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Prove that $ V=(operatornamenull T^n-2) oplus (operatornamerange T^n-2)$
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Suppose $T in mathcalL(V) $ and $3$ and $8$ are eigenvalues of $T$ . Let $ n =dim (V)$
Prove that $V=(operatornamenull T^n-2) oplus (operatornamerange T^n-2) $
Answer:
Since the eigen values of $T$ are $3,8$ and it has no zero eigenvale.
Thus $T$ is non-singular.
Clerarly $T$ is diagonalisable and the diagonal matrix is $beginpmatrix 3 & 0 \ 0 & 8 endpmatrix$
Thus the eigenvalues of $T^n-2$ are $3^n-2 , 8^n-2$ and these are distinct.
So $T^n-2$ is non-singular.
Thus $ dim(operatornamenull T^n-2)=0$.
But now what would be $ dim(operatornamerange T^n-2)$?
If I can show that $dim V=dim (operatornamenull T^n-2) oplus dim (operatornamerange T^n-2) $
This implies $V=operatornamenull(T^n-2) oplus operatornamerange (T^n-2)$
Help me doing this.
linear-algebra
edited Jul 25 at 11:22
mechanodroid
22.2k52041
asked Jul 25 at 10:50
yourmath
1,7951617
add a comment |Â
up vote
1
down vote
favorite
Suppose $T in mathcalL(V) $ and $3$ and $8$ are eigenvalues of $T$ . Let $ n =dim (V)$
Prove that $V=(operatornamenull T^n-2) oplus (operatornamerange T^n-2) $
Answer:
Since the eigen values of $T$ are $3,8$ and it has no zero eigenvale.
Thus $T$ is non-singular.
Clerarly $T$ is diagonalisable and the diagonal matrix is $beginpmatrix 3 & 0 \ 0 & 8 endpmatrix$
Thus the eigenvalues of $T^n-2$ are $3^n-2 , 8^n-2$ and these are distinct.
So $T^n-2$ is non-singular.
Thus $ dim(operatornamenull T^n-2)=0$.
But now what would be $ dim(operatornamerange T^n-2)$?
If I can show that $dim V=dim (operatornamenull T^n-2) oplus dim (operatornamerange T^n-2) $
This implies $V=operatornamenull(T^n-2) oplus operatornamerange (T^n-2)$
Help me doing this.
linear-algebra
edited Jul 25 at 11:22
mechanodroid
22.2k52041
asked Jul 25 at 10:50
yourmath
1,7951617
$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
yes It my mistake ?
– yourmath
Jul 25 at 10:56
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $T in mathcalL(V) $ and $3$ and $8$ are eigenvalues of $T$ . Let $ n =dim (V)$
Prove that $V=(operatornamenull T^n-2) oplus (operatornamerange T^n-2) $
Answer:
Since the eigen values of $T$ are $3,8$ and it has no zero eigenvale.
Thus $T$ is non-singular.
Clerarly $T$ is diagonalisable and the diagonal matrix is $beginpmatrix 3 & 0 \ 0 & 8 endpmatrix$
Thus the eigenvalues of $T^n-2$ are $3^n-2 , 8^n-2$ and these are distinct.
So $T^n-2$ is non-singular.
Thus $ dim(operatornamenull T^n-2)=0$.
But now what would be $ dim(operatornamerange T^n-2)$?
If I can show that $dim V=dim (operatornamenull T^n-2) oplus dim (operatornamerange T^n-2) $
This implies $V=operatornamenull(T^n-2) oplus operatornamerange (T^n-2)$
Help me doing this.
linear-algebra
edited Jul 25 at 11:22
mechanodroid
22.2k52041
asked Jul 25 at 10:50
yourmath
1,7951617
Suppose $T in mathcalL(V) $ and $3$ and $8$ are eigenvalues of $T$ . Let $ n =dim (V)$
Prove that $V=(operatornamenull T^n-2) oplus (operatornamerange T^n-2) $
Answer:
Since the eigen values of $T$ are $3,8$ and it has no zero eigenvale.
Thus $T$ is non-singular.
Clerarly $T$ is diagonalisable and the diagonal matrix is $beginpmatrix 3 & 0 \ 0 & 8 endpmatrix$
Thus the eigenvalues of $T^n-2$ are $3^n-2 , 8^n-2$ and these are distinct.
So $T^n-2$ is non-singular.
Thus $ dim(operatornamenull T^n-2)=0$.
But now what would be $ dim(operatornamerange T^n-2)$?
If I can show that $dim V=dim (operatornamenull T^n-2) oplus dim (operatornamerange T^n-2) $
This implies $V=operatornamenull(T^n-2) oplus operatornamerange (T^n-2)$
Help me doing this.
linear-algebra
edited Jul 25 at 11:22
mechanodroid
22.2k52041
asked Jul 25 at 10:50
yourmath
1,7951617
edited Jul 25 at 11:22
mechanodroid
22.2k52041
edited Jul 25 at 11:22
mechanodroid
22.2k52041
edited Jul 25 at 11:22
mechanodroid
22.2k52041
22.2k52041
asked Jul 25 at 10:50
yourmath
1,7951617
asked Jul 25 at 10:50
yourmath
1,7951617
asked Jul 25 at 10:50
yourmath
1,7951617
1,7951617
$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
yes It my mistake ?
– yourmath
Jul 25 at 10:56
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57
add a comment |Â
$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
yes It my mistake ?
– yourmath
Jul 25 at 10:56
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57
$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
yes It my mistake ?
– yourmath
Jul 25 at 10:56
yes It my mistake ?
– yourmath
Jul 25 at 10:56
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as
$$m_T(x) = x^k(x-3)(x-8)q(x)$$
for some $q in mathbbC[x]$ and $0 le k le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$operatornamenull T^n-2 = operatornamenull T^n-1 = operatornamenull T = cdots$$
$$operatornamerange T^n-2 = operatornamerange T^n-1 = operatornamerange T = cdots$$
Assume $y in operatornamenull T^n-2 cap operatornamerange T^n-2$. Hence $y = T^n-2x$ for some $x in V$ and then $0 = T^n-2y = T^2n-4x$ which implies $xin operatornamenull T^2n-4 = operatornamenull T^n-2$ (note that $2n-4 ge n-2$ because $n ge 2$). Hence $y = T^n-2x = 0$ so we conclude $operatornamenull T^n-2 cap operatornamerange T^n-2 = 0$.
On the other hand for any $x in V$ we have $T^n-2x in operatornamerange T^n-2 = operatornamerange T^2n-4$ so $T^n-2x = T^2n-4z$ for some $z in V$. We have
$$x = underbrace(x - T^n-2z)_in operatornamenull T^n-2 + underbraceT^n-2z_inoperatornamerangeT^n-2$$
because $T^n-2(x - T^n-2z) = T^n-2x - T^2n-4z = 0$.
We conclude $$operatornamenull T^n-2 oplus operatornamerange T^n-2 = V$$
answered Jul 25 at 11:15
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form
$$
J = pmatrixJ_1&0&0\0&J_2&0\0&0&J_3
$$
where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) times (n-2)$.
It now suffices to note that any $k times k$ matrix $A$ satisfies
$$
Bbb C^k = operatornamenull A^k oplus operatornamerange A^k
$$
answered Jul 25 at 11:05
Omnomnomnom
121k784170
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as
$$m_T(x) = x^k(x-3)(x-8)q(x)$$
for some $q in mathbbC[x]$ and $0 le k le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$operatornamenull T^n-2 = operatornamenull T^n-1 = operatornamenull T = cdots$$
$$operatornamerange T^n-2 = operatornamerange T^n-1 = operatornamerange T = cdots$$
Assume $y in operatornamenull T^n-2 cap operatornamerange T^n-2$. Hence $y = T^n-2x$ for some $x in V$ and then $0 = T^n-2y = T^2n-4x$ which implies $xin operatornamenull T^2n-4 = operatornamenull T^n-2$ (note that $2n-4 ge n-2$ because $n ge 2$). Hence $y = T^n-2x = 0$ so we conclude $operatornamenull T^n-2 cap operatornamerange T^n-2 = 0$.
On the other hand for any $x in V$ we have $T^n-2x in operatornamerange T^n-2 = operatornamerange T^2n-4$ so $T^n-2x = T^2n-4z$ for some $z in V$. We have
$$x = underbrace(x - T^n-2z)_in operatornamenull T^n-2 + underbraceT^n-2z_inoperatornamerangeT^n-2$$
because $T^n-2(x - T^n-2z) = T^n-2x - T^2n-4z = 0$.
We conclude $$operatornamenull T^n-2 oplus operatornamerange T^n-2 = V$$
answered Jul 25 at 11:15
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as
$$m_T(x) = x^k(x-3)(x-8)q(x)$$
for some $q in mathbbC[x]$ and $0 le k le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$operatornamenull T^n-2 = operatornamenull T^n-1 = operatornamenull T = cdots$$
$$operatornamerange T^n-2 = operatornamerange T^n-1 = operatornamerange T = cdots$$
Assume $y in operatornamenull T^n-2 cap operatornamerange T^n-2$. Hence $y = T^n-2x$ for some $x in V$ and then $0 = T^n-2y = T^2n-4x$ which implies $xin operatornamenull T^2n-4 = operatornamenull T^n-2$ (note that $2n-4 ge n-2$ because $n ge 2$). Hence $y = T^n-2x = 0$ so we conclude $operatornamenull T^n-2 cap operatornamerange T^n-2 = 0$.
On the other hand for any $x in V$ we have $T^n-2x in operatornamerange T^n-2 = operatornamerange T^2n-4$ so $T^n-2x = T^2n-4z$ for some $z in V$. We have
$$x = underbrace(x - T^n-2z)_in operatornamenull T^n-2 + underbraceT^n-2z_inoperatornamerangeT^n-2$$
because $T^n-2(x - T^n-2z) = T^n-2x - T^2n-4z = 0$.
We conclude $$operatornamenull T^n-2 oplus operatornamerange T^n-2 = V$$
answered Jul 25 at 11:15
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as
$$m_T(x) = x^k(x-3)(x-8)q(x)$$
for some $q in mathbbC[x]$ and $0 le k le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$operatornamenull T^n-2 = operatornamenull T^n-1 = operatornamenull T = cdots$$
$$operatornamerange T^n-2 = operatornamerange T^n-1 = operatornamerange T = cdots$$
Assume $y in operatornamenull T^n-2 cap operatornamerange T^n-2$. Hence $y = T^n-2x$ for some $x in V$ and then $0 = T^n-2y = T^2n-4x$ which implies $xin operatornamenull T^2n-4 = operatornamenull T^n-2$ (note that $2n-4 ge n-2$ because $n ge 2$). Hence $y = T^n-2x = 0$ so we conclude $operatornamenull T^n-2 cap operatornamerange T^n-2 = 0$.
On the other hand for any $x in V$ we have $T^n-2x in operatornamerange T^n-2 = operatornamerange T^2n-4$ so $T^n-2x = T^2n-4z$ for some $z in V$. We have
$$x = underbrace(x - T^n-2z)_in operatornamenull T^n-2 + underbraceT^n-2z_inoperatornamerangeT^n-2$$
because $T^n-2(x - T^n-2z) = T^n-2x - T^2n-4z = 0$.
We conclude $$operatornamenull T^n-2 oplus operatornamerange T^n-2 = V$$
answered Jul 25 at 11:15
mechanodroid
22.2k52041
Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as
$$m_T(x) = x^k(x-3)(x-8)q(x)$$
for some $q in mathbbC[x]$ and $0 le k le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$operatornamenull T^n-2 = operatornamenull T^n-1 = operatornamenull T = cdots$$
$$operatornamerange T^n-2 = operatornamerange T^n-1 = operatornamerange T = cdots$$
Assume $y in operatornamenull T^n-2 cap operatornamerange T^n-2$. Hence $y = T^n-2x$ for some $x in V$ and then $0 = T^n-2y = T^2n-4x$ which implies $xin operatornamenull T^2n-4 = operatornamenull T^n-2$ (note that $2n-4 ge n-2$ because $n ge 2$). Hence $y = T^n-2x = 0$ so we conclude $operatornamenull T^n-2 cap operatornamerange T^n-2 = 0$.
On the other hand for any $x in V$ we have $T^n-2x in operatornamerange T^n-2 = operatornamerange T^2n-4$ so $T^n-2x = T^2n-4z$ for some $z in V$. We have
$$x = underbrace(x - T^n-2z)_in operatornamenull T^n-2 + underbraceT^n-2z_inoperatornamerangeT^n-2$$
because $T^n-2(x - T^n-2z) = T^n-2x - T^2n-4z = 0$.
We conclude $$operatornamenull T^n-2 oplus operatornamerange T^n-2 = V$$
answered Jul 25 at 11:15
mechanodroid
22.2k52041
answered Jul 25 at 11:15
mechanodroid
22.2k52041
answered Jul 25 at 11:15
mechanodroid
22.2k52041
answered Jul 25 at 11:15
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
1
down vote
One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form
$$
J = pmatrixJ_1&0&0\0&J_2&0\0&0&J_3
$$
where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) times (n-2)$.
It now suffices to note that any $k times k$ matrix $A$ satisfies
$$
Bbb C^k = operatornamenull A^k oplus operatornamerange A^k
$$
answered Jul 25 at 11:05
Omnomnomnom
121k784170
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
add a comment |Â
up vote
1
down vote
One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form
$$
J = pmatrixJ_1&0&0\0&J_2&0\0&0&J_3
$$
where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) times (n-2)$.
It now suffices to note that any $k times k$ matrix $A$ satisfies
$$
Bbb C^k = operatornamenull A^k oplus operatornamerange A^k
$$
answered Jul 25 at 11:05
Omnomnomnom
121k784170
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form
$$
J = pmatrixJ_1&0&0\0&J_2&0\0&0&J_3
$$
where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) times (n-2)$.
It now suffices to note that any $k times k$ matrix $A$ satisfies
$$
Bbb C^k = operatornamenull A^k oplus operatornamerange A^k
$$
answered Jul 25 at 11:05
Omnomnomnom
121k784170
One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form
$$
J = pmatrixJ_1&0&0\0&J_2&0\0&0&J_3
$$
where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) times (n-2)$.
It now suffices to note that any $k times k$ matrix $A$ satisfies
$$
Bbb C^k = operatornamenull A^k oplus operatornamerange A^k
$$
answered Jul 25 at 11:05
Omnomnomnom
121k784170
answered Jul 25 at 11:05
Omnomnomnom
121k784170
answered Jul 25 at 11:05
Omnomnomnom
121k784170
answered Jul 25 at 11:05
Omnomnomnom
121k784170
121k784170
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
add a comment |Â
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
I did not get you on the last part of your answer. If I can show that $ mathbbC^k=null A^k oplus range A^k $ , then what would I get?
– yourmath
Jul 25 at 11:28
add a comment |Â
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$T$ is not necessarily non-singular or diagonalizable when $n geq 2$.
– Omnomnomnom
Jul 25 at 10:53
yes It my mistake ?
– yourmath
Jul 25 at 10:56
Yes I forgot that dim (V)=n. . It my mistake. please help me
– yourmath
Jul 25 at 10:57