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Symmetric matrices containing $1, 2, 3, …$ in each line
There exists a real number $c$ such that $A+cI$ is positive when $A$ is symmetricProof skew-symmetric matrix $A$ exhibits $vecx^T A vecx = 0$bases for hermitian matricesFor all square matrices $A$ and $B$ of the same size, it is true that $(A+B)^2 = A^2 + 2AB + B^2$?How to encode matrices uniquelyLinear algebra: diagonalisation of antisymmetrisationIs there a mathematical notation of indexing a matrix?Existence and uniqueness of $Ain M_2(mathbb R)$ such that $(A^-1 - 3 I_2)^t = 2 beginpmatrix -1 & -2 \ 1 & -5 endpmatrix$Skew-symmetric square roots of symmetric matrixProperties of matrices changing with the parity of matrix dimension
$begingroup$
I am a bit curious about an exercise.
I was supposed to prove that there is no matrix that has $1,2,3,4$ in each line and is symmetric. I did, by examination of all such matrices.
Now, there is a such a matrix for $1,2,3$, namely
$$left(beginarrayccc
1&3&2\
3&2&1\
2&1&3\
endarrayright)$$
How does the result generalize? Is it a odd vs even thing? Or $n gt 4$ is always impossible?
Proofs using linear algebra are especially welcome!
linear-algebra combinatorics symmetric-matrices
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
$endgroup$
add a comment |
$begingroup$
I am a bit curious about an exercise.
I was supposed to prove that there is no matrix that has $1,2,3,4$ in each line and is symmetric. I did, by examination of all such matrices.
Now, there is a such a matrix for $1,2,3$, namely
$$left(beginarrayccc
1&3&2\
3&2&1\
2&1&3\
endarrayright)$$
How does the result generalize? Is it a odd vs even thing? Or $n gt 4$ is always impossible?
Proofs using linear algebra are especially welcome!
linear-algebra combinatorics symmetric-matrices
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
$endgroup$
$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago
add a comment |
$begingroup$
I am a bit curious about an exercise.
I was supposed to prove that there is no matrix that has $1,2,3,4$ in each line and is symmetric. I did, by examination of all such matrices.
Now, there is a such a matrix for $1,2,3$, namely
$$left(beginarrayccc
1&3&2\
3&2&1\
2&1&3\
endarrayright)$$
How does the result generalize? Is it a odd vs even thing? Or $n gt 4$ is always impossible?
Proofs using linear algebra are especially welcome!
linear-algebra combinatorics symmetric-matrices
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
$endgroup$
I am a bit curious about an exercise.
I was supposed to prove that there is no matrix that has $1,2,3,4$ in each line and is symmetric. I did, by examination of all such matrices.
Now, there is a such a matrix for $1,2,3$, namely
$$left(beginarrayccc
1&3&2\
3&2&1\
2&1&3\
endarrayright)$$
How does the result generalize? Is it a odd vs even thing? Or $n gt 4$ is always impossible?
Proofs using linear algebra are especially welcome!
linear-algebra combinatorics symmetric-matrices
linear-algebra combinatorics symmetric-matrices
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
edited 19 mins ago
Alexander Gruber♦
20.2k26104175
20.2k26104175
asked 3 hours ago
josinalvojosinalvo
1,226810
asked 3 hours ago
josinalvojosinalvo
1,226810
asked 3 hours ago
josinalvojosinalvo
1,226810
1,226810
$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago
add a comment |
$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago
$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago
$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
On the contrary... it is possible for every $n$.
Let $A_i,j = i+jpmodn$
By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
$endgroup$
add a comment |
$begingroup$
There are 2 other ways for building such a matrix, for most dimensions.
1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf
2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).
Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).
If moreover, this group is commutative, it provides an answer to the question.
Let us take an example : the multiplicative group of the finite field $mathbbF_p$ where $p$ is a prime number ; we can build a matrix $(p-1) times (p-1)$ in the following way
$$A_i,j=i*j textmod p$$
(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $overline1,overline2,cdots overlinep-1$ an abelian group structure.
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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active
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votes
$begingroup$
On the contrary... it is possible for every $n$.
Let $A_i,j = i+jpmodn$
By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
$endgroup$
add a comment |
$begingroup$
On the contrary... it is possible for every $n$.
Let $A_i,j = i+jpmodn$
By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
$endgroup$
add a comment |
$begingroup$
On the contrary... it is possible for every $n$.
Let $A_i,j = i+jpmodn$
By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
$endgroup$
On the contrary... it is possible for every $n$.
Let $A_i,j = i+jpmodn$
By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
answered 3 hours ago
JMoravitzJMoravitz
50.4k44092
50.4k44092
add a comment |
add a comment |
$begingroup$
There are 2 other ways for building such a matrix, for most dimensions.
1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf
2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).
Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).
If moreover, this group is commutative, it provides an answer to the question.
Let us take an example : the multiplicative group of the finite field $mathbbF_p$ where $p$ is a prime number ; we can build a matrix $(p-1) times (p-1)$ in the following way
$$A_i,j=i*j textmod p$$
(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $overline1,overline2,cdots overlinep-1$ an abelian group structure.
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
$endgroup$
add a comment |
$begingroup$
There are 2 other ways for building such a matrix, for most dimensions.
1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf
2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).
Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).
If moreover, this group is commutative, it provides an answer to the question.
Let us take an example : the multiplicative group of the finite field $mathbbF_p$ where $p$ is a prime number ; we can build a matrix $(p-1) times (p-1)$ in the following way
$$A_i,j=i*j textmod p$$
(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $overline1,overline2,cdots overlinep-1$ an abelian group structure.
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
$endgroup$
add a comment |
$begingroup$
There are 2 other ways for building such a matrix, for most dimensions.
1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf
2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).
Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).
If moreover, this group is commutative, it provides an answer to the question.
Let us take an example : the multiplicative group of the finite field $mathbbF_p$ where $p$ is a prime number ; we can build a matrix $(p-1) times (p-1)$ in the following way
$$A_i,j=i*j textmod p$$
(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $overline1,overline2,cdots overlinep-1$ an abelian group structure.
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
$endgroup$
There are 2 other ways for building such a matrix, for most dimensions.
1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf
2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).
Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).
If moreover, this group is commutative, it provides an answer to the question.
Let us take an example : the multiplicative group of the finite field $mathbbF_p$ where $p$ is a prime number ; we can build a matrix $(p-1) times (p-1)$ in the following way
$$A_i,j=i*j textmod p$$
(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $overline1,overline2,cdots overlinep-1$ an abelian group structure.
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
answered 2 hours ago
Jean MarieJean Marie
32.9k42357
32.9k42357
add a comment |
add a comment |
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$begingroup$
It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row.
$endgroup$
– Morgan Rodgers
3 hours ago