Nilpotent Matrix is Similar to a block diagonal matrix
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Prove that any nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1's along the first super diagonal and 0's elsewhere.
I'm not sure where to start exactly. Any guidance would be helpful!
linear-algebra matrices jordan-normal-form nilpotence
asked Mar 6 '17 at 4:54
bmmcutet12
1739
add a comment |Â
up vote
2
down vote
favorite
Prove that any nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1's along the first super diagonal and 0's elsewhere.
I'm not sure where to start exactly. Any guidance would be helpful!
linear-algebra matrices jordan-normal-form nilpotence
asked Mar 6 '17 at 4:54
bmmcutet12
1739
I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
2
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
2
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that any nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1's along the first super diagonal and 0's elsewhere.
I'm not sure where to start exactly. Any guidance would be helpful!
linear-algebra matrices jordan-normal-form nilpotence
asked Mar 6 '17 at 4:54
bmmcutet12
1739
Prove that any nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1's along the first super diagonal and 0's elsewhere.
I'm not sure where to start exactly. Any guidance would be helpful!
linear-algebra matrices jordan-normal-form nilpotence
asked Mar 6 '17 at 4:54
bmmcutet12
1739
asked Mar 6 '17 at 4:54
bmmcutet12
1739
asked Mar 6 '17 at 4:54
bmmcutet12
1739
asked Mar 6 '17 at 4:54
bmmcutet12
1739
1739
I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
2
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
2
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44
add a comment |Â
I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
2
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
2
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44
I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
2
2
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
2
2
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44
add a comment |Â
1 Answer
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My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."
Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⻹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.
answered Aug 11 at 20:27
Maurice P
1,1601630
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."
Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⻹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.
answered Aug 11 at 20:27
Maurice P
1,1601630
add a comment |Â
up vote
0
down vote
My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."
Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⻹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.
answered Aug 11 at 20:27
Maurice P
1,1601630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."
Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⻹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.
answered Aug 11 at 20:27
Maurice P
1,1601630
My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."
Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⻹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.
answered Aug 11 at 20:27
Maurice P
1,1601630
answered Aug 11 at 20:27
Maurice P
1,1601630
answered Aug 11 at 20:27
Maurice P
1,1601630
answered Aug 11 at 20:27
Maurice P
1,1601630
1,1601630
add a comment |Â
add a comment |Â
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I am assuming over $mathbb C$?
– Couchy311
Mar 6 '17 at 4:58
2
Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$.
– D_S
Mar 6 '17 at 5:08
Do you know about quotient spaces?
– Omnomnomnom
Mar 6 '17 at 5:26
2
I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $xnot=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $x,Nx,N^2x, dots,N^k-1x$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat.
– ancientmathematician
Mar 6 '17 at 9:43
Check Daniel's answer here: math.stackexchange.com/questions/809473/…
– Daniel
Mar 6 '17 at 21:44