Mutually commuting matrix with $A_i^2=0$
up vote
5
down vote
favorite
Let $A_1,dots,A_n$ be mutually commuting $mtimes m$ matrices such that $A_i^2=0$ for all $1le i le n$. If $m<2^n$, prove that $A_1 A_2cdots A_n=0$
Since $A_i^2=0$ So $operatorname{Im}(A)subset ker(A) implies dim(ker(A))ge frac m 2$ I think we have to use it to prove ..but don't know how go through ...
linear-algebra matrices nilpotence
edited Nov 23 at 18:13
Zvi
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
|
show 1 more comment
up vote
5
down vote
favorite
Let $A_1,dots,A_n$ be mutually commuting $mtimes m$ matrices such that $A_i^2=0$ for all $1le i le n$. If $m<2^n$, prove that $A_1 A_2cdots A_n=0$
Since $A_i^2=0$ So $operatorname{Im}(A)subset ker(A) implies dim(ker(A))ge frac m 2$ I think we have to use it to prove ..but don't know how go through ...
linear-algebra matrices nilpotence
edited Nov 23 at 18:13
Zvi
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
2
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
2
yes right ......
– RAM_3R
Nov 23 at 15:05
1
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55
|
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $A_1,dots,A_n$ be mutually commuting $mtimes m$ matrices such that $A_i^2=0$ for all $1le i le n$. If $m<2^n$, prove that $A_1 A_2cdots A_n=0$
Since $A_i^2=0$ So $operatorname{Im}(A)subset ker(A) implies dim(ker(A))ge frac m 2$ I think we have to use it to prove ..but don't know how go through ...
linear-algebra matrices nilpotence
edited Nov 23 at 18:13
Zvi
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
Let $A_1,dots,A_n$ be mutually commuting $mtimes m$ matrices such that $A_i^2=0$ for all $1le i le n$. If $m<2^n$, prove that $A_1 A_2cdots A_n=0$
Since $A_i^2=0$ So $operatorname{Im}(A)subset ker(A) implies dim(ker(A))ge frac m 2$ I think we have to use it to prove ..but don't know how go through ...
linear-algebra matrices nilpotence
linear-algebra matrices nilpotence
edited Nov 23 at 18:13
Zvi
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
edited Nov 23 at 18:13
Zvi
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
edited Nov 23 at 18:13
Zvi
4,125328
edited Nov 23 at 18:13
Zvi
4,125328
edited Nov 23 at 18:13
Zvi
4,125328
4,125328
asked Nov 23 at 14:12
RAM_3R
527214
asked Nov 23 at 14:12
RAM_3R
527214
asked Nov 23 at 14:12
RAM_3R
527214
527214
2
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
2
yes right ......
– RAM_3R
Nov 23 at 15:05
1
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55
|
show 1 more comment
2
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
2
yes right ......
– RAM_3R
Nov 23 at 15:05
1
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55
2
2
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
2
2
yes right ......
– RAM_3R
Nov 23 at 15:05
yes right ......
– RAM_3R
Nov 23 at 15:05
1
1
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Let $xin K^m$.
Then $A_1xin Im(A_1)$, vector space of dimension $leq m/2$.
In the sequel, $tilde{U}$ denotes a restriction of $U$.
$tilde{A_2}:Im(A_1)rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1xin Im(tilde{A_2})$,
vector space of dimension $leq dim(Im(A_1))/2leq m/2^2$, and so on $tilde{A_3}:Im(tilde{A_2})rightarrow Im(tilde{A_2})$ is $2$-nilpotent,....
$tilde{A_n}:Im(tilde{A_{n-1}})rightarrow Im(tilde{A_{n-1}})$
is $2$-nilpotent. Then $A_ncdots A_1xin Im(tilde{A_n})$, vector space of dimension $leq dim(Im(tilde{A_{n-1}}))/2leq m/2^n<1$.
Finally $Im(tilde{A_n})={0}$ and we are done.
answered Nov 23 at 17:29
loup blanc
22.4k21750
add a comment |
up vote
2
down vote
Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.
Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$begin{bmatrix}
0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}1 & 3 \
0 & 0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
end{bmatrix}$
has first minidiagonal (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)
Multiplication $A_1A_2 dots A_n$ means that the result matrix has shift at least $nm/2$ what is
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.
For $n=3$ we have limit for $m$ equal 7. Result shift $12$
For $n=4$ we have limit for $m$ equal 15. Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
answered Nov 23 at 16:13
Widawensen
4,40121445
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Let $xin K^m$.
Then $A_1xin Im(A_1)$, vector space of dimension $leq m/2$.
In the sequel, $tilde{U}$ denotes a restriction of $U$.
$tilde{A_2}:Im(A_1)rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1xin Im(tilde{A_2})$,
vector space of dimension $leq dim(Im(A_1))/2leq m/2^2$, and so on $tilde{A_3}:Im(tilde{A_2})rightarrow Im(tilde{A_2})$ is $2$-nilpotent,....
$tilde{A_n}:Im(tilde{A_{n-1}})rightarrow Im(tilde{A_{n-1}})$
is $2$-nilpotent. Then $A_ncdots A_1xin Im(tilde{A_n})$, vector space of dimension $leq dim(Im(tilde{A_{n-1}}))/2leq m/2^n<1$.
Finally $Im(tilde{A_n})={0}$ and we are done.
answered Nov 23 at 17:29
loup blanc
22.4k21750
add a comment |
up vote
7
down vote
accepted
Let $xin K^m$.
Then $A_1xin Im(A_1)$, vector space of dimension $leq m/2$.
In the sequel, $tilde{U}$ denotes a restriction of $U$.
$tilde{A_2}:Im(A_1)rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1xin Im(tilde{A_2})$,
vector space of dimension $leq dim(Im(A_1))/2leq m/2^2$, and so on $tilde{A_3}:Im(tilde{A_2})rightarrow Im(tilde{A_2})$ is $2$-nilpotent,....
$tilde{A_n}:Im(tilde{A_{n-1}})rightarrow Im(tilde{A_{n-1}})$
is $2$-nilpotent. Then $A_ncdots A_1xin Im(tilde{A_n})$, vector space of dimension $leq dim(Im(tilde{A_{n-1}}))/2leq m/2^n<1$.
Finally $Im(tilde{A_n})={0}$ and we are done.
answered Nov 23 at 17:29
loup blanc
22.4k21750
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Let $xin K^m$.
Then $A_1xin Im(A_1)$, vector space of dimension $leq m/2$.
In the sequel, $tilde{U}$ denotes a restriction of $U$.
$tilde{A_2}:Im(A_1)rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1xin Im(tilde{A_2})$,
vector space of dimension $leq dim(Im(A_1))/2leq m/2^2$, and so on $tilde{A_3}:Im(tilde{A_2})rightarrow Im(tilde{A_2})$ is $2$-nilpotent,....
$tilde{A_n}:Im(tilde{A_{n-1}})rightarrow Im(tilde{A_{n-1}})$
is $2$-nilpotent. Then $A_ncdots A_1xin Im(tilde{A_n})$, vector space of dimension $leq dim(Im(tilde{A_{n-1}}))/2leq m/2^n<1$.
Finally $Im(tilde{A_n})={0}$ and we are done.
answered Nov 23 at 17:29
loup blanc
22.4k21750
Let $xin K^m$.
Then $A_1xin Im(A_1)$, vector space of dimension $leq m/2$.
In the sequel, $tilde{U}$ denotes a restriction of $U$.
$tilde{A_2}:Im(A_1)rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1xin Im(tilde{A_2})$,
vector space of dimension $leq dim(Im(A_1))/2leq m/2^2$, and so on $tilde{A_3}:Im(tilde{A_2})rightarrow Im(tilde{A_2})$ is $2$-nilpotent,....
$tilde{A_n}:Im(tilde{A_{n-1}})rightarrow Im(tilde{A_{n-1}})$
is $2$-nilpotent. Then $A_ncdots A_1xin Im(tilde{A_n})$, vector space of dimension $leq dim(Im(tilde{A_{n-1}}))/2leq m/2^n<1$.
Finally $Im(tilde{A_n})={0}$ and we are done.
answered Nov 23 at 17:29
loup blanc
22.4k21750
edited Nov 23 at 17:34
answered Nov 23 at 17:29
loup blanc
22.4k21750
answered Nov 23 at 17:29
loup blanc
22.4k21750
answered Nov 23 at 17:29
loup blanc
22.4k21750
22.4k21750
add a comment |
add a comment |
up vote
2
down vote
Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.
Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$begin{bmatrix}
0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}1 & 3 \
0 & 0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
end{bmatrix}$
has first minidiagonal (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)
Multiplication $A_1A_2 dots A_n$ means that the result matrix has shift at least $nm/2$ what is
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.
For $n=3$ we have limit for $m$ equal 7. Result shift $12$
For $n=4$ we have limit for $m$ equal 15. Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
answered Nov 23 at 16:13
Widawensen
4,40121445
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
add a comment |
up vote
2
down vote
Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.
Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$begin{bmatrix}
0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}1 & 3 \
0 & 0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
end{bmatrix}$
has first minidiagonal (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)
Multiplication $A_1A_2 dots A_n$ means that the result matrix has shift at least $nm/2$ what is
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.
For $n=3$ we have limit for $m$ equal 7. Result shift $12$
For $n=4$ we have limit for $m$ equal 15. Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
answered Nov 23 at 16:13
Widawensen
4,40121445
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
add a comment |
up vote
2
down vote
up vote
2
down vote
Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.
Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$begin{bmatrix}
0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}1 & 3 \
0 & 0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
end{bmatrix}$
has first minidiagonal (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)
Multiplication $A_1A_2 dots A_n$ means that the result matrix has shift at least $nm/2$ what is
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.
For $n=3$ we have limit for $m$ equal 7. Result shift $12$
For $n=4$ we have limit for $m$ equal 15. Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
answered Nov 23 at 16:13
Widawensen
4,40121445
Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.
Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$begin{bmatrix}
0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}1 & 3 \
0 & 0color{red}rightarrow & 0color{red}rightarrow & 0color{red}rightarrow & color{red}2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
end{bmatrix}$
has first minidiagonal (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)
Multiplication $A_1A_2 dots A_n$ means that the result matrix has shift at least $nm/2$ what is
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.
For $n=3$ we have limit for $m$ equal 7. Result shift $12$
For $n=4$ we have limit for $m$ equal 15. Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
answered Nov 23 at 16:13
Widawensen
4,40121445
edited Nov 23 at 17:29
answered Nov 23 at 16:13
Widawensen
4,40121445
answered Nov 23 at 16:13
Widawensen
4,40121445
answered Nov 23 at 16:13
Widawensen
4,40121445
4,40121445
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
add a comment |
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
1
1
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices.
– loup blanc
Nov 23 at 16:36
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
@loupblanc So would it be only a partial answer for particular form of nilpotent matrices?
– Widawensen
Nov 23 at 16:45
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
I think that it is difficult to show that we can put our matrices in the form you consider.
– loup blanc
Nov 23 at 17:32
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
@loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you..
– Widawensen
Nov 23 at 17:34
add a comment |
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2
I guess you mean $dim(ker(A_i)) geq m/2$!? Probably it is useful to get a bound for $dim(ker(A_{n-1}A_n))$ and iterate
– Stockfish
Nov 23 at 14:39
2
yes right ......
– RAM_3R
Nov 23 at 15:05
1
@Stockfish probably yes but i don't understand how relate this $m<2^n$??
– RAM_3R
Nov 23 at 15:41
Not a new question. That follows is a post with $m=n$. cf. math.stackexchange.com/questions/880429/…
– loup blanc
Nov 23 at 16:27
But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent
– Vincent
Nov 23 at 16:55