Mixing
A linear map $T: mathbb{R^3 to mathbb{R^3}}$ has a two dimensional invariant subspace.
$begingroup$
Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$
I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.
I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$
In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.
linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form
asked Jan 10 at 11:17
user371231user371231
826511
$endgroup$
add a comment |
$begingroup$
Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$
I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.
I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$
In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.
linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form
asked Jan 10 at 11:17
user371231user371231
826511
$endgroup$
add a comment |
$begingroup$
Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$
I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.
I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$
In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.
linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form
asked Jan 10 at 11:17
user371231user371231
826511
$endgroup$
Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$
I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.
I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$
In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.
linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form
linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form
asked Jan 10 at 11:17
user371231user371231
826511
asked Jan 10 at 11:17
user371231user371231
826511
asked Jan 10 at 11:17
user371231user371231
826511
asked Jan 10 at 11:17
user371231user371231
826511
asked Jan 10 at 11:17
user371231user371231
826511
826511
add a comment |
add a comment |
1 Answer
1
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If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
add a comment |
$begingroup$
If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
add a comment |
$begingroup$
If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
edited Jan 10 at 11:42
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 10 at 11:36
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
add a comment |
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
How can it be a diagonal matrix ? also second matrix how do you get ?
$endgroup$
– user371231
Jan 10 at 11:41
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
$begingroup$
I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
$endgroup$
– José Carlos Santos
Jan 10 at 11:43
add a comment |
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