 & 0 & 0 & -4 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 5 \
0 & 0 & 1 & 0 \
end{array}
right)
$$





  • $m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

  • $m=1$, $text{tr}(A)=0$, hence $c_{4-1}=c_3=0$


  • $m=2$, $text{tr}(A^2)=10$
    $$
    c_{4-2}=frac{(-1)^2}{2!}left|
    begin{array}{cc}
    0 & 2-1 \ 10 & 0
    end{array}
    right|=frac{1}{2}(-10)=-5
    $$


  • $m=3$, $text{tr}(A^3)=0$
    $$
    c_{4-3}=frac{(-1)^3}{3!}left|
    begin{array}{ccc}
    0 & 3-1 & 0 \ 10 & 0 & 3-2 \
    0 & 10 & 0
    end{array}
    right|=frac{-1}{6}(0)=0
    $$


  • $m=4$, $text{tr}(A^4)=34$
    $$
    c_{4-4}=frac{(-1)^4}{4!}left|
    begin{array}{cccc}
    0 & 4-1 & 0 & 0 \
    10 & 0 & 4-2 & 0 \
    0 & 10 & 0 & 4-3 \
    34 & 0 & 10 & 0 \
    end{array}
    right|=frac{1}{24}(96)=4
    $$



Your characteristic polynomial is
$$
P(lambda)=4-5lambda^2+lambda^4
$$



which is hopefully the same answer as the "companion matrix" identification trick.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Probably overkill in this example.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:35










  • $begingroup$
    @LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
    $endgroup$
    – Picaud Vincent
    Jan 17 at 6:38












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm





In your peculiar case however, your matrix has the form of a companion matrix:



$$
A=begin{bmatrix}0&0&dots &0&-c_{0}\1&0&dots &0&-c_{1}\0&1&dots &0&-c_{2}\vdots &vdots &ddots &vdots &vdots \0&0&dots &1&-c_{{n-1}}end{bmatrix}
$$



and by identification the characteristic polynomial is
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$





In the general case if you want to play the game of not using $det(lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:



For a $A$ a $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$



However there are a lot of computations and it is easier to directly compute $det(lambda I -A)$.



Anyway, here is the details for your example:
$$
A=left(
begin{array}{cccc}
0 & 0 & 0 & -4 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 5 \
0 & 0 & 1 & 0 \
end{array}
right)
$$





  • $m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

  • $m=1$, $text{tr}(A)=0$, hence $c_{4-1}=c_3=0$


  • $m=2$, $text{tr}(A^2)=10$
    $$
    c_{4-2}=frac{(-1)^2}{2!}left|
    begin{array}{cc}
    0 & 2-1 \ 10 & 0
    end{array}
    right|=frac{1}{2}(-10)=-5
    $$


  • $m=3$, $text{tr}(A^3)=0$
    $$
    c_{4-3}=frac{(-1)^3}{3!}left|
    begin{array}{ccc}
    0 & 3-1 & 0 \ 10 & 0 & 3-2 \
    0 & 10 & 0
    end{array}
    right|=frac{-1}{6}(0)=0
    $$


  • $m=4$, $text{tr}(A^4)=34$
    $$
    c_{4-4}=frac{(-1)^4}{4!}left|
    begin{array}{cccc}
    0 & 4-1 & 0 & 0 \
    10 & 0 & 4-2 & 0 \
    0 & 10 & 0 & 4-3 \
    34 & 0 & 10 & 0 \
    end{array}
    right|=frac{1}{24}(96)=4
    $$



Your characteristic polynomial is
$$
P(lambda)=4-5lambda^2+lambda^4
$$



which is hopefully the same answer as the "companion matrix" identification trick.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Probably overkill in this example.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:35










  • $begingroup$
    @LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
    $endgroup$
    – Picaud Vincent
    Jan 17 at 6:38
















1












$begingroup$

To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm





In your peculiar case however, your matrix has the form of a companion matrix:



$$
A=begin{bmatrix}0&0&dots &0&-c_{0}\1&0&dots &0&-c_{1}\0&1&dots &0&-c_{2}\vdots &vdots &ddots &vdots &vdots \0&0&dots &1&-c_{{n-1}}end{bmatrix}
$$



and by identification the characteristic polynomial is
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$





In the general case if you want to play the game of not using $det(lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:



For a $A$ a $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$



However there are a lot of computations and it is easier to directly compute $det(lambda I -A)$.



Anyway, here is the details for your example:
$$
A=left(
begin{array}{cccc}
0 & 0 & 0 & -4 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 5 \
0 & 0 & 1 & 0 \
end{array}
right)
$$





  • $m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

  • $m=1$, $text{tr}(A)=0$, hence $c_{4-1}=c_3=0$


  • $m=2$, $text{tr}(A^2)=10$
    $$
    c_{4-2}=frac{(-1)^2}{2!}left|
    begin{array}{cc}
    0 & 2-1 \ 10 & 0
    end{array}
    right|=frac{1}{2}(-10)=-5
    $$


  • $m=3$, $text{tr}(A^3)=0$
    $$
    c_{4-3}=frac{(-1)^3}{3!}left|
    begin{array}{ccc}
    0 & 3-1 & 0 \ 10 & 0 & 3-2 \
    0 & 10 & 0
    end{array}
    right|=frac{-1}{6}(0)=0
    $$


  • $m=4$, $text{tr}(A^4)=34$
    $$
    c_{4-4}=frac{(-1)^4}{4!}left|
    begin{array}{cccc}
    0 & 4-1 & 0 & 0 \
    10 & 0 & 4-2 & 0 \
    0 & 10 & 0 & 4-3 \
    34 & 0 & 10 & 0 \
    end{array}
    right|=frac{1}{24}(96)=4
    $$



Your characteristic polynomial is
$$
P(lambda)=4-5lambda^2+lambda^4
$$



which is hopefully the same answer as the "companion matrix" identification trick.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Probably overkill in this example.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:35










  • $begingroup$
    @LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
    $endgroup$
    – Picaud Vincent
    Jan 17 at 6:38














1












1








1





$begingroup$

To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm





In your peculiar case however, your matrix has the form of a companion matrix:



$$
A=begin{bmatrix}0&0&dots &0&-c_{0}\1&0&dots &0&-c_{1}\0&1&dots &0&-c_{2}\vdots &vdots &ddots &vdots &vdots \0&0&dots &1&-c_{{n-1}}end{bmatrix}
$$



and by identification the characteristic polynomial is
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$





In the general case if you want to play the game of not using $det(lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:



For a $A$ a $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$



However there are a lot of computations and it is easier to directly compute $det(lambda I -A)$.



Anyway, here is the details for your example:
$$
A=left(
begin{array}{cccc}
0 & 0 & 0 & -4 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 5 \
0 & 0 & 1 & 0 \
end{array}
right)
$$





  • $m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

  • $m=1$, $text{tr}(A)=0$, hence $c_{4-1}=c_3=0$


  • $m=2$, $text{tr}(A^2)=10$
    $$
    c_{4-2}=frac{(-1)^2}{2!}left|
    begin{array}{cc}
    0 & 2-1 \ 10 & 0
    end{array}
    right|=frac{1}{2}(-10)=-5
    $$


  • $m=3$, $text{tr}(A^3)=0$
    $$
    c_{4-3}=frac{(-1)^3}{3!}left|
    begin{array}{ccc}
    0 & 3-1 & 0 \ 10 & 0 & 3-2 \
    0 & 10 & 0
    end{array}
    right|=frac{-1}{6}(0)=0
    $$


  • $m=4$, $text{tr}(A^4)=34$
    $$
    c_{4-4}=frac{(-1)^4}{4!}left|
    begin{array}{cccc}
    0 & 4-1 & 0 & 0 \
    10 & 0 & 4-2 & 0 \
    0 & 10 & 0 & 4-3 \
    34 & 0 & 10 & 0 \
    end{array}
    right|=frac{1}{24}(96)=4
    $$



Your characteristic polynomial is
$$
P(lambda)=4-5lambda^2+lambda^4
$$



which is hopefully the same answer as the "companion matrix" identification trick.






share|cite|improve this answer











$endgroup$



To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm





In your peculiar case however, your matrix has the form of a companion matrix:



$$
A=begin{bmatrix}0&0&dots &0&-c_{0}\1&0&dots &0&-c_{1}\0&1&dots &0&-c_{2}\vdots &vdots &ddots &vdots &vdots \0&0&dots &1&-c_{{n-1}}end{bmatrix}
$$



and by identification the characteristic polynomial is
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$





In the general case if you want to play the game of not using $det(lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:



For a $A$ a $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$



However there are a lot of computations and it is easier to directly compute $det(lambda I -A)$.



Anyway, here is the details for your example:
$$
A=left(
begin{array}{cccc}
0 & 0 & 0 & -4 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 5 \
0 & 0 & 1 & 0 \
end{array}
right)
$$





  • $m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

  • $m=1$, $text{tr}(A)=0$, hence $c_{4-1}=c_3=0$


  • $m=2$, $text{tr}(A^2)=10$
    $$
    c_{4-2}=frac{(-1)^2}{2!}left|
    begin{array}{cc}
    0 & 2-1 \ 10 & 0
    end{array}
    right|=frac{1}{2}(-10)=-5
    $$


  • $m=3$, $text{tr}(A^3)=0$
    $$
    c_{4-3}=frac{(-1)^3}{3!}left|
    begin{array}{ccc}
    0 & 3-1 & 0 \ 10 & 0 & 3-2 \
    0 & 10 & 0
    end{array}
    right|=frac{-1}{6}(0)=0
    $$


  • $m=4$, $text{tr}(A^4)=34$
    $$
    c_{4-4}=frac{(-1)^4}{4!}left|
    begin{array}{cccc}
    0 & 4-1 & 0 & 0 \
    10 & 0 & 4-2 & 0 \
    0 & 10 & 0 & 4-3 \
    34 & 0 & 10 & 0 \
    end{array}
    right|=frac{1}{24}(96)=4
    $$



Your characteristic polynomial is
$$
P(lambda)=4-5lambda^2+lambda^4
$$



which is hopefully the same answer as the "companion matrix" identification trick.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 9:44

























answered Jan 17 at 6:33









Picaud VincentPicaud Vincent

1,434310




1,434310








  • 1




    $begingroup$
    Probably overkill in this example.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:35










  • $begingroup$
    @LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
    $endgroup$
    – Picaud Vincent
    Jan 17 at 6:38














  • 1




    $begingroup$
    Probably overkill in this example.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 6:35










  • $begingroup$
    @LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
    $endgroup$
    – Picaud Vincent
    Jan 17 at 6:38








1




1




$begingroup$
Probably overkill in this example.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:35




$begingroup$
Probably overkill in this example.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 6:35












$begingroup$
@LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
$endgroup$
– Picaud Vincent
Jan 17 at 6:38




$begingroup$
@LordSharktheUnknown yes, however I m writing a more compact & detailed example using the Jacobi formula, hold on
$endgroup$
– Picaud Vincent
Jan 17 at 6:38


















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