Upper bound of the spectral norm of a matrix power

Clash Royale CLAN TAG#URR8PPP

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Let $AinmathbbC^ntimes n$ be a complex valued square matrix which can be written as $A=PUP$ in which $P$ is a projector and $U$ is a unitary matrix. The interesting case is $P$ and $U$ do not commute.

The matrix $A$ is not normal, and I would like to upper bound the 2-norm of $A^n-1$. It is the case that $|A|=1$, so an trivial upper bound is $|A^n-1|leq|A|^n-1=1$. Is it possible to find a bound which is strictly less than 1?

I tried to read through some articles, e.g., this thesis, in particular, Chapter 3, but did not find an obvious approach.

spectral-radius matrix-norms

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edited Jul 18 at 0:52

asked Jul 17 at 20:16

user50394

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
  – amsmath
  Jul 17 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
  – user50394
  Jul 17 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
  – amsmath
  Jul 17 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
  – user50394
  Jul 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are they orthogonal projectors?
  – RHowe
  Jul 17 at 23:48
  

  
  
  
  

 | 
show 8 more comments

up vote
0
down vote

favorite

1

Let $AinmathbbC^ntimes n$ be a complex valued square matrix which can be written as $A=PUP$ in which $P$ is a projector and $U$ is a unitary matrix. The interesting case is $P$ and $U$ do not commute.

The matrix $A$ is not normal, and I would like to upper bound the 2-norm of $A^n-1$. It is the case that $|A|=1$, so an trivial upper bound is $|A^n-1|leq|A|^n-1=1$. Is it possible to find a bound which is strictly less than 1?

I tried to read through some articles, e.g., this thesis, in particular, Chapter 3, but did not find an obvious approach.

spectral-radius matrix-norms

share|cite|improve this question

edited Jul 18 at 0:52

asked Jul 17 at 20:16

user50394

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
  – amsmath
  Jul 17 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
  – user50394
  Jul 17 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
  – amsmath
  Jul 17 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
  – user50394
  Jul 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are they orthogonal projectors?
  – RHowe
  Jul 17 at 23:48
  

  
  
  
  

 | 
show 8 more comments

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let $AinmathbbC^ntimes n$ be a complex valued square matrix which can be written as $A=PUP$ in which $P$ is a projector and $U$ is a unitary matrix. The interesting case is $P$ and $U$ do not commute.

The matrix $A$ is not normal, and I would like to upper bound the 2-norm of $A^n-1$. It is the case that $|A|=1$, so an trivial upper bound is $|A^n-1|leq|A|^n-1=1$. Is it possible to find a bound which is strictly less than 1?

I tried to read through some articles, e.g., this thesis, in particular, Chapter 3, but did not find an obvious approach.

spectral-radius matrix-norms

share|cite|improve this question

edited Jul 18 at 0:52

asked Jul 17 at 20:16

user50394

11

Let $AinmathbbC^ntimes n$ be a complex valued square matrix which can be written as $A=PUP$ in which $P$ is a projector and $U$ is a unitary matrix. The interesting case is $P$ and $U$ do not commute.

The matrix $A$ is not normal, and I would like to upper bound the 2-norm of $A^n-1$. It is the case that $|A|=1$, so an trivial upper bound is $|A^n-1|leq|A|^n-1=1$. Is it possible to find a bound which is strictly less than 1?

I tried to read through some articles, e.g., this thesis, in particular, Chapter 3, but did not find an obvious approach.

spectral-radius matrix-norms

share|cite|improve this question

edited Jul 18 at 0:52

asked Jul 17 at 20:16

user50394

11

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 0:52

edited Jul 18 at 0:52

edited Jul 18 at 0:52

asked Jul 17 at 20:16

user50394

11

asked Jul 17 at 20:16

user50394

11

asked Jul 17 at 20:16

user50394

11

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
  – amsmath
  Jul 17 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
  – user50394
  Jul 17 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
  – amsmath
  Jul 17 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
  – user50394
  Jul 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are they orthogonal projectors?
  – RHowe
  Jul 17 at 23:48
  

  
  
  
  

 | 
show 8 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
  – amsmath
  Jul 17 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
  – user50394
  Jul 17 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
  – amsmath
  Jul 17 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
  – user50394
  Jul 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are they orthogonal projectors?
  – RHowe
  Jul 17 at 23:48
  

  
  
  
  

If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
– amsmath
Jul 17 at 20:42

If the spectral radius of $A$ is $1$ (as you write), then there is an eigenvalue of $A$ on the unit circle. This then also holds for any $A^k$. Hence also $rho(A^n-1)=1$ and thus $|A^n-1|=1$.
– amsmath
Jul 17 at 20:42

It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
– user50394
Jul 17 at 22:24

It is not necessarily the case $|A^n-1|=1$; we may take the answer by @Omnomnomnom as an example.
– user50394
Jul 17 at 22:24

So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
– amsmath
Jul 17 at 22:29

So, why do you write $rho(A)=1$ in your question? This is not the case in Omnomnom's example.
– amsmath
Jul 17 at 22:29

Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
– user50394
Jul 17 at 22:58

Oh I see. But did you claim that $rho(A)^k=|A^k|$? I am not sure if that is the case when $A$ is not normal.
– user50394
Jul 17 at 22:58

Are they orthogonal projectors?
– RHowe
Jul 17 at 23:48

Are they orthogonal projectors?
– RHowe
Jul 17 at 23:48

 | 
show 8 more comments

1 Answer
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Taking
$$
P = pmatrix1&0\0&0, quad U = pmatrixsqrt1 - epsilon^2 & -epsilon \ epsilon & sqrt1 - epsilon^2
$$
is enough for us to see that $|(PUP)^n-1|$ can be made arbitrarily close to $1$ in the case of $n = 2$. A similar example works for arbitrary $n$. That is, it's enough to take
$$
pmatrix1&0\0&0_(n-1)times(n-1), quad pmatrixU&0\0&I_n-2
$$
for sufficiently small $epsilon > 0$.

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
  – Omnomnomnom
  Jul 17 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
  – user50394
  Jul 17 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
  – Omnomnomnom
  Jul 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
  – user50394
  Jul 17 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
  – Omnomnomnom
  Jul 18 at 16:43
  

  
  
  
  

 | 
show 1 more comment

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

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oldest

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active

oldest

votes

up vote
0
down vote

Taking
$$
P = pmatrix1&0\0&0, quad U = pmatrixsqrt1 - epsilon^2 & -epsilon \ epsilon & sqrt1 - epsilon^2
$$
is enough for us to see that $|(PUP)^n-1|$ can be made arbitrarily close to $1$ in the case of $n = 2$. A similar example works for arbitrary $n$. That is, it's enough to take
$$
pmatrix1&0\0&0_(n-1)times(n-1), quad pmatrixU&0\0&I_n-2
$$
for sufficiently small $epsilon > 0$.

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
  – Omnomnomnom
  Jul 17 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
  – user50394
  Jul 17 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
  – Omnomnomnom
  Jul 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
  – user50394
  Jul 17 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
  – Omnomnomnom
  Jul 18 at 16:43
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

Taking
$$
P = pmatrix1&0\0&0, quad U = pmatrixsqrt1 - epsilon^2 & -epsilon \ epsilon & sqrt1 - epsilon^2
$$
is enough for us to see that $|(PUP)^n-1|$ can be made arbitrarily close to $1$ in the case of $n = 2$. A similar example works for arbitrary $n$. That is, it's enough to take
$$
pmatrix1&0\0&0_(n-1)times(n-1), quad pmatrixU&0\0&I_n-2
$$
for sufficiently small $epsilon > 0$.

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
  – Omnomnomnom
  Jul 17 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
  – user50394
  Jul 17 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
  – Omnomnomnom
  Jul 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
  – user50394
  Jul 17 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
  – Omnomnomnom
  Jul 18 at 16:43
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

up vote
0
down vote

Taking
$$
P = pmatrix1&0\0&0, quad U = pmatrixsqrt1 - epsilon^2 & -epsilon \ epsilon & sqrt1 - epsilon^2
$$
is enough for us to see that $|(PUP)^n-1|$ can be made arbitrarily close to $1$ in the case of $n = 2$. A similar example works for arbitrary $n$. That is, it's enough to take
$$
pmatrix1&0\0&0_(n-1)times(n-1), quad pmatrixU&0\0&I_n-2
$$
for sufficiently small $epsilon > 0$.

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

Taking
$$
P = pmatrix1&0\0&0, quad U = pmatrixsqrt1 - epsilon^2 & -epsilon \ epsilon & sqrt1 - epsilon^2
$$
is enough for us to see that $|(PUP)^n-1|$ can be made arbitrarily close to $1$ in the case of $n = 2$. A similar example works for arbitrary $n$. That is, it's enough to take
$$
pmatrix1&0\0&0_(n-1)times(n-1), quad pmatrixU&0\0&I_n-2
$$
for sufficiently small $epsilon > 0$.

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 20:26

Omnomnomnom

121k784170

answered Jul 17 at 20:26

Omnomnomnom

121k784170

answered Jul 17 at 20:26

Omnomnomnom

121k784170

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
  – Omnomnomnom
  Jul 17 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
  – user50394
  Jul 17 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
  – Omnomnomnom
  Jul 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
  – user50394
  Jul 17 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
  – Omnomnomnom
  Jul 18 at 16:43
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
  – Omnomnomnom
  Jul 17 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
  – user50394
  Jul 17 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
  – Omnomnomnom
  Jul 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
  – user50394
  Jul 17 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
  – Omnomnomnom
  Jul 18 at 16:43
  

  
  
  
  

I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
– Omnomnomnom
Jul 17 at 20:27

I would suspect, however, that it is possible to come up with a non-trivial bound which depends on $|PU - UP|$.
– Omnomnomnom
Jul 17 at 20:27

Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
– user50394
Jul 17 at 20:46

Thanks, @Omnomnomnom. I think I would view it as that if the projector is of rank one, then the upper bound is trivial. Perhaps a more interesting case to me is that if $P$ is of rank say $n-1$ then can we say something about the upper bound?
– user50394
Jul 17 at 20:46

@user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
– Omnomnomnom
Jul 17 at 20:57

@user50394 the answer is still no. If $tilde P$ denotes the bigger projection matrix, consider $I - tilde P$
– Omnomnomnom
Jul 17 at 20:57

Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
– user50394
Jul 17 at 22:30

Not sure if I understand it. Your answer seems to claim that $|((I-P)U(I-P))^n-1|=|(PUP)^n-1|$ but I don't see why that should be the case.
– user50394
Jul 17 at 22:30

You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
– Omnomnomnom
Jul 18 at 16:43

You should find that $(I-P)U(I-P)$ is a diagonal matrix. Take it from there.
– Omnomnomnom
Jul 18 at 16:43

 | 
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