$operatorname{rank}(A^2)+operatorname{rank}(B^2)geq2operatorname{rank}(AB)$ whenever $AB=BA$?












8















Let $A,B$ be $ntimes n$ matrices. If $AB=BA$, then $operatorname{rank}(A^2)+operatorname{rank}(B^2)geq2operatorname{rank}(AB)$.




Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $operatorname{rank}(B^2)$ is not tractable.



Any hints will be appreciated!






linear-algebra matrices inequality matrix-rank







share|cite|improve this question


















  • 2




    I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
    – Diger
    Nov 26 at 13:57












  • What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
    – Diger
    Nov 26 at 14:36








  • 1




    @Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
    – Colescu
    Nov 26 at 14:43
















8















Let $A,B$ be $ntimes n$ matrices. If $AB=BA$, then $operatorname{rank}(A^2)+operatorname{rank}(B^2)geq2operatorname{rank}(AB)$.




Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $operatorname{rank}(B^2)$ is not tractable.



Any hints will be appreciated!






linear-algebra matrices inequality matrix-rank







share|cite|improve this question


















  • 2




    I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
    – Diger
    Nov 26 at 13:57












  • What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
    – Diger
    Nov 26 at 14:36








  • 1




    @Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
    – Colescu
    Nov 26 at 14:43














8












8








8


3






Let $A,B$ be $ntimes n$ matrices. If $AB=BA$, then $operatorname{rank}(A^2)+operatorname{rank}(B^2)geq2operatorname{rank}(AB)$.




Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $operatorname{rank}(B^2)$ is not tractable.



Any hints will be appreciated!






linear-algebra matrices inequality matrix-rank







share|cite|improve this question














Let $A,B$ be $ntimes n$ matrices. If $AB=BA$, then $operatorname{rank}(A^2)+operatorname{rank}(B^2)geq2operatorname{rank}(AB)$.




Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $operatorname{rank}(B^2)$ is not tractable.



Any hints will be appreciated!






linear-algebra matrices inequality matrix-rank




linear-algebra matrices inequality matrix-rank






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 13:40









Colescu

3,0751736




3,0751736








  • 2




    I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
    – Diger
    Nov 26 at 13:57












  • What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
    – Diger
    Nov 26 at 14:36








  • 1




    @Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
    – Colescu
    Nov 26 at 14:43














  • 2




    I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
    – Diger
    Nov 26 at 13:57












  • What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
    – Diger
    Nov 26 at 14:36








  • 1




    @Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
    – Colescu
    Nov 26 at 14:43








2




2




I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
– Diger
Nov 26 at 13:57






I don't know but it's funny, that this looks very much alike $$ {rm rank}(A^2 + B^2 - 2AB) = {rm rank}((A-B)^2) geq 0 $$ if $AB=BA$ ;) .
– Diger
Nov 26 at 13:57














What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
– Diger
Nov 26 at 14:36






What about $${rm rank}(A^2 + B^2 - 2AB) leq {rm rank}(A^2) + {rm rank}(B^2) + {rm rank}(-2AB) , ? $$
– Diger
Nov 26 at 14:36






1




1




@Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
– Colescu
Nov 26 at 14:43




@Diger You got me! It was inspired by the famous inequality $a^2+b^2geq2ab$. :)
– Colescu
Nov 26 at 14:43










1 Answer
1






active

oldest

votes


















3














Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=begin{pmatrix}0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0end{pmatrix},quad B=begin{pmatrix}0&1&1&0\0&0&0&1\0&0&0&1\0&0&0&0end{pmatrix}.$$ Sorry for misleading. :(






share|cite|improve this answer





















  • Thanks anyway.I've been doing this a whole day.
    – Oolong milk tea
    Nov 27 at 12:57










  • Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
    – Widawensen
    Nov 27 at 13:05












  • We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
    – Widawensen
    Nov 27 at 14:12













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1 Answer
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3














Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=begin{pmatrix}0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0end{pmatrix},quad B=begin{pmatrix}0&1&1&0\0&0&0&1\0&0&0&1\0&0&0&0end{pmatrix}.$$ Sorry for misleading. :(






share|cite|improve this answer





















  • Thanks anyway.I've been doing this a whole day.
    – Oolong milk tea
    Nov 27 at 12:57










  • Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
    – Widawensen
    Nov 27 at 13:05












  • We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
    – Widawensen
    Nov 27 at 14:12


















3














Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=begin{pmatrix}0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0end{pmatrix},quad B=begin{pmatrix}0&1&1&0\0&0&0&1\0&0&0&1\0&0&0&0end{pmatrix}.$$ Sorry for misleading. :(






share|cite|improve this answer





















  • Thanks anyway.I've been doing this a whole day.
    – Oolong milk tea
    Nov 27 at 12:57










  • Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
    – Widawensen
    Nov 27 at 13:05












  • We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
    – Widawensen
    Nov 27 at 14:12
















3












3








3






Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=begin{pmatrix}0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0end{pmatrix},quad B=begin{pmatrix}0&1&1&0\0&0&0&1\0&0&0&1\0&0&0&0end{pmatrix}.$$ Sorry for misleading. :(






share|cite|improve this answer












Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=begin{pmatrix}0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0end{pmatrix},quad B=begin{pmatrix}0&1&1&0\0&0&0&1\0&0&0&1\0&0&0&0end{pmatrix}.$$ Sorry for misleading. :(







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 12:08









Colescu

3,0751736




3,0751736












  • Thanks anyway.I've been doing this a whole day.
    – Oolong milk tea
    Nov 27 at 12:57










  • Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
    – Widawensen
    Nov 27 at 13:05












  • We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
    – Widawensen
    Nov 27 at 14:12




















  • Thanks anyway.I've been doing this a whole day.
    – Oolong milk tea
    Nov 27 at 12:57










  • Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
    – Widawensen
    Nov 27 at 13:05












  • We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
    – Widawensen
    Nov 27 at 14:12


















Thanks anyway.I've been doing this a whole day.
– Oolong milk tea
Nov 27 at 12:57




Thanks anyway.I've been doing this a whole day.
– Oolong milk tea
Nov 27 at 12:57












Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
– Widawensen
Nov 27 at 13:05






Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA ne 0$ then counterexample is ready..
– Widawensen
Nov 27 at 13:05














We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
– Widawensen
Nov 27 at 14:12






We see that the counterexample is of the form $AB=begin{bmatrix}N & I \ 0 & N end{bmatrix}begin{bmatrix}N & 0 \ 0 & N end{bmatrix}=begin{bmatrix}0 & N \ 0 & 0 end{bmatrix}=BA$ with $begin{bmatrix}N & I \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 2N \ 0 & 0 end{bmatrix}$ and $begin{bmatrix}N & 0 \ 0 & N end{bmatrix}^2=begin{bmatrix}0 & 0 \ 0 & 0 end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question.
– Widawensen
Nov 27 at 14:12




















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