Unipotent matrix similar to an upper-triangular matrix

"Any unipotent matrix is similar to an upper-triangular matrix with 1's on the diagonal"...

This is usually alleged, but I have no idea how to demonstrate that, starting with the definition : $A$ is unipotent if and only if there is $kin mathbb{N}$ so that $(A-I_n)^k=0$.

And I browsed Internet for hints but found nothing useful. I am not looking here for a ready-made solution, but I would like to understand what is the procedure, what are the steps one has to make, in order to proceed from definition to the result I stated above.

Thanks in advance, if someone is able to detail the path to do it.

asked Jan 21 at 21:43

Andrew

163

1

$begingroup$
What's your ground field? Are you familiar with Jordan normal form? You can look this up on Wikipedia.
$endgroup$
– jflipp
Jan 21 at 22:00

$begingroup$
Unfortunately not...
$endgroup$
– Andrew
Jan 22 at 9:48

add a comment |

"Any unipotent matrix is similar to an upper-triangular matrix with 1's on the diagonal"...

This is usually alleged, but I have no idea how to demonstrate that, starting with the definition : $A$ is unipotent if and only if there is $kin mathbb{N}$ so that $(A-I_n)^k=0$.

Thanks in advance, if someone is able to detail the path to do it.

asked Jan 21 at 21:43

Andrew

163

1

$begingroup$
What's your ground field? Are you familiar with Jordan normal form? You can look this up on Wikipedia.
$endgroup$
– jflipp
Jan 21 at 22:00

$begingroup$
Unfortunately not...
$endgroup$
– Andrew
Jan 22 at 9:48

add a comment |

"Any unipotent matrix is similar to an upper-triangular matrix with 1's on the diagonal"...

This is usually alleged, but I have no idea how to demonstrate that, starting with the definition : $A$ is unipotent if and only if there is $kin mathbb{N}$ so that $(A-I_n)^k=0$.

Thanks in advance, if someone is able to detail the path to do it.

asked Jan 21 at 21:43

Andrew

163

"Any unipotent matrix is similar to an upper-triangular matrix with 1's on the diagonal"...

This is usually alleged, but I have no idea how to demonstrate that, starting with the definition : $A$ is unipotent if and only if there is $kin mathbb{N}$ so that $(A-I_n)^k=0$.

Thanks in advance, if someone is able to detail the path to do it.

unipotent-matrices

asked Jan 21 at 21:43

Andrew

163

asked Jan 21 at 21:43

Andrew

163

asked Jan 21 at 21:43

Andrew

163

asked Jan 21 at 21:43

Andrew

163

asked Jan 21 at 21:43

Andrew

163

1

$begingroup$
What's your ground field? Are you familiar with Jordan normal form? You can look this up on Wikipedia.
$endgroup$
– jflipp
Jan 21 at 22:00

$begingroup$
Unfortunately not...
$endgroup$
– Andrew
Jan 22 at 9:48

add a comment |

1

$begingroup$
What's your ground field? Are you familiar with Jordan normal form? You can look this up on Wikipedia.
$endgroup$
– jflipp
Jan 21 at 22:00

$begingroup$
Unfortunately not...
$endgroup$
– Andrew
Jan 22 at 9:48

What's your ground field? Are you familiar with Jordan normal form? You can look this up on Wikipedia.

– jflipp
Jan 21 at 22:00

Unfortunately not...

– Andrew
Jan 22 at 9:48

add a comment |

2 Answers
2

active

oldest

votes

Here I work over the complex field $Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A vec x = lambda vec x = lambda I_n vec x, ; vec x ne 0, tag 2$

then

$(A - I_n) vec x = (lambda I_n - I_n) vec x = (lambda - 1)I_n vec x = (lambda - 1) vec x, tag 3$

from which we find

$(lambda - 1)^k vec x = (A - I_n)^k vec x = 0; tag 4$

now since $vec x ne 0$ we infer that

$(lambda - 1)^k = 0 Longrightarrow lambda = 1. tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, tag 7$

which is the requisite result. $OEDelta$.

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

1

$begingroup$
In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?
$endgroup$
– J. W. Tanner
Jan 21 at 22:21

$begingroup$
$lambda - 1$. Thanks for the catch. Will edit.
$endgroup$
– Robert Lewis
Jan 21 at 22:23

$begingroup$
Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.
$endgroup$
– Andrew
Jan 22 at 9:55

add a comment |

You have already given us that a matrix $A$ is unipotent $Longleftrightarrow$ $(A-I)^k=0$ for some $k in mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A in mathbb C ^{ntext{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

answered Jan 21 at 22:25

Joel Biffin

1017

$begingroup$
Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?
$endgroup$
– Andrew
Jan 22 at 10:01

$begingroup$
@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.
$endgroup$
– Joel Biffin
yesterday

add a comment |

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2 Answers
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active

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2 Answers
2

active

oldest

votes

Here I work over the complex field $Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A vec x = lambda vec x = lambda I_n vec x, ; vec x ne 0, tag 2$

then

$(A - I_n) vec x = (lambda I_n - I_n) vec x = (lambda - 1)I_n vec x = (lambda - 1) vec x, tag 3$

from which we find

$(lambda - 1)^k vec x = (A - I_n)^k vec x = 0; tag 4$

now since $vec x ne 0$ we infer that

$(lambda - 1)^k = 0 Longrightarrow lambda = 1. tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, tag 7$

which is the requisite result. $OEDelta$.

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

1

$begingroup$
In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?
$endgroup$
– J. W. Tanner
Jan 21 at 22:21

$begingroup$
$lambda - 1$. Thanks for the catch. Will edit.
$endgroup$
– Robert Lewis
Jan 21 at 22:23

$begingroup$
Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.
$endgroup$
– Andrew
Jan 22 at 9:55

add a comment |

Here I work over the complex field $Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A vec x = lambda vec x = lambda I_n vec x, ; vec x ne 0, tag 2$

then

$(A - I_n) vec x = (lambda I_n - I_n) vec x = (lambda - 1)I_n vec x = (lambda - 1) vec x, tag 3$

from which we find

$(lambda - 1)^k vec x = (A - I_n)^k vec x = 0; tag 4$

now since $vec x ne 0$ we infer that

$(lambda - 1)^k = 0 Longrightarrow lambda = 1. tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, tag 7$

which is the requisite result. $OEDelta$.

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

1

$begingroup$
In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?
$endgroup$
– J. W. Tanner
Jan 21 at 22:21

$begingroup$
$lambda - 1$. Thanks for the catch. Will edit.
$endgroup$
– Robert Lewis
Jan 21 at 22:23

$begingroup$
Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.
$endgroup$
– Andrew
Jan 22 at 9:55

add a comment |

Here I work over the complex field $Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A vec x = lambda vec x = lambda I_n vec x, ; vec x ne 0, tag 2$

then

$(A - I_n) vec x = (lambda I_n - I_n) vec x = (lambda - 1)I_n vec x = (lambda - 1) vec x, tag 3$

from which we find

$(lambda - 1)^k vec x = (A - I_n)^k vec x = 0; tag 4$

now since $vec x ne 0$ we infer that

$(lambda - 1)^k = 0 Longrightarrow lambda = 1. tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, tag 7$

which is the requisite result. $OEDelta$.

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

Here I work over the complex field $Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A vec x = lambda vec x = lambda I_n vec x, ; vec x ne 0, tag 2$

then

$(A - I_n) vec x = (lambda I_n - I_n) vec x = (lambda - 1)I_n vec x = (lambda - 1) vec x, tag 3$

from which we find

$(lambda - 1)^k vec x = (A - I_n)^k vec x = 0; tag 4$

now since $vec x ne 0$ we infer that

$(lambda - 1)^k = 0 Longrightarrow lambda = 1. tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, tag 7$

which is the requisite result. $OEDelta$.

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

edited Jan 21 at 22:23

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

answered Jan 21 at 22:15

Robert Lewis

47.8k23067

1

$begingroup$
In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?
$endgroup$
– J. W. Tanner
Jan 21 at 22:21

$begingroup$
$lambda - 1$. Thanks for the catch. Will edit.
$endgroup$
– Robert Lewis
Jan 21 at 22:23

$begingroup$
Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.
$endgroup$
– Andrew
Jan 22 at 9:55

add a comment |

1

$begingroup$
In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?
$endgroup$
– J. W. Tanner
Jan 21 at 22:21

$begingroup$
$lambda - 1$. Thanks for the catch. Will edit.
$endgroup$
– Robert Lewis
Jan 21 at 22:23

$begingroup$
Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.
$endgroup$
– Andrew
Jan 22 at 9:55

In line (4), did you mean $(lambda I_n - I_n)$ and/or $(lambda - 1)$ ?

– J. W. Tanner
Jan 21 at 22:21

$lambda - 1$. Thanks for the catch. Will edit.

– Robert Lewis
Jan 21 at 22:23

Thanks Robert. I am not familiar with Jordan forms, But your reasoning looks clear. I will look into it in detail and I'm sure I will manage to deal with it.

– Andrew
Jan 22 at 9:55

add a comment |

You have already given us that a matrix $A$ is unipotent $Longleftrightarrow$ $(A-I)^k=0$ for some $k in mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A in mathbb C ^{ntext{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

answered Jan 21 at 22:25

Joel Biffin

1017

$begingroup$
Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?
$endgroup$
– Andrew
Jan 22 at 10:01

$begingroup$
@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.
$endgroup$
– Joel Biffin
yesterday

add a comment |

You have already given us that a matrix $A$ is unipotent $Longleftrightarrow$ $(A-I)^k=0$ for some $k in mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A in mathbb C ^{ntext{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

answered Jan 21 at 22:25

Joel Biffin

1017

$begingroup$
Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?
$endgroup$
– Andrew
Jan 22 at 10:01

$begingroup$
@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.
$endgroup$
– Joel Biffin
yesterday

add a comment |

You have already given us that a matrix $A$ is unipotent $Longleftrightarrow$ $(A-I)^k=0$ for some $k in mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A in mathbb C ^{ntext{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

answered Jan 21 at 22:25

Joel Biffin

1017

You have already given us that a matrix $A$ is unipotent $Longleftrightarrow$ $(A-I)^k=0$ for some $k in mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A in mathbb C ^{ntext{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

answered Jan 21 at 22:25

Joel Biffin

1017

answered Jan 21 at 22:25

Joel Biffin

1017

answered Jan 21 at 22:25

Joel Biffin

1017

answered Jan 21 at 22:25

Joel Biffin

1017

$begingroup$
Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?
$endgroup$
– Andrew
Jan 22 at 10:01

$begingroup$
@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.
$endgroup$
– Joel Biffin
yesterday

add a comment |

$begingroup$
Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?
$endgroup$
– Andrew
Jan 22 at 10:01

$begingroup$
@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.
$endgroup$
– Joel Biffin
yesterday

Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ?

– Andrew
Jan 22 at 10:01

@Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^*$ then you have the reverse similarity transformation $T=VAV^*$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A.

– Joel Biffin
yesterday

add a comment |

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