Mixing
Proof that in any power of a matrix, two entries on the main diagonal will be the same
$begingroup$
Say you are given a 3x3 matrix, where two elements on the diagonal of the matrix are the same, how would you go about proving that with any power of the matrix, two entries on the diagonal will still be the same?
I started by trying to use induction, but couldn't get anywhere.
I then read up about calculating the characteristic polynomial, eigenvectors and eigenvalues but again couldn't get anywhere with that.
I haven't included the exact matrix in the question since I'd prefer people to guide me to the answer rather than give me the answer straight up.
Any help is appreciated.
EDIT. Let $Ain M_3(K)$. We say that $A$ satisfies the property $mathcal{P}$ iff, for every $rgeq 1$, the $3$ entries of the diagonal of $A^r$ are not all distinct.
QUESTION. Can we explicitly find a positive integer $k$ and a procedure involving a finite number of actions on the entries of $A,cdots,A^k$, and allowing to conclude whether or not $A$ satisfies $mathcal{P}$ ?
linear-algebra matrices eigenvalues-eigenvectors induction
edited Mar 19 at 15:31
N. F. Taussig
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
$endgroup$
|
show 1 more comment
$begingroup$
Say you are given a 3x3 matrix, where two elements on the diagonal of the matrix are the same, how would you go about proving that with any power of the matrix, two entries on the diagonal will still be the same?
I started by trying to use induction, but couldn't get anywhere.
I then read up about calculating the characteristic polynomial, eigenvectors and eigenvalues but again couldn't get anywhere with that.
I haven't included the exact matrix in the question since I'd prefer people to guide me to the answer rather than give me the answer straight up.
Any help is appreciated.
EDIT. Let $Ain M_3(K)$. We say that $A$ satisfies the property $mathcal{P}$ iff, for every $rgeq 1$, the $3$ entries of the diagonal of $A^r$ are not all distinct.
QUESTION. Can we explicitly find a positive integer $k$ and a procedure involving a finite number of actions on the entries of $A,cdots,A^k$, and allowing to conclude whether or not $A$ satisfies $mathcal{P}$ ?
linear-algebra matrices eigenvalues-eigenvectors induction
edited Mar 19 at 15:31
N. F. Taussig
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
$endgroup$
$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25
|
show 1 more comment
$begingroup$
Say you are given a 3x3 matrix, where two elements on the diagonal of the matrix are the same, how would you go about proving that with any power of the matrix, two entries on the diagonal will still be the same?
I started by trying to use induction, but couldn't get anywhere.
I then read up about calculating the characteristic polynomial, eigenvectors and eigenvalues but again couldn't get anywhere with that.
I haven't included the exact matrix in the question since I'd prefer people to guide me to the answer rather than give me the answer straight up.
Any help is appreciated.
EDIT. Let $Ain M_3(K)$. We say that $A$ satisfies the property $mathcal{P}$ iff, for every $rgeq 1$, the $3$ entries of the diagonal of $A^r$ are not all distinct.
QUESTION. Can we explicitly find a positive integer $k$ and a procedure involving a finite number of actions on the entries of $A,cdots,A^k$, and allowing to conclude whether or not $A$ satisfies $mathcal{P}$ ?
linear-algebra matrices eigenvalues-eigenvectors induction
edited Mar 19 at 15:31
N. F. Taussig
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
$endgroup$
Say you are given a 3x3 matrix, where two elements on the diagonal of the matrix are the same, how would you go about proving that with any power of the matrix, two entries on the diagonal will still be the same?
I started by trying to use induction, but couldn't get anywhere.
I then read up about calculating the characteristic polynomial, eigenvectors and eigenvalues but again couldn't get anywhere with that.
I haven't included the exact matrix in the question since I'd prefer people to guide me to the answer rather than give me the answer straight up.
Any help is appreciated.
EDIT. Let $Ain M_3(K)$. We say that $A$ satisfies the property $mathcal{P}$ iff, for every $rgeq 1$, the $3$ entries of the diagonal of $A^r$ are not all distinct.
QUESTION. Can we explicitly find a positive integer $k$ and a procedure involving a finite number of actions on the entries of $A,cdots,A^k$, and allowing to conclude whether or not $A$ satisfies $mathcal{P}$ ?
linear-algebra matrices eigenvalues-eigenvectors induction
linear-algebra matrices eigenvalues-eigenvectors induction
edited Mar 19 at 15:31
N. F. Taussig
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
edited Mar 19 at 15:31
N. F. Taussig
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
edited Mar 19 at 15:31
N. F. Taussig
45k103358
edited Mar 19 at 15:31
N. F. Taussig
45k103358
edited Mar 19 at 15:31
N. F. Taussig
45k103358
45k103358
asked Jan 30 at 18:32
Student_1996Student_1996
644
asked Jan 30 at 18:32
Student_1996Student_1996
644
asked Jan 30 at 18:32
Student_1996Student_1996
644
644
$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25
|
show 1 more comment
$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25
$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Induction seems to be the way to go. If you can prove it for two general matrices with this property, then you're done. Given that you have a specific problem, I would recommend just calculating the product of the two matrices.
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
add a comment |
$begingroup$
$textbf{Conjecture (Quasi Theorem)}.$ If, for $kleq 6$, the 3 entries of the diagonal of $A^k$ are not all distinct, then $A$ satisfies $mathcal{P}$.
$textbf{60% of the proof}.$ We know that $A^3=aI+bA+cA^2$.
For $kleq 6$ we associate the pattern $(1,2)$, (resp. $(1,3)$ or $(2,3))$ when ${A^k}_{1,1}={A^k}_{2,2}$ (resp.$cdots$) (each $A^k$ has one or three associated patterns). In the sequel, we show that we can associate a pattern to the $(A^r)_{r>6}$.
We consider the first occurrence of $2$ equal patterns in the sequence of powers of $A$. There are $5$ non-trivial cases.
i) $A^2,A^4$, ii) $A,A^3$, iii) $A^2,A^3$, iv) $A,A^4$, v) $A^3,A^4$.
Indeed, if $A,A^2$ has same pattern $p$, then $A^r=a_rI+b_rA+c_rA^2$ has a pattern and it's $p$.
$textbf{Case i)}.$ The first $4$ patterns are in the form : $p_1,p_2,p_3,p_2$. The system $I,A^2,A^4$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^4$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^4in span(I,A^2)$. Thus, for every even $k$, $A^k$ has a pattern and it's $p_2$.
$A^5in span(A,A^3)$; if $A^5=uA$ or $A^5=vA^3$, then the $(A^r)$ have patterns and they are periodic and we are done; otherwise, $A^5$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Finally, $A^4,A^5,A^6$ have pattern $p_2$; then $A^7=aA^4+bA^5+cA^6$ has pattern $p_2$. In fact, this case is degenerate because $A^7in span(A^3,A^5)$ and cannot have pattern $p_2$ except if $A^7=alpha A^5$ and we are done.
$textbf{Case ii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_1$. The system $I,A,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^3$ and have same pattern.
If $A=alpha I$, we are done and we may assume that $A^3in span(I,A),A^3=alpha I +beta A$.
$A^4in span(A,A^2)$; the cases $A^4=uA,A^4=vA^2$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^6in span(A^3,A^4)$; the cases $A^6=uA^3,A^6=vA^4$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Yet, $A^6=alpha^2I+beta^2A^2+2alphabeta A$ and its pattern cannot be $p_2$ except if $alphabeta=0$, an easy case.
That follows is an example where $A^3=65A$.
$$Aapprox begin{pmatrix}2& -1.066666667-2.421202640i& -1\ -2.285714286+5.188291371i & 2& 5\-3& 7& -4end{pmatrix}$$
$textbf{Case iii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_2$. The system $I,A^2,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A^2,A^3$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^3in span(I,A^2),A^3=alpha A^2+beta I$.
$A^4in span(A,A^3)$; the cases $A^4=uA,A^4=vA^3$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^5in span(A^2,A^4)$; the cases $A^5=uA^2,A^5=vA^4$ are easy; otherwise, $A^5$ has a pattern that cannot be $p_2,p_3$ and, therefore, has pattern $p_1$.
$A^6in span(A^3,A^5)$; the cases $A^6=uA^3,A^6=vA^5$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_2,p_1$ and, therefore, has pattern $p_3$. Yet, $A^6=alpha^2 A^4+beta^2 I+2alphabeta A^2$ and its pattern cannot be $p_3$ except if $alphabeta=0$, an easy case.
TO BE CONTINUED...
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
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2 Answers
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2 Answers
2
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$begingroup$
Induction seems to be the way to go. If you can prove it for two general matrices with this property, then you're done. Given that you have a specific problem, I would recommend just calculating the product of the two matrices.
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
add a comment |
$begingroup$
Induction seems to be the way to go. If you can prove it for two general matrices with this property, then you're done. Given that you have a specific problem, I would recommend just calculating the product of the two matrices.
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
add a comment |
$begingroup$
Induction seems to be the way to go. If you can prove it for two general matrices with this property, then you're done. Given that you have a specific problem, I would recommend just calculating the product of the two matrices.
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
$endgroup$
Induction seems to be the way to go. If you can prove it for two general matrices with this property, then you're done. Given that you have a specific problem, I would recommend just calculating the product of the two matrices.
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
answered Jan 30 at 18:37
OldGodzillaOldGodzilla
58227
58227
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
add a comment |
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
$begingroup$
Product of which two matrices? The question only has one matrix.
$endgroup$
– Student_1996
Jan 30 at 18:40
add a comment |
$begingroup$
$textbf{Conjecture (Quasi Theorem)}.$ If, for $kleq 6$, the 3 entries of the diagonal of $A^k$ are not all distinct, then $A$ satisfies $mathcal{P}$.
$textbf{60% of the proof}.$ We know that $A^3=aI+bA+cA^2$.
For $kleq 6$ we associate the pattern $(1,2)$, (resp. $(1,3)$ or $(2,3))$ when ${A^k}_{1,1}={A^k}_{2,2}$ (resp.$cdots$) (each $A^k$ has one or three associated patterns). In the sequel, we show that we can associate a pattern to the $(A^r)_{r>6}$.
We consider the first occurrence of $2$ equal patterns in the sequence of powers of $A$. There are $5$ non-trivial cases.
i) $A^2,A^4$, ii) $A,A^3$, iii) $A^2,A^3$, iv) $A,A^4$, v) $A^3,A^4$.
Indeed, if $A,A^2$ has same pattern $p$, then $A^r=a_rI+b_rA+c_rA^2$ has a pattern and it's $p$.
$textbf{Case i)}.$ The first $4$ patterns are in the form : $p_1,p_2,p_3,p_2$. The system $I,A^2,A^4$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^4$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^4in span(I,A^2)$. Thus, for every even $k$, $A^k$ has a pattern and it's $p_2$.
$A^5in span(A,A^3)$; if $A^5=uA$ or $A^5=vA^3$, then the $(A^r)$ have patterns and they are periodic and we are done; otherwise, $A^5$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Finally, $A^4,A^5,A^6$ have pattern $p_2$; then $A^7=aA^4+bA^5+cA^6$ has pattern $p_2$. In fact, this case is degenerate because $A^7in span(A^3,A^5)$ and cannot have pattern $p_2$ except if $A^7=alpha A^5$ and we are done.
$textbf{Case ii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_1$. The system $I,A,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^3$ and have same pattern.
If $A=alpha I$, we are done and we may assume that $A^3in span(I,A),A^3=alpha I +beta A$.
$A^4in span(A,A^2)$; the cases $A^4=uA,A^4=vA^2$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^6in span(A^3,A^4)$; the cases $A^6=uA^3,A^6=vA^4$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Yet, $A^6=alpha^2I+beta^2A^2+2alphabeta A$ and its pattern cannot be $p_2$ except if $alphabeta=0$, an easy case.
That follows is an example where $A^3=65A$.
$$Aapprox begin{pmatrix}2& -1.066666667-2.421202640i& -1\ -2.285714286+5.188291371i & 2& 5\-3& 7& -4end{pmatrix}$$
$textbf{Case iii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_2$. The system $I,A^2,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A^2,A^3$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^3in span(I,A^2),A^3=alpha A^2+beta I$.
$A^4in span(A,A^3)$; the cases $A^4=uA,A^4=vA^3$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^5in span(A^2,A^4)$; the cases $A^5=uA^2,A^5=vA^4$ are easy; otherwise, $A^5$ has a pattern that cannot be $p_2,p_3$ and, therefore, has pattern $p_1$.
$A^6in span(A^3,A^5)$; the cases $A^6=uA^3,A^6=vA^5$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_2,p_1$ and, therefore, has pattern $p_3$. Yet, $A^6=alpha^2 A^4+beta^2 I+2alphabeta A^2$ and its pattern cannot be $p_3$ except if $alphabeta=0$, an easy case.
TO BE CONTINUED...
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
$endgroup$
add a comment |
$begingroup$
$textbf{Conjecture (Quasi Theorem)}.$ If, for $kleq 6$, the 3 entries of the diagonal of $A^k$ are not all distinct, then $A$ satisfies $mathcal{P}$.
$textbf{60% of the proof}.$ We know that $A^3=aI+bA+cA^2$.
For $kleq 6$ we associate the pattern $(1,2)$, (resp. $(1,3)$ or $(2,3))$ when ${A^k}_{1,1}={A^k}_{2,2}$ (resp.$cdots$) (each $A^k$ has one or three associated patterns). In the sequel, we show that we can associate a pattern to the $(A^r)_{r>6}$.
We consider the first occurrence of $2$ equal patterns in the sequence of powers of $A$. There are $5$ non-trivial cases.
i) $A^2,A^4$, ii) $A,A^3$, iii) $A^2,A^3$, iv) $A,A^4$, v) $A^3,A^4$.
Indeed, if $A,A^2$ has same pattern $p$, then $A^r=a_rI+b_rA+c_rA^2$ has a pattern and it's $p$.
$textbf{Case i)}.$ The first $4$ patterns are in the form : $p_1,p_2,p_3,p_2$. The system $I,A^2,A^4$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^4$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^4in span(I,A^2)$. Thus, for every even $k$, $A^k$ has a pattern and it's $p_2$.
$A^5in span(A,A^3)$; if $A^5=uA$ or $A^5=vA^3$, then the $(A^r)$ have patterns and they are periodic and we are done; otherwise, $A^5$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Finally, $A^4,A^5,A^6$ have pattern $p_2$; then $A^7=aA^4+bA^5+cA^6$ has pattern $p_2$. In fact, this case is degenerate because $A^7in span(A^3,A^5)$ and cannot have pattern $p_2$ except if $A^7=alpha A^5$ and we are done.
$textbf{Case ii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_1$. The system $I,A,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^3$ and have same pattern.
If $A=alpha I$, we are done and we may assume that $A^3in span(I,A),A^3=alpha I +beta A$.
$A^4in span(A,A^2)$; the cases $A^4=uA,A^4=vA^2$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^6in span(A^3,A^4)$; the cases $A^6=uA^3,A^6=vA^4$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Yet, $A^6=alpha^2I+beta^2A^2+2alphabeta A$ and its pattern cannot be $p_2$ except if $alphabeta=0$, an easy case.
That follows is an example where $A^3=65A$.
$$Aapprox begin{pmatrix}2& -1.066666667-2.421202640i& -1\ -2.285714286+5.188291371i & 2& 5\-3& 7& -4end{pmatrix}$$
$textbf{Case iii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_2$. The system $I,A^2,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A^2,A^3$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^3in span(I,A^2),A^3=alpha A^2+beta I$.
$A^4in span(A,A^3)$; the cases $A^4=uA,A^4=vA^3$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^5in span(A^2,A^4)$; the cases $A^5=uA^2,A^5=vA^4$ are easy; otherwise, $A^5$ has a pattern that cannot be $p_2,p_3$ and, therefore, has pattern $p_1$.
$A^6in span(A^3,A^5)$; the cases $A^6=uA^3,A^6=vA^5$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_2,p_1$ and, therefore, has pattern $p_3$. Yet, $A^6=alpha^2 A^4+beta^2 I+2alphabeta A^2$ and its pattern cannot be $p_3$ except if $alphabeta=0$, an easy case.
TO BE CONTINUED...
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
$endgroup$
add a comment |
$begingroup$
$textbf{Conjecture (Quasi Theorem)}.$ If, for $kleq 6$, the 3 entries of the diagonal of $A^k$ are not all distinct, then $A$ satisfies $mathcal{P}$.
$textbf{60% of the proof}.$ We know that $A^3=aI+bA+cA^2$.
For $kleq 6$ we associate the pattern $(1,2)$, (resp. $(1,3)$ or $(2,3))$ when ${A^k}_{1,1}={A^k}_{2,2}$ (resp.$cdots$) (each $A^k$ has one or three associated patterns). In the sequel, we show that we can associate a pattern to the $(A^r)_{r>6}$.
We consider the first occurrence of $2$ equal patterns in the sequence of powers of $A$. There are $5$ non-trivial cases.
i) $A^2,A^4$, ii) $A,A^3$, iii) $A^2,A^3$, iv) $A,A^4$, v) $A^3,A^4$.
Indeed, if $A,A^2$ has same pattern $p$, then $A^r=a_rI+b_rA+c_rA^2$ has a pattern and it's $p$.
$textbf{Case i)}.$ The first $4$ patterns are in the form : $p_1,p_2,p_3,p_2$. The system $I,A^2,A^4$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^4$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^4in span(I,A^2)$. Thus, for every even $k$, $A^k$ has a pattern and it's $p_2$.
$A^5in span(A,A^3)$; if $A^5=uA$ or $A^5=vA^3$, then the $(A^r)$ have patterns and they are periodic and we are done; otherwise, $A^5$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Finally, $A^4,A^5,A^6$ have pattern $p_2$; then $A^7=aA^4+bA^5+cA^6$ has pattern $p_2$. In fact, this case is degenerate because $A^7in span(A^3,A^5)$ and cannot have pattern $p_2$ except if $A^7=alpha A^5$ and we are done.
$textbf{Case ii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_1$. The system $I,A,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^3$ and have same pattern.
If $A=alpha I$, we are done and we may assume that $A^3in span(I,A),A^3=alpha I +beta A$.
$A^4in span(A,A^2)$; the cases $A^4=uA,A^4=vA^2$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^6in span(A^3,A^4)$; the cases $A^6=uA^3,A^6=vA^4$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Yet, $A^6=alpha^2I+beta^2A^2+2alphabeta A$ and its pattern cannot be $p_2$ except if $alphabeta=0$, an easy case.
That follows is an example where $A^3=65A$.
$$Aapprox begin{pmatrix}2& -1.066666667-2.421202640i& -1\ -2.285714286+5.188291371i & 2& 5\-3& 7& -4end{pmatrix}$$
$textbf{Case iii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_2$. The system $I,A^2,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A^2,A^3$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^3in span(I,A^2),A^3=alpha A^2+beta I$.
$A^4in span(A,A^3)$; the cases $A^4=uA,A^4=vA^3$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^5in span(A^2,A^4)$; the cases $A^5=uA^2,A^5=vA^4$ are easy; otherwise, $A^5$ has a pattern that cannot be $p_2,p_3$ and, therefore, has pattern $p_1$.
$A^6in span(A^3,A^5)$; the cases $A^6=uA^3,A^6=vA^5$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_2,p_1$ and, therefore, has pattern $p_3$. Yet, $A^6=alpha^2 A^4+beta^2 I+2alphabeta A^2$ and its pattern cannot be $p_3$ except if $alphabeta=0$, an easy case.
TO BE CONTINUED...
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
$endgroup$
$textbf{Conjecture (Quasi Theorem)}.$ If, for $kleq 6$, the 3 entries of the diagonal of $A^k$ are not all distinct, then $A$ satisfies $mathcal{P}$.
$textbf{60% of the proof}.$ We know that $A^3=aI+bA+cA^2$.
For $kleq 6$ we associate the pattern $(1,2)$, (resp. $(1,3)$ or $(2,3))$ when ${A^k}_{1,1}={A^k}_{2,2}$ (resp.$cdots$) (each $A^k$ has one or three associated patterns). In the sequel, we show that we can associate a pattern to the $(A^r)_{r>6}$.
We consider the first occurrence of $2$ equal patterns in the sequence of powers of $A$. There are $5$ non-trivial cases.
i) $A^2,A^4$, ii) $A,A^3$, iii) $A^2,A^3$, iv) $A,A^4$, v) $A^3,A^4$.
Indeed, if $A,A^2$ has same pattern $p$, then $A^r=a_rI+b_rA+c_rA^2$ has a pattern and it's $p$.
$textbf{Case i)}.$ The first $4$ patterns are in the form : $p_1,p_2,p_3,p_2$. The system $I,A^2,A^4$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^4$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^4in span(I,A^2)$. Thus, for every even $k$, $A^k$ has a pattern and it's $p_2$.
$A^5in span(A,A^3)$; if $A^5=uA$ or $A^5=vA^3$, then the $(A^r)$ have patterns and they are periodic and we are done; otherwise, $A^5$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Finally, $A^4,A^5,A^6$ have pattern $p_2$; then $A^7=aA^4+bA^5+cA^6$ has pattern $p_2$. In fact, this case is degenerate because $A^7in span(A^3,A^5)$ and cannot have pattern $p_2$ except if $A^7=alpha A^5$ and we are done.
$textbf{Case ii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_1$. The system $I,A,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A,A^3$ and have same pattern.
If $A=alpha I$, we are done and we may assume that $A^3in span(I,A),A^3=alpha I +beta A$.
$A^4in span(A,A^2)$; the cases $A^4=uA,A^4=vA^2$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^6in span(A^3,A^4)$; the cases $A^6=uA^3,A^6=vA^4$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_1,p_3$ and, therefore, has pattern $p_2$. Yet, $A^6=alpha^2I+beta^2A^2+2alphabeta A$ and its pattern cannot be $p_2$ except if $alphabeta=0$, an easy case.
That follows is an example where $A^3=65A$.
$$Aapprox begin{pmatrix}2& -1.066666667-2.421202640i& -1\ -2.285714286+5.188291371i & 2& 5\-3& 7& -4end{pmatrix}$$
$textbf{Case iii).}$ The first $3$ patterns are in the form : $p_1,p_2,p_2$. The system $I,A^2,A^3$ is linearly dependent; otherwise, $I,A,A^2$ are linear combination of $I,A^2,A^3$ and have same pattern.
If $A^2=alpha I$, we are done and we may assume that $A^3in span(I,A^2),A^3=alpha A^2+beta I$.
$A^4in span(A,A^3)$; the cases $A^4=uA,A^4=vA^3$ are easy; otherwise, $A^4$ has a pattern that cannot be $p_1,p_2$ and, therefore, has pattern $p_3$.
$A^5in span(A^2,A^4)$; the cases $A^5=uA^2,A^5=vA^4$ are easy; otherwise, $A^5$ has a pattern that cannot be $p_2,p_3$ and, therefore, has pattern $p_1$.
$A^6in span(A^3,A^5)$; the cases $A^6=uA^3,A^6=vA^5$ are easy; otherwise, $A^6$ has a pattern that cannot be $p_2,p_1$ and, therefore, has pattern $p_3$. Yet, $A^6=alpha^2 A^4+beta^2 I+2alphabeta A^2$ and its pattern cannot be $p_3$ except if $alphabeta=0$, an easy case.
TO BE CONTINUED...
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
answered Jan 31 at 17:00
loup blancloup blanc
24.2k21851
24.2k21851
add a comment |
add a comment |
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$begingroup$
This question as posed is not likely to get good answers. For one thing, the matrix in question, which I will write as $A$, has other properties in addition, that guarantee that $A^k$ has two diagonal entries the same. The responses given really should have recognized this.
$endgroup$
– Mike
Jan 30 at 22:24
$begingroup$
There are 3-by-3 matrices $M$ that satisfy the property that 2 of the diagonals are the same but the diagonals of $M^2$ are distinct.
$endgroup$
– Mike
Jan 30 at 22:34
$begingroup$
@Mike ; I rewrote the question. Do you agree ?
$endgroup$
– loup blanc
Jan 30 at 23:12
$begingroup$
So what you want is an algorithm?
$endgroup$
– Mike
Jan 31 at 2:52
$begingroup$
@Mike , no, I don't want anything; however, I think the question in this form is interesting - (in any case, more than 80% of the questions asked on MSE). In particular, I think, but I'm not sure, that if $A,cdots,A^5$ satisfy the imposed condition, then the $(A^r)_{r>5}$ satisfy it too.
$endgroup$
– loup blanc
Jan 31 at 12:25