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How to compute the characteristic polynomial of a companion matrix to a polynomial with matrix-valued...
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Consider we have a polynomial $p = z^m + b_{m-1}z^{m-1} + dotsb + b_0$ with matrix coefficients $b_i in M_n(mathbb{C})$. Then we might consider the companion matrix
$$T = left[
begin{matrix}
0_n & 0_n &dots & b_0 \
I_n & 0_n &dotsb & b_1 \
& ddots && vdots \
&&I_n & b_{m-1}
end{matrix}
right],
$$
where $I_n$ is the identity matrix, and $0_n$ is the zero matrix in $M_n(mathbb{C})$.
$T$ is a block matrix, so we might consider it as a complex-valued matrix of dimension $mcdot n times m cdot n$, and I want to show that its characteristic polynomial $chi_T(z) = det(zcdot I_{ncdot m} - T)$ equals $det(p(z))$, where we consider $p(z)$ as a matrix with polynomial-valued entries.
If we naively compute the characteristic polynomial of $T$ over the ring of matrices we simply obtain $p$, but I'm not shure if I can even apply the Laplace expansion theorem over non-commutative rings. And even then, computing block-wise determinants does and then the determinant of the resulting matrix does in general not yield the determinant of the matrix one started with.
Applying Laplace expansion directly to $zcdot I_{nm} - T$ yields rather ugly mixed terms that I don't know how to handle.
I also found this nice answer on how to show that the characteristic polynomial of the companion matrix equals the polynomial you started with, but I don't see if this generalizes: The claim that $chi_T = mu_T$ ($mu_T$ is the minimal polynomial) is wrong in my situation. For example if $b_i = 0$ for all $i$, then $T^m = 0$, so in general the minimal polynomial has lower degree.
linear-algebra matrices block-matrices
asked 15 hours ago
red_trumpet
682218
add a comment |
up vote
2
down vote
favorite
Consider we have a polynomial $p = z^m + b_{m-1}z^{m-1} + dotsb + b_0$ with matrix coefficients $b_i in M_n(mathbb{C})$. Then we might consider the companion matrix
$$T = left[
begin{matrix}
0_n & 0_n &dots & b_0 \
I_n & 0_n &dotsb & b_1 \
& ddots && vdots \
&&I_n & b_{m-1}
end{matrix}
right],
$$
where $I_n$ is the identity matrix, and $0_n$ is the zero matrix in $M_n(mathbb{C})$.
$T$ is a block matrix, so we might consider it as a complex-valued matrix of dimension $mcdot n times m cdot n$, and I want to show that its characteristic polynomial $chi_T(z) = det(zcdot I_{ncdot m} - T)$ equals $det(p(z))$, where we consider $p(z)$ as a matrix with polynomial-valued entries.
If we naively compute the characteristic polynomial of $T$ over the ring of matrices we simply obtain $p$, but I'm not shure if I can even apply the Laplace expansion theorem over non-commutative rings. And even then, computing block-wise determinants does and then the determinant of the resulting matrix does in general not yield the determinant of the matrix one started with.
Applying Laplace expansion directly to $zcdot I_{nm} - T$ yields rather ugly mixed terms that I don't know how to handle.
I also found this nice answer on how to show that the characteristic polynomial of the companion matrix equals the polynomial you started with, but I don't see if this generalizes: The claim that $chi_T = mu_T$ ($mu_T$ is the minimal polynomial) is wrong in my situation. For example if $b_i = 0$ for all $i$, then $T^m = 0$, so in general the minimal polynomial has lower degree.
linear-algebra matrices block-matrices
asked 15 hours ago
red_trumpet
682218
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider we have a polynomial $p = z^m + b_{m-1}z^{m-1} + dotsb + b_0$ with matrix coefficients $b_i in M_n(mathbb{C})$. Then we might consider the companion matrix
$$T = left[
begin{matrix}
0_n & 0_n &dots & b_0 \
I_n & 0_n &dotsb & b_1 \
& ddots && vdots \
&&I_n & b_{m-1}
end{matrix}
right],
$$
where $I_n$ is the identity matrix, and $0_n$ is the zero matrix in $M_n(mathbb{C})$.
$T$ is a block matrix, so we might consider it as a complex-valued matrix of dimension $mcdot n times m cdot n$, and I want to show that its characteristic polynomial $chi_T(z) = det(zcdot I_{ncdot m} - T)$ equals $det(p(z))$, where we consider $p(z)$ as a matrix with polynomial-valued entries.
If we naively compute the characteristic polynomial of $T$ over the ring of matrices we simply obtain $p$, but I'm not shure if I can even apply the Laplace expansion theorem over non-commutative rings. And even then, computing block-wise determinants does and then the determinant of the resulting matrix does in general not yield the determinant of the matrix one started with.
Applying Laplace expansion directly to $zcdot I_{nm} - T$ yields rather ugly mixed terms that I don't know how to handle.
I also found this nice answer on how to show that the characteristic polynomial of the companion matrix equals the polynomial you started with, but I don't see if this generalizes: The claim that $chi_T = mu_T$ ($mu_T$ is the minimal polynomial) is wrong in my situation. For example if $b_i = 0$ for all $i$, then $T^m = 0$, so in general the minimal polynomial has lower degree.
linear-algebra matrices block-matrices
asked 15 hours ago
red_trumpet
682218
Consider we have a polynomial $p = z^m + b_{m-1}z^{m-1} + dotsb + b_0$ with matrix coefficients $b_i in M_n(mathbb{C})$. Then we might consider the companion matrix
$$T = left[
begin{matrix}
0_n & 0_n &dots & b_0 \
I_n & 0_n &dotsb & b_1 \
& ddots && vdots \
&&I_n & b_{m-1}
end{matrix}
right],
$$
where $I_n$ is the identity matrix, and $0_n$ is the zero matrix in $M_n(mathbb{C})$.
$T$ is a block matrix, so we might consider it as a complex-valued matrix of dimension $mcdot n times m cdot n$, and I want to show that its characteristic polynomial $chi_T(z) = det(zcdot I_{ncdot m} - T)$ equals $det(p(z))$, where we consider $p(z)$ as a matrix with polynomial-valued entries.
If we naively compute the characteristic polynomial of $T$ over the ring of matrices we simply obtain $p$, but I'm not shure if I can even apply the Laplace expansion theorem over non-commutative rings. And even then, computing block-wise determinants does and then the determinant of the resulting matrix does in general not yield the determinant of the matrix one started with.
Applying Laplace expansion directly to $zcdot I_{nm} - T$ yields rather ugly mixed terms that I don't know how to handle.
I also found this nice answer on how to show that the characteristic polynomial of the companion matrix equals the polynomial you started with, but I don't see if this generalizes: The claim that $chi_T = mu_T$ ($mu_T$ is the minimal polynomial) is wrong in my situation. For example if $b_i = 0$ for all $i$, then $T^m = 0$, so in general the minimal polynomial has lower degree.
linear-algebra matrices block-matrices
linear-algebra matrices block-matrices
asked 15 hours ago
red_trumpet
682218
asked 15 hours ago
red_trumpet
682218
asked 15 hours ago
red_trumpet
682218
asked 15 hours ago
red_trumpet
682218
asked 15 hours ago
red_trumpet
682218
682218
add a comment |
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