Let T be a linear operator of an n-dimensional vector space V over a field F. Assume that T is nilpotent....
$begingroup$
Let T be a linear operator of an n-dimensional vector space V over a field F. Assume that T is nilpotent. Show that $T^n = 0$.
I've seen people prove this with an argument involving minimal polynomials. Can someone prove this without such an argument?
linear-algebra
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
$endgroup$
add a comment |
$begingroup$
Let T be a linear operator of an n-dimensional vector space V over a field F. Assume that T is nilpotent. Show that $T^n = 0$.
I've seen people prove this with an argument involving minimal polynomials. Can someone prove this without such an argument?
linear-algebra
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
$endgroup$
2
$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58
add a comment |
$begingroup$
Let T be a linear operator of an n-dimensional vector space V over a field F. Assume that T is nilpotent. Show that $T^n = 0$.
I've seen people prove this with an argument involving minimal polynomials. Can someone prove this without such an argument?
linear-algebra
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
$endgroup$
Let T be a linear operator of an n-dimensional vector space V over a field F. Assume that T is nilpotent. Show that $T^n = 0$.
I've seen people prove this with an argument involving minimal polynomials. Can someone prove this without such an argument?
linear-algebra
linear-algebra
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
edited Dec 7 '18 at 3:50
Key Flex
8,33261233
8,33261233
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
asked Dec 7 '18 at 3:48
IUissoprettyIUissopretty
516
516
2
$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58
add a comment |
2
$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58
2
2
$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58
$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Suppose $T^n ne 0$. Then there exists $xne 0$ such that $T^n x ne 0$. Hence,
${ x,Tx,T^2x,cdots,T^n x}$ is a linearly independent set of vectors. To see why, suppose
$$
alpha_0 x+alpha_1 Tx+cdots + alpha_n T^n x = 0,
$$
and apply $T^k$ to the above, where $T^kxne 0$ and $T^{k+1}x=0$. Conclude that $alpha_0=0$. Continue in orer to conclude that $alpha_0=alpha_1=cdots=alpha_n=0$. Hence ${x,Tx,T^2x,cdots,T^nx}$ is a linearly-independent set, which forces $n < dim{V}$. This is true for every $x$. Therefore $T^{dim(V)}=0$.
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
$endgroup$
add a comment |
$begingroup$
Note first that if $T$ fixes any nonzero $Wsubset V$, then $T(W)< W$; otherwise, $Tvert W$ is an isomorphism, and so $T^knot = 0$ for all $k$. Now consider the chain of inclusions $Vsupset T(V) supset T^2(V) supset cdots$.
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
$endgroup$
add a comment |
$begingroup$
Consider the subspaces $V_i:=text{Im}(T^i)$. If $V_i=V_{i+1}$, then by definition $V_i=V_j,forall jge i$. Since $T$ si nilpotent, $V_i$ is descending to $left{0right}$. Thus if $V_i=V_{i+1}$ for some $i$, then $V_i=left{0right}$. Also if $V_ine V_{i+1}$ then $text{dim}(V_i)getext{dim}(V_{i+1})+1$. As a consequence $V_n=left{0right}$.
Also, one can use the pigeon-hole principle: the dimensions of $V_i$ are in the set $left{1,2,cdots,nright}$, and there are $n+1$ elements in $left{text{dim}(V_i)mid i=0,1,cdots,nright}$, so $text{dim}(V_i)=text{dim}(V_{i+1})$ for some $i=0,1,cdots,n$, which implies our desired result as $V_isupseteq V_{i+1},forall i$.
Hope this helps.
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
$endgroup$
add a comment |
$begingroup$
For a more computational approach, every matrix can be upper triangularized. It suffices to prove that if $g$ is a nilpotent $n$ by $n$ upper triangular matrix, then $g^n = 0$. You can check that all the diagonal entries of $g$ are zero, and then the claim easily follows.
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $T^n ne 0$. Then there exists $xne 0$ such that $T^n x ne 0$. Hence,
${ x,Tx,T^2x,cdots,T^n x}$ is a linearly independent set of vectors. To see why, suppose
$$
alpha_0 x+alpha_1 Tx+cdots + alpha_n T^n x = 0,
$$
and apply $T^k$ to the above, where $T^kxne 0$ and $T^{k+1}x=0$. Conclude that $alpha_0=0$. Continue in orer to conclude that $alpha_0=alpha_1=cdots=alpha_n=0$. Hence ${x,Tx,T^2x,cdots,T^nx}$ is a linearly-independent set, which forces $n < dim{V}$. This is true for every $x$. Therefore $T^{dim(V)}=0$.
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
$endgroup$
add a comment |
$begingroup$
Suppose $T^n ne 0$. Then there exists $xne 0$ such that $T^n x ne 0$. Hence,
${ x,Tx,T^2x,cdots,T^n x}$ is a linearly independent set of vectors. To see why, suppose
$$
alpha_0 x+alpha_1 Tx+cdots + alpha_n T^n x = 0,
$$
and apply $T^k$ to the above, where $T^kxne 0$ and $T^{k+1}x=0$. Conclude that $alpha_0=0$. Continue in orer to conclude that $alpha_0=alpha_1=cdots=alpha_n=0$. Hence ${x,Tx,T^2x,cdots,T^nx}$ is a linearly-independent set, which forces $n < dim{V}$. This is true for every $x$. Therefore $T^{dim(V)}=0$.
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
$endgroup$
add a comment |
$begingroup$
Suppose $T^n ne 0$. Then there exists $xne 0$ such that $T^n x ne 0$. Hence,
${ x,Tx,T^2x,cdots,T^n x}$ is a linearly independent set of vectors. To see why, suppose
$$
alpha_0 x+alpha_1 Tx+cdots + alpha_n T^n x = 0,
$$
and apply $T^k$ to the above, where $T^kxne 0$ and $T^{k+1}x=0$. Conclude that $alpha_0=0$. Continue in orer to conclude that $alpha_0=alpha_1=cdots=alpha_n=0$. Hence ${x,Tx,T^2x,cdots,T^nx}$ is a linearly-independent set, which forces $n < dim{V}$. This is true for every $x$. Therefore $T^{dim(V)}=0$.
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
$endgroup$
Suppose $T^n ne 0$. Then there exists $xne 0$ such that $T^n x ne 0$. Hence,
${ x,Tx,T^2x,cdots,T^n x}$ is a linearly independent set of vectors. To see why, suppose
$$
alpha_0 x+alpha_1 Tx+cdots + alpha_n T^n x = 0,
$$
and apply $T^k$ to the above, where $T^kxne 0$ and $T^{k+1}x=0$. Conclude that $alpha_0=0$. Continue in orer to conclude that $alpha_0=alpha_1=cdots=alpha_n=0$. Hence ${x,Tx,T^2x,cdots,T^nx}$ is a linearly-independent set, which forces $n < dim{V}$. This is true for every $x$. Therefore $T^{dim(V)}=0$.
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
answered Dec 7 '18 at 4:52
DisintegratingByPartsDisintegratingByParts
59.6k42581
59.6k42581
add a comment |
add a comment |
$begingroup$
Note first that if $T$ fixes any nonzero $Wsubset V$, then $T(W)< W$; otherwise, $Tvert W$ is an isomorphism, and so $T^knot = 0$ for all $k$. Now consider the chain of inclusions $Vsupset T(V) supset T^2(V) supset cdots$.
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
$endgroup$
add a comment |
$begingroup$
Note first that if $T$ fixes any nonzero $Wsubset V$, then $T(W)< W$; otherwise, $Tvert W$ is an isomorphism, and so $T^knot = 0$ for all $k$. Now consider the chain of inclusions $Vsupset T(V) supset T^2(V) supset cdots$.
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
$endgroup$
add a comment |
$begingroup$
Note first that if $T$ fixes any nonzero $Wsubset V$, then $T(W)< W$; otherwise, $Tvert W$ is an isomorphism, and so $T^knot = 0$ for all $k$. Now consider the chain of inclusions $Vsupset T(V) supset T^2(V) supset cdots$.
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
$endgroup$
Note first that if $T$ fixes any nonzero $Wsubset V$, then $T(W)< W$; otherwise, $Tvert W$ is an isomorphism, and so $T^knot = 0$ for all $k$. Now consider the chain of inclusions $Vsupset T(V) supset T^2(V) supset cdots$.
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
answered Dec 7 '18 at 3:56
anomalyanomaly
17.7k42666
17.7k42666
add a comment |
add a comment |
$begingroup$
Consider the subspaces $V_i:=text{Im}(T^i)$. If $V_i=V_{i+1}$, then by definition $V_i=V_j,forall jge i$. Since $T$ si nilpotent, $V_i$ is descending to $left{0right}$. Thus if $V_i=V_{i+1}$ for some $i$, then $V_i=left{0right}$. Also if $V_ine V_{i+1}$ then $text{dim}(V_i)getext{dim}(V_{i+1})+1$. As a consequence $V_n=left{0right}$.
Also, one can use the pigeon-hole principle: the dimensions of $V_i$ are in the set $left{1,2,cdots,nright}$, and there are $n+1$ elements in $left{text{dim}(V_i)mid i=0,1,cdots,nright}$, so $text{dim}(V_i)=text{dim}(V_{i+1})$ for some $i=0,1,cdots,n$, which implies our desired result as $V_isupseteq V_{i+1},forall i$.
Hope this helps.
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
$endgroup$
add a comment |
$begingroup$
Consider the subspaces $V_i:=text{Im}(T^i)$. If $V_i=V_{i+1}$, then by definition $V_i=V_j,forall jge i$. Since $T$ si nilpotent, $V_i$ is descending to $left{0right}$. Thus if $V_i=V_{i+1}$ for some $i$, then $V_i=left{0right}$. Also if $V_ine V_{i+1}$ then $text{dim}(V_i)getext{dim}(V_{i+1})+1$. As a consequence $V_n=left{0right}$.
Also, one can use the pigeon-hole principle: the dimensions of $V_i$ are in the set $left{1,2,cdots,nright}$, and there are $n+1$ elements in $left{text{dim}(V_i)mid i=0,1,cdots,nright}$, so $text{dim}(V_i)=text{dim}(V_{i+1})$ for some $i=0,1,cdots,n$, which implies our desired result as $V_isupseteq V_{i+1},forall i$.
Hope this helps.
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
$endgroup$
add a comment |
$begingroup$
Consider the subspaces $V_i:=text{Im}(T^i)$. If $V_i=V_{i+1}$, then by definition $V_i=V_j,forall jge i$. Since $T$ si nilpotent, $V_i$ is descending to $left{0right}$. Thus if $V_i=V_{i+1}$ for some $i$, then $V_i=left{0right}$. Also if $V_ine V_{i+1}$ then $text{dim}(V_i)getext{dim}(V_{i+1})+1$. As a consequence $V_n=left{0right}$.
Also, one can use the pigeon-hole principle: the dimensions of $V_i$ are in the set $left{1,2,cdots,nright}$, and there are $n+1$ elements in $left{text{dim}(V_i)mid i=0,1,cdots,nright}$, so $text{dim}(V_i)=text{dim}(V_{i+1})$ for some $i=0,1,cdots,n$, which implies our desired result as $V_isupseteq V_{i+1},forall i$.
Hope this helps.
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
$endgroup$
Consider the subspaces $V_i:=text{Im}(T^i)$. If $V_i=V_{i+1}$, then by definition $V_i=V_j,forall jge i$. Since $T$ si nilpotent, $V_i$ is descending to $left{0right}$. Thus if $V_i=V_{i+1}$ for some $i$, then $V_i=left{0right}$. Also if $V_ine V_{i+1}$ then $text{dim}(V_i)getext{dim}(V_{i+1})+1$. As a consequence $V_n=left{0right}$.
Also, one can use the pigeon-hole principle: the dimensions of $V_i$ are in the set $left{1,2,cdots,nright}$, and there are $n+1$ elements in $left{text{dim}(V_i)mid i=0,1,cdots,nright}$, so $text{dim}(V_i)=text{dim}(V_{i+1})$ for some $i=0,1,cdots,n$, which implies our desired result as $V_isupseteq V_{i+1},forall i$.
Hope this helps.
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
edited Dec 7 '18 at 4:28
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
answered Dec 7 '18 at 4:03
awllowerawllower
10.4k42671
10.4k42671
add a comment |
add a comment |
$begingroup$
For a more computational approach, every matrix can be upper triangularized. It suffices to prove that if $g$ is a nilpotent $n$ by $n$ upper triangular matrix, then $g^n = 0$. You can check that all the diagonal entries of $g$ are zero, and then the claim easily follows.
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
$endgroup$
add a comment |
$begingroup$
For a more computational approach, every matrix can be upper triangularized. It suffices to prove that if $g$ is a nilpotent $n$ by $n$ upper triangular matrix, then $g^n = 0$. You can check that all the diagonal entries of $g$ are zero, and then the claim easily follows.
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
$endgroup$
add a comment |
$begingroup$
For a more computational approach, every matrix can be upper triangularized. It suffices to prove that if $g$ is a nilpotent $n$ by $n$ upper triangular matrix, then $g^n = 0$. You can check that all the diagonal entries of $g$ are zero, and then the claim easily follows.
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
$endgroup$
For a more computational approach, every matrix can be upper triangularized. It suffices to prove that if $g$ is a nilpotent $n$ by $n$ upper triangular matrix, then $g^n = 0$. You can check that all the diagonal entries of $g$ are zero, and then the claim easily follows.
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
answered Dec 7 '18 at 4:41
D_SD_S
13.7k61552
13.7k61552
add a comment |
add a comment |
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$begingroup$
They're looking to prove that the degree of nilpotency is bounded by the dimension of the vector space.
$endgroup$
– Ashwin Trisal
Dec 7 '18 at 3:58