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Characteristic polynomial of $T:M_n(mathbb{F}) rightarrow M_n(mathbb{F}) , TX = AX (Ain M_n(mathbb{F}))$
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Question - how would I proceed to find the characteristic polynomial of $T:M_n(mathbb{F}) rightarrow M_n(mathbb{F}) , TX = AX (Ain M_n(mathbb{F}))$ ?
What I've been trying:
Given the the standard base ${E_{11}, E_{12}, dots, E_{nn}}$ of $M_n(mathbb{F})$ in which ($E_{ij})_{kl} =left{begin{matrix}
1,& k=i and l=j \ 0, &otherwise
end{matrix}right.$
$T$ can be represented by the following $n^2times n^2$ matrix:
$$[T] = begin{pmatrix}
(A)_{11}I_n&(A)_{12}I_n&cdots&(A)_{1n}I_n\
(A)_{21}I_n&(A)_{22}I_n&cdots&(A)_{2n}I_n\
vdots&vdots&ddots&vdots\
(A)_{n1}I_n&(A)_{n2}I_n&cdots&(A)_{nn}I_n
end{pmatrix} $$
Now, from from here I'd like to calculate $det([T]-tI_{n^2})$, and this is the point where I got stuck.
I'd be glad for ideas on how to proceed from here, or ideas for other ways to tackle the problem.
linear-algebra matrices
edited 17 hours ago
red_trumpet
682218
asked Apr 14 '17 at 14:15
j3M
604516
add a comment |
up vote
6
down vote
favorite
Question - how would I proceed to find the characteristic polynomial of $T:M_n(mathbb{F}) rightarrow M_n(mathbb{F}) , TX = AX (Ain M_n(mathbb{F}))$ ?
What I've been trying:
Given the the standard base ${E_{11}, E_{12}, dots, E_{nn}}$ of $M_n(mathbb{F})$ in which ($E_{ij})_{kl} =left{begin{matrix}
1,& k=i and l=j \ 0, &otherwise
end{matrix}right.$
$T$ can be represented by the following $n^2times n^2$ matrix:
$$[T] = begin{pmatrix}
(A)_{11}I_n&(A)_{12}I_n&cdots&(A)_{1n}I_n\
(A)_{21}I_n&(A)_{22}I_n&cdots&(A)_{2n}I_n\
vdots&vdots&ddots&vdots\
(A)_{n1}I_n&(A)_{n2}I_n&cdots&(A)_{nn}I_n
end{pmatrix} $$
Now, from from here I'd like to calculate $det([T]-tI_{n^2})$, and this is the point where I got stuck.
I'd be glad for ideas on how to proceed from here, or ideas for other ways to tackle the problem.
linear-algebra matrices
edited 17 hours ago
red_trumpet
682218
asked Apr 14 '17 at 14:15
j3M
604516
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Question - how would I proceed to find the characteristic polynomial of $T:M_n(mathbb{F}) rightarrow M_n(mathbb{F}) , TX = AX (Ain M_n(mathbb{F}))$ ?
What I've been trying:
Given the the standard base ${E_{11}, E_{12}, dots, E_{nn}}$ of $M_n(mathbb{F})$ in which ($E_{ij})_{kl} =left{begin{matrix}
1,& k=i and l=j \ 0, &otherwise
end{matrix}right.$
$T$ can be represented by the following $n^2times n^2$ matrix:
$$[T] = begin{pmatrix}
(A)_{11}I_n&(A)_{12}I_n&cdots&(A)_{1n}I_n\
(A)_{21}I_n&(A)_{22}I_n&cdots&(A)_{2n}I_n\
vdots&vdots&ddots&vdots\
(A)_{n1}I_n&(A)_{n2}I_n&cdots&(A)_{nn}I_n
end{pmatrix} $$
Now, from from here I'd like to calculate $det([T]-tI_{n^2})$, and this is the point where I got stuck.
I'd be glad for ideas on how to proceed from here, or ideas for other ways to tackle the problem.
linear-algebra matrices
edited 17 hours ago
red_trumpet
682218
asked Apr 14 '17 at 14:15
j3M
604516
Question - how would I proceed to find the characteristic polynomial of $T:M_n(mathbb{F}) rightarrow M_n(mathbb{F}) , TX = AX (Ain M_n(mathbb{F}))$ ?
What I've been trying:
Given the the standard base ${E_{11}, E_{12}, dots, E_{nn}}$ of $M_n(mathbb{F})$ in which ($E_{ij})_{kl} =left{begin{matrix}
1,& k=i and l=j \ 0, &otherwise
end{matrix}right.$
$T$ can be represented by the following $n^2times n^2$ matrix:
$$[T] = begin{pmatrix}
(A)_{11}I_n&(A)_{12}I_n&cdots&(A)_{1n}I_n\
(A)_{21}I_n&(A)_{22}I_n&cdots&(A)_{2n}I_n\
vdots&vdots&ddots&vdots\
(A)_{n1}I_n&(A)_{n2}I_n&cdots&(A)_{nn}I_n
end{pmatrix} $$
Now, from from here I'd like to calculate $det([T]-tI_{n^2})$, and this is the point where I got stuck.
I'd be glad for ideas on how to proceed from here, or ideas for other ways to tackle the problem.
linear-algebra matrices
linear-algebra matrices
edited 17 hours ago
red_trumpet
682218
asked Apr 14 '17 at 14:15
j3M
604516
edited 17 hours ago
red_trumpet
682218
asked Apr 14 '17 at 14:15
j3M
604516
edited 17 hours ago
red_trumpet
682218
edited 17 hours ago
red_trumpet
682218
edited 17 hours ago
red_trumpet
682218
682218
asked Apr 14 '17 at 14:15
j3M
604516
asked Apr 14 '17 at 14:15
j3M
604516
asked Apr 14 '17 at 14:15
j3M
604516
604516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
You have $(AE_{i,j})_{k,l}= 0$ if $lneq j$ and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $chi_T(X)=chi_A(X)^n$.
answered Apr 14 '17 at 14:46
Yoël
451110
add a comment |
up vote
1
down vote
You can view $M_n(Bbb{F})$ as the direct sum of $n$ copies of $Bbb{F}^n$, say $V_1,V_2,ldots,V_n$, where $V_i$ represents the $i$th column. So, basically, $T:bigoplus_{i=1}^nV_ito bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,ldots,v_n)mapsto (Av_1,Av_2,ldots,Av_n).$$ So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_ito V_i$ is the same as the linear operator $A$. We can use a more general result below to prove that $chi_T(t)=big(chi_A(t)big)^n$.
Let $U_1,U_2,ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_ito U_i$ be linear maps. Then, the direct sum $S=bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,ldots,S_k$ is the linear transformation $S:bigoplus_{i=1}^kU_ito bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,ldots, u_k)=(S_1u_1,S_2u_2,ldots,S_ku_k).$$
Then, the characteristic polynomial $chi_S(t)$ of $S$ is the product of the characteristic polynomials $chi_{S_i}(t)$ of each $S_i$. Id est,
$$chi_{bigoplus_{i=1}^kS_i}(t)=prod_{i=1}^kchi_{S_i}(t).$$ (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
answered 14 hours ago
Zvi
3,165221
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You have $(AE_{i,j})_{k,l}= 0$ if $lneq j$ and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $chi_T(X)=chi_A(X)^n$.
answered Apr 14 '17 at 14:46
Yoël
451110
add a comment |
up vote
5
down vote
accepted
You have $(AE_{i,j})_{k,l}= 0$ if $lneq j$ and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $chi_T(X)=chi_A(X)^n$.
answered Apr 14 '17 at 14:46
Yoël
451110
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You have $(AE_{i,j})_{k,l}= 0$ if $lneq j$ and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $chi_T(X)=chi_A(X)^n$.
answered Apr 14 '17 at 14:46
Yoël
451110
You have $(AE_{i,j})_{k,l}= 0$ if $lneq j$ and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $chi_T(X)=chi_A(X)^n$.
answered Apr 14 '17 at 14:46
Yoël
451110
edited Apr 14 '17 at 14:54
answered Apr 14 '17 at 14:46
Yoël
451110
answered Apr 14 '17 at 14:46
Yoël
451110
answered Apr 14 '17 at 14:46
Yoël
451110
451110
add a comment |
add a comment |
up vote
1
down vote
You can view $M_n(Bbb{F})$ as the direct sum of $n$ copies of $Bbb{F}^n$, say $V_1,V_2,ldots,V_n$, where $V_i$ represents the $i$th column. So, basically, $T:bigoplus_{i=1}^nV_ito bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,ldots,v_n)mapsto (Av_1,Av_2,ldots,Av_n).$$ So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_ito V_i$ is the same as the linear operator $A$. We can use a more general result below to prove that $chi_T(t)=big(chi_A(t)big)^n$.
Let $U_1,U_2,ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_ito U_i$ be linear maps. Then, the direct sum $S=bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,ldots,S_k$ is the linear transformation $S:bigoplus_{i=1}^kU_ito bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,ldots, u_k)=(S_1u_1,S_2u_2,ldots,S_ku_k).$$
Then, the characteristic polynomial $chi_S(t)$ of $S$ is the product of the characteristic polynomials $chi_{S_i}(t)$ of each $S_i$. Id est,
$$chi_{bigoplus_{i=1}^kS_i}(t)=prod_{i=1}^kchi_{S_i}(t).$$ (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
answered 14 hours ago
Zvi
3,165221
add a comment |
up vote
1
down vote
You can view $M_n(Bbb{F})$ as the direct sum of $n$ copies of $Bbb{F}^n$, say $V_1,V_2,ldots,V_n$, where $V_i$ represents the $i$th column. So, basically, $T:bigoplus_{i=1}^nV_ito bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,ldots,v_n)mapsto (Av_1,Av_2,ldots,Av_n).$$ So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_ito V_i$ is the same as the linear operator $A$. We can use a more general result below to prove that $chi_T(t)=big(chi_A(t)big)^n$.
Let $U_1,U_2,ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_ito U_i$ be linear maps. Then, the direct sum $S=bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,ldots,S_k$ is the linear transformation $S:bigoplus_{i=1}^kU_ito bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,ldots, u_k)=(S_1u_1,S_2u_2,ldots,S_ku_k).$$
Then, the characteristic polynomial $chi_S(t)$ of $S$ is the product of the characteristic polynomials $chi_{S_i}(t)$ of each $S_i$. Id est,
$$chi_{bigoplus_{i=1}^kS_i}(t)=prod_{i=1}^kchi_{S_i}(t).$$ (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
answered 14 hours ago
Zvi
3,165221
add a comment |
up vote
1
down vote
up vote
1
down vote
You can view $M_n(Bbb{F})$ as the direct sum of $n$ copies of $Bbb{F}^n$, say $V_1,V_2,ldots,V_n$, where $V_i$ represents the $i$th column. So, basically, $T:bigoplus_{i=1}^nV_ito bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,ldots,v_n)mapsto (Av_1,Av_2,ldots,Av_n).$$ So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_ito V_i$ is the same as the linear operator $A$. We can use a more general result below to prove that $chi_T(t)=big(chi_A(t)big)^n$.
Let $U_1,U_2,ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_ito U_i$ be linear maps. Then, the direct sum $S=bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,ldots,S_k$ is the linear transformation $S:bigoplus_{i=1}^kU_ito bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,ldots, u_k)=(S_1u_1,S_2u_2,ldots,S_ku_k).$$
Then, the characteristic polynomial $chi_S(t)$ of $S$ is the product of the characteristic polynomials $chi_{S_i}(t)$ of each $S_i$. Id est,
$$chi_{bigoplus_{i=1}^kS_i}(t)=prod_{i=1}^kchi_{S_i}(t).$$ (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
answered 14 hours ago
Zvi
3,165221
You can view $M_n(Bbb{F})$ as the direct sum of $n$ copies of $Bbb{F}^n$, say $V_1,V_2,ldots,V_n$, where $V_i$ represents the $i$th column. So, basically, $T:bigoplus_{i=1}^nV_ito bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,ldots,v_n)mapsto (Av_1,Av_2,ldots,Av_n).$$ So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_ito V_i$ is the same as the linear operator $A$. We can use a more general result below to prove that $chi_T(t)=big(chi_A(t)big)^n$.
Let $U_1,U_2,ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_ito U_i$ be linear maps. Then, the direct sum $S=bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,ldots,S_k$ is the linear transformation $S:bigoplus_{i=1}^kU_ito bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,ldots, u_k)=(S_1u_1,S_2u_2,ldots,S_ku_k).$$
Then, the characteristic polynomial $chi_S(t)$ of $S$ is the product of the characteristic polynomials $chi_{S_i}(t)$ of each $S_i$. Id est,
$$chi_{bigoplus_{i=1}^kS_i}(t)=prod_{i=1}^kchi_{S_i}(t).$$ (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
answered 14 hours ago
Zvi
3,165221
answered 14 hours ago
Zvi
3,165221
answered 14 hours ago
Zvi
3,165221
answered 14 hours ago
Zvi
3,165221
3,165221
add a comment |
add a comment |
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