Prove that $v, Tv, T^2v, … , T^{m-1}v$ is linearly independent
Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that
$(T^{m-1})v neq 0$,
and
$(T^m)v = 0$.
Prove that $[v, Tv, T^2v, ... , T^{m-1}v]$ is linearly independent
I get that $(T^j)v neq 0 forall j < m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal.
However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all.
Any help is appreciated!
linear-algebra matrices linear-transformations
asked Apr 19 '17 at 18:49
Student_514
906
add a comment |
Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that
$(T^{m-1})v neq 0$,
and
$(T^m)v = 0$.
Prove that $[v, Tv, T^2v, ... , T^{m-1}v]$ is linearly independent
I get that $(T^j)v neq 0 forall j < m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal.
However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all.
Any help is appreciated!
linear-algebra matrices linear-transformations
asked Apr 19 '17 at 18:49
Student_514
906
add a comment |
Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that
$(T^{m-1})v neq 0$,
and
$(T^m)v = 0$.
Prove that $[v, Tv, T^2v, ... , T^{m-1}v]$ is linearly independent
I get that $(T^j)v neq 0 forall j < m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal.
However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all.
Any help is appreciated!
linear-algebra matrices linear-transformations
asked Apr 19 '17 at 18:49
Student_514
906
Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that
$(T^{m-1})v neq 0$,
and
$(T^m)v = 0$.
Prove that $[v, Tv, T^2v, ... , T^{m-1}v]$ is linearly independent
I get that $(T^j)v neq 0 forall j < m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal.
However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all.
Any help is appreciated!
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Apr 19 '17 at 18:49
Student_514
906
asked Apr 19 '17 at 18:49
Student_514
906
edited Apr 19 '17 at 20:28
asked Apr 19 '17 at 18:49
Student_514
906
asked Apr 19 '17 at 18:49
Student_514
906
asked Apr 19 '17 at 18:49
Student_514
906
906
add a comment |
add a comment |
2 Answers
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It is simple to do it directly:
If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.
answered Apr 19 '17 at 18:57
lhf
162k10166386
add a comment |
The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.
Can you do something similar for $n=3$? General $n$?
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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It is simple to do it directly:
If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.
answered Apr 19 '17 at 18:57
lhf
162k10166386
add a comment |
It is simple to do it directly:
If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.
answered Apr 19 '17 at 18:57
lhf
162k10166386
add a comment |
It is simple to do it directly:
If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.
answered Apr 19 '17 at 18:57
lhf
162k10166386
It is simple to do it directly:
If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.
answered Apr 19 '17 at 18:57
lhf
162k10166386
answered Apr 19 '17 at 18:57
lhf
162k10166386
answered Apr 19 '17 at 18:57
lhf
162k10166386
answered Apr 19 '17 at 18:57
lhf
162k10166386
162k10166386
add a comment |
add a comment |
The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.
Can you do something similar for $n=3$? General $n$?
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
add a comment |
The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.
Can you do something similar for $n=3$? General $n$?
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
add a comment |
The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.
Can you do something similar for $n=3$? General $n$?
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.
Can you do something similar for $n=3$? General $n$?
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
answered Apr 19 '17 at 18:53
Lord Shark the Unknown
101k958132
101k958132
add a comment |
add a comment |
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