Transformation restriction to commutative matrices

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Let $T: Bbb M_3x3(Bbb R) rightarrow Bbb M_3x3(Bbb R)$ be a linear transformation such that $T(B) = AB$ where:

$$ A=
beginpmatrix
2 & -1 & -1 \
-1 & 2 & -1 \
-1 & -1 & 2 \
endpmatrix
$$

now let $W = AB=BA$ be the group of all matrices that commute with $A$.

I want to

• find the minimal polyonimal for $T|_w$.


• prove that $W$ is the direct sum of two spaces $W = W_1 oplus W_2$

---

my idea so far was:

• Let P be any matrix:$$ P =beginpmatrix
  a & b & c \
  d & e & f \
  g & h & i \
  endpmatrix $$
  then we could look at the transformation as $Bbb R^9 rightarrow Bbb R^9$ like that:
  $$ Tbeginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix = beginpmatrix
  beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & & \
  & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & \
  & & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix \
  endpmatrix beginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix $$




• Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o=
  beginpmatrix
  0 & 0 & 0 \
  0 & 3 & 0 \
  0 & 0 & 3 \
  endpmatrix
  $$

which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$

my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...

Can you help me to better understand the restriction?
Thanks..

linear-algebra matrices eigenvalues-eigenvectors linear-transformations invariant-subspace

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edited Jul 31 at 9:25

asked Jul 31 at 7:33

Limitless

986

This question has an open bounty worth +50
reputation from Limitless ending ending at 2018-08-14 10:08:20Z">in 5 days.

This question has not received enough attention.

I am looking for a detailed walkthrough to the solution of this question

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up vote
1
down vote

favorite

Let $T: Bbb M_3x3(Bbb R) rightarrow Bbb M_3x3(Bbb R)$ be a linear transformation such that $T(B) = AB$ where:

$$ A=
beginpmatrix
2 & -1 & -1 \
-1 & 2 & -1 \
-1 & -1 & 2 \
endpmatrix
$$

now let $W = AB=BA$ be the group of all matrices that commute with $A$.

I want to

• find the minimal polyonimal for $T|_w$.


• prove that $W$ is the direct sum of two spaces $W = W_1 oplus W_2$

---

my idea so far was:

• Let P be any matrix:$$ P =beginpmatrix
  a & b & c \
  d & e & f \
  g & h & i \
  endpmatrix $$
  then we could look at the transformation as $Bbb R^9 rightarrow Bbb R^9$ like that:
  $$ Tbeginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix = beginpmatrix
  beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & & \
  & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & \
  & & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix \
  endpmatrix beginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix $$




• Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o=
  beginpmatrix
  0 & 0 & 0 \
  0 & 3 & 0 \
  0 & 0 & 3 \
  endpmatrix
  $$

which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$

my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...

Can you help me to better understand the restriction?
Thanks..

linear-algebra matrices eigenvalues-eigenvectors linear-transformations invariant-subspace

share|cite|improve this question

edited Jul 31 at 9:25

asked Jul 31 at 7:33

Limitless

986

This question has an open bounty worth +50
reputation from Limitless ending ending at 2018-08-14 10:08:20Z">in 5 days.

This question has not received enough attention.

I am looking for a detailed walkthrough to the solution of this question

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $T: Bbb M_3x3(Bbb R) rightarrow Bbb M_3x3(Bbb R)$ be a linear transformation such that $T(B) = AB$ where:

$$ A=
beginpmatrix
2 & -1 & -1 \
-1 & 2 & -1 \
-1 & -1 & 2 \
endpmatrix
$$

now let $W = AB=BA$ be the group of all matrices that commute with $A$.

I want to

• find the minimal polyonimal for $T|_w$.


• prove that $W$ is the direct sum of two spaces $W = W_1 oplus W_2$

---

my idea so far was:

• Let P be any matrix:$$ P =beginpmatrix
  a & b & c \
  d & e & f \
  g & h & i \
  endpmatrix $$
  then we could look at the transformation as $Bbb R^9 rightarrow Bbb R^9$ like that:
  $$ Tbeginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix = beginpmatrix
  beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & & \
  & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & \
  & & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix \
  endpmatrix beginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix $$




• Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o=
  beginpmatrix
  0 & 0 & 0 \
  0 & 3 & 0 \
  0 & 0 & 3 \
  endpmatrix
  $$

which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$

my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...

Can you help me to better understand the restriction?
Thanks..

linear-algebra matrices eigenvalues-eigenvectors linear-transformations invariant-subspace

share|cite|improve this question

edited Jul 31 at 9:25

asked Jul 31 at 7:33

Limitless

986

Let $T: Bbb M_3x3(Bbb R) rightarrow Bbb M_3x3(Bbb R)$ be a linear transformation such that $T(B) = AB$ where:

$$ A=
beginpmatrix
2 & -1 & -1 \
-1 & 2 & -1 \
-1 & -1 & 2 \
endpmatrix
$$

now let $W = AB=BA$ be the group of all matrices that commute with $A$.

I want to

• find the minimal polyonimal for $T|_w$.


• prove that $W$ is the direct sum of two spaces $W = W_1 oplus W_2$

---

my idea so far was:

• Let P be any matrix:$$ P =beginpmatrix
  a & b & c \
  d & e & f \
  g & h & i \
  endpmatrix $$
  then we could look at the transformation as $Bbb R^9 rightarrow Bbb R^9$ like that:
  $$ Tbeginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix = beginpmatrix
  beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & & \
  & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix & \
  & & beginpmatrix
  2 & -1 & -1 \
  -1 & 2 & -1 \
  -1 & -1 & 2 \
  endpmatrix \
  endpmatrix beginpmatrix
  a \
  d \
  g \
  b \
  e \
  h \
  c \
  f \
  i \
  endpmatrix $$




• Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o=
  beginpmatrix
  0 & 0 & 0 \
  0 & 3 & 0 \
  0 & 0 & 3 \
  endpmatrix
  $$

which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$

my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...

Can you help me to better understand the restriction?
Thanks..

linear-algebra matrices eigenvalues-eigenvectors linear-transformations invariant-subspace

share|cite|improve this question

edited Jul 31 at 9:25

asked Jul 31 at 7:33

Limitless

986

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 9:25

edited Jul 31 at 9:25

edited Jul 31 at 9:25

asked Jul 31 at 7:33

Limitless

986

asked Jul 31 at 7:33

Limitless

986

asked Jul 31 at 7:33

Limitless

986

986

This question has an open bounty worth +50
reputation from Limitless ending ending at 2018-08-14 10:08:20Z">in 5 days.

This question has not received enough attention.

I am looking for a detailed walkthrough to the solution of this question

This question has an open bounty worth +50
reputation from Limitless ending ending at 2018-08-14 10:08:20Z">in 5 days.

This question has not received enough attention.

I am looking for a detailed walkthrough to the solution of this question

add a comment | 

add a comment | 

1 Answer
1

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up vote
0
down vote

$W$ is a vector space and not a group. Note that $T=Aotimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)subset W$. Moreover $spectrum(T)=0,0,0,3,3,3,3,3,3$; cf.

https://en.wikipedia.org/wiki/Kronecker_product

We may assume that $A=diag(0,3,3)$ -because $(Potimes I)(Aotimes I)(Potimes I)^-1=(PAP^-1)otimes I$-.

Thus $W=diag(a,R);ainmathbbR,Rin M_2$ has dimension $5$. Moreover $W=W_1oplus W_2$ where $W_1=span(E_1,1),W_2=span(E_2,2,E_2,3,E_3,2,E_3,3)$ and $T_W_1=0,T_W_2=3I$.

Finally, $T_$ has $2$ eigenvalues $0,3$ and $spectrum(T_)=0,3,3,3,3$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

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1 Answer
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1 Answer
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$W$ is a vector space and not a group. Note that $T=Aotimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)subset W$. Moreover $spectrum(T)=0,0,0,3,3,3,3,3,3$; cf.

https://en.wikipedia.org/wiki/Kronecker_product

We may assume that $A=diag(0,3,3)$ -because $(Potimes I)(Aotimes I)(Potimes I)^-1=(PAP^-1)otimes I$-.

Thus $W=diag(a,R);ainmathbbR,Rin M_2$ has dimension $5$. Moreover $W=W_1oplus W_2$ where $W_1=span(E_1,1),W_2=span(E_2,2,E_2,3,E_3,2,E_3,3)$ and $T_W_1=0,T_W_2=3I$.

Finally, $T_$ has $2$ eigenvalues $0,3$ and $spectrum(T_)=0,3,3,3,3$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

add a comment | 

up vote
0
down vote

$W$ is a vector space and not a group. Note that $T=Aotimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)subset W$. Moreover $spectrum(T)=0,0,0,3,3,3,3,3,3$; cf.

https://en.wikipedia.org/wiki/Kronecker_product

We may assume that $A=diag(0,3,3)$ -because $(Potimes I)(Aotimes I)(Potimes I)^-1=(PAP^-1)otimes I$-.

Thus $W=diag(a,R);ainmathbbR,Rin M_2$ has dimension $5$. Moreover $W=W_1oplus W_2$ where $W_1=span(E_1,1),W_2=span(E_2,2,E_2,3,E_3,2,E_3,3)$ and $T_W_1=0,T_W_2=3I$.

Finally, $T_$ has $2$ eigenvalues $0,3$ and $spectrum(T_)=0,3,3,3,3$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

add a comment | 

up vote
0
down vote

up vote
0
down vote

$W$ is a vector space and not a group. Note that $T=Aotimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)subset W$. Moreover $spectrum(T)=0,0,0,3,3,3,3,3,3$; cf.

https://en.wikipedia.org/wiki/Kronecker_product

We may assume that $A=diag(0,3,3)$ -because $(Potimes I)(Aotimes I)(Potimes I)^-1=(PAP^-1)otimes I$-.

Thus $W=diag(a,R);ainmathbbR,Rin M_2$ has dimension $5$. Moreover $W=W_1oplus W_2$ where $W_1=span(E_1,1),W_2=span(E_2,2,E_2,3,E_3,2,E_3,3)$ and $T_W_1=0,T_W_2=3I$.

Finally, $T_$ has $2$ eigenvalues $0,3$ and $spectrum(T_)=0,3,3,3,3$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

$W$ is a vector space and not a group. Note that $T=Aotimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)subset W$. Moreover $spectrum(T)=0,0,0,3,3,3,3,3,3$; cf.

https://en.wikipedia.org/wiki/Kronecker_product

We may assume that $A=diag(0,3,3)$ -because $(Potimes I)(Aotimes I)(Potimes I)^-1=(PAP^-1)otimes I$-.

Thus $W=diag(a,R);ainmathbbR,Rin M_2$ has dimension $5$. Moreover $W=W_1oplus W_2$ where $W_1=span(E_1,1),W_2=span(E_2,2,E_2,3,E_3,2,E_3,3)$ and $T_W_1=0,T_W_2=3I$.

Finally, $T_$ has $2$ eigenvalues $0,3$ and $spectrum(T_)=0,3,3,3,3$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

loup blanc

20.3k21549

answered 21 hours ago

loup blanc

20.3k21549

answered 21 hours ago

loup blanc

20.3k21549

20.3k21549

add a comment | 

add a comment | 

 

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