Mixing
No nontrivial subspaces implies irreducibility of characteristic polynomial
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Let $k$ be a field and let $V$ be a $k$-vector space. Let $T:Vto V$ be a linear transformation, and suppose $T$ has no invariant subspaces in $V$ other than $0$ and $V$ itself. Does it follow that the characteristic polynomial of $T$ is irreducible?
It's easy to show that the minimal polynomial $mu_T$ must be irreducible: if $mu_T=fg$ Is a nontrivial factorization then $W=ker(f(T))$ is a nontrivial subspace of $V$. Indeed if $W=0$ then $mu_T$ divides $g$, and if $W=V$ then the minimal polynomial divides $f$, a contradiction.
linear-algebra
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
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add a comment |
$begingroup$
Let $k$ be a field and let $V$ be a $k$-vector space. Let $T:Vto V$ be a linear transformation, and suppose $T$ has no invariant subspaces in $V$ other than $0$ and $V$ itself. Does it follow that the characteristic polynomial of $T$ is irreducible?
It's easy to show that the minimal polynomial $mu_T$ must be irreducible: if $mu_T=fg$ Is a nontrivial factorization then $W=ker(f(T))$ is a nontrivial subspace of $V$. Indeed if $W=0$ then $mu_T$ divides $g$, and if $W=V$ then the minimal polynomial divides $f$, a contradiction.
linear-algebra
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and let $V$ be a $k$-vector space. Let $T:Vto V$ be a linear transformation, and suppose $T$ has no invariant subspaces in $V$ other than $0$ and $V$ itself. Does it follow that the characteristic polynomial of $T$ is irreducible?
It's easy to show that the minimal polynomial $mu_T$ must be irreducible: if $mu_T=fg$ Is a nontrivial factorization then $W=ker(f(T))$ is a nontrivial subspace of $V$. Indeed if $W=0$ then $mu_T$ divides $g$, and if $W=V$ then the minimal polynomial divides $f$, a contradiction.
linear-algebra
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
$endgroup$
Let $k$ be a field and let $V$ be a $k$-vector space. Let $T:Vto V$ be a linear transformation, and suppose $T$ has no invariant subspaces in $V$ other than $0$ and $V$ itself. Does it follow that the characteristic polynomial of $T$ is irreducible?
It's easy to show that the minimal polynomial $mu_T$ must be irreducible: if $mu_T=fg$ Is a nontrivial factorization then $W=ker(f(T))$ is a nontrivial subspace of $V$. Indeed if $W=0$ then $mu_T$ divides $g$, and if $W=V$ then the minimal polynomial divides $f$, a contradiction.
linear-algebra
linear-algebra
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
asked Jan 16 at 20:32
leibnewtzleibnewtz
2,5861717
2,5861717
add a comment |
add a comment |
2 Answers
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I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $begin{pmatrix}A&B\0&Cend{pmatrix}$, and therefore, its characteristic polynomial $chi_T=chi_Achi_C$ is reducible.
ii) Conversely, assume that $chi_T$ is reducible; then, there is a vector $vin Vsetminus {0}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,cdots)$ is a proper $T$-invariant space.
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
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Why does such a vector exist?
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– leibnewtz
Jan 18 at 3:27
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See, for example, math.stackexchange.com/questions/414536/…
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– loup blanc
Jan 18 at 11:21
add a comment |
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I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible.
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $begin{pmatrix}A&B\0&Cend{pmatrix}$, and therefore, its characteristic polynomial $chi_T=chi_Achi_C$ is reducible.
ii) Conversely, assume that $chi_T$ is reducible; then, there is a vector $vin Vsetminus {0}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,cdots)$ is a proper $T$-invariant space.
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
$endgroup$
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
add a comment |
$begingroup$
I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $begin{pmatrix}A&B\0&Cend{pmatrix}$, and therefore, its characteristic polynomial $chi_T=chi_Achi_C$ is reducible.
ii) Conversely, assume that $chi_T$ is reducible; then, there is a vector $vin Vsetminus {0}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,cdots)$ is a proper $T$-invariant space.
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
$endgroup$
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
add a comment |
$begingroup$
I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $begin{pmatrix}A&B\0&Cend{pmatrix}$, and therefore, its characteristic polynomial $chi_T=chi_Achi_C$ is reducible.
ii) Conversely, assume that $chi_T$ is reducible; then, there is a vector $vin Vsetminus {0}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,cdots)$ is a proper $T$-invariant space.
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
$endgroup$
I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $begin{pmatrix}A&B\0&Cend{pmatrix}$, and therefore, its characteristic polynomial $chi_T=chi_Achi_C$ is reducible.
ii) Conversely, assume that $chi_T$ is reducible; then, there is a vector $vin Vsetminus {0}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,cdots)$ is a proper $T$-invariant space.
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
edited Jan 17 at 11:30
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
answered Jan 17 at 11:05
loup blancloup blanc
23.4k21851
23.4k21851
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
add a comment |
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
Why does such a vector exist?
$endgroup$
– leibnewtz
Jan 18 at 3:27
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
$begingroup$
See, for example, math.stackexchange.com/questions/414536/…
$endgroup$
– loup blanc
Jan 18 at 11:21
add a comment |
$begingroup$
I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible.
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
$endgroup$
add a comment |
$begingroup$
I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible.
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
$endgroup$
add a comment |
$begingroup$
I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible.
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
$endgroup$
I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible.
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
answered Jan 17 at 5:11
leibnewtzleibnewtz
2,5861717
2,5861717
add a comment |
add a comment |
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