Property of a map












1












$begingroup$


Let $F_n$ be the set of real integrable function on $left[1,nright]$ and $M_nleft(mathbb{R}right)$ the set of square matrix of range $n$.



For fun, I've defined the application $xi : F_n rightarrow M_nleft(mathbb{R}right)$ that makes $f$ correspond to a matrix $M=left(m_{i,j}right)$ which coefficients are
$$
m_{i,j}=int_{i}^{j}fleft(tright)text{d}t
$$




Hence we have the following properties : $xi$ is linear and
$$
m_{i,i}=0, m_{i,j}=-m_{j,i} text{ and }m_{i,j}+m_{j,k}=m_{i,k}
$$



I would like to know if we could find the image of $xi$ or find its dimension knowing that it is included in the set of antisymmetric matrix.
I think that because $m_{i,j}+m_{j,k}=m_{i,k}$ we cannot say that $xi$ is surjective ( that would lead to $text{Im}left(xiright)=A_nleft(mathbb{R}right)$ ) which would be surprising.



I've also shown that $x mapsto sinleft(2pi xright) in text{Ker}left(xiright)$ so it is not injective.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
    $endgroup$
    – Saucy O'Path
    Jan 1 at 22:37












  • $begingroup$
    Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:06










  • $begingroup$
    It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:28
















1












$begingroup$


Let $F_n$ be the set of real integrable function on $left[1,nright]$ and $M_nleft(mathbb{R}right)$ the set of square matrix of range $n$.



For fun, I've defined the application $xi : F_n rightarrow M_nleft(mathbb{R}right)$ that makes $f$ correspond to a matrix $M=left(m_{i,j}right)$ which coefficients are
$$
m_{i,j}=int_{i}^{j}fleft(tright)text{d}t
$$




Hence we have the following properties : $xi$ is linear and
$$
m_{i,i}=0, m_{i,j}=-m_{j,i} text{ and }m_{i,j}+m_{j,k}=m_{i,k}
$$



I would like to know if we could find the image of $xi$ or find its dimension knowing that it is included in the set of antisymmetric matrix.
I think that because $m_{i,j}+m_{j,k}=m_{i,k}$ we cannot say that $xi$ is surjective ( that would lead to $text{Im}left(xiright)=A_nleft(mathbb{R}right)$ ) which would be surprising.



I've also shown that $x mapsto sinleft(2pi xright) in text{Ker}left(xiright)$ so it is not injective.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
    $endgroup$
    – Saucy O'Path
    Jan 1 at 22:37












  • $begingroup$
    Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:06










  • $begingroup$
    It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:28














1












1








1





$begingroup$


Let $F_n$ be the set of real integrable function on $left[1,nright]$ and $M_nleft(mathbb{R}right)$ the set of square matrix of range $n$.



For fun, I've defined the application $xi : F_n rightarrow M_nleft(mathbb{R}right)$ that makes $f$ correspond to a matrix $M=left(m_{i,j}right)$ which coefficients are
$$
m_{i,j}=int_{i}^{j}fleft(tright)text{d}t
$$




Hence we have the following properties : $xi$ is linear and
$$
m_{i,i}=0, m_{i,j}=-m_{j,i} text{ and }m_{i,j}+m_{j,k}=m_{i,k}
$$



I would like to know if we could find the image of $xi$ or find its dimension knowing that it is included in the set of antisymmetric matrix.
I think that because $m_{i,j}+m_{j,k}=m_{i,k}$ we cannot say that $xi$ is surjective ( that would lead to $text{Im}left(xiright)=A_nleft(mathbb{R}right)$ ) which would be surprising.



I've also shown that $x mapsto sinleft(2pi xright) in text{Ker}left(xiright)$ so it is not injective.






linear-algebra matrices







share|cite|improve this question











$endgroup$




Let $F_n$ be the set of real integrable function on $left[1,nright]$ and $M_nleft(mathbb{R}right)$ the set of square matrix of range $n$.



For fun, I've defined the application $xi : F_n rightarrow M_nleft(mathbb{R}right)$ that makes $f$ correspond to a matrix $M=left(m_{i,j}right)$ which coefficients are
$$
m_{i,j}=int_{i}^{j}fleft(tright)text{d}t
$$




Hence we have the following properties : $xi$ is linear and
$$
m_{i,i}=0, m_{i,j}=-m_{j,i} text{ and }m_{i,j}+m_{j,k}=m_{i,k}
$$



I would like to know if we could find the image of $xi$ or find its dimension knowing that it is included in the set of antisymmetric matrix.
I think that because $m_{i,j}+m_{j,k}=m_{i,k}$ we cannot say that $xi$ is surjective ( that would lead to $text{Im}left(xiright)=A_nleft(mathbb{R}right)$ ) which would be surprising.



I've also shown that $x mapsto sinleft(2pi xright) in text{Ker}left(xiright)$ so it is not injective.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 23:50







Atmos

















asked Jan 1 at 22:27









AtmosAtmos

4,732419




4,732419








  • 1




    $begingroup$
    The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
    $endgroup$
    – Saucy O'Path
    Jan 1 at 22:37












  • $begingroup$
    Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:06










  • $begingroup$
    It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:28














  • 1




    $begingroup$
    The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
    $endgroup$
    – Saucy O'Path
    Jan 1 at 22:37












  • $begingroup$
    Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:06










  • $begingroup$
    It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
    $endgroup$
    – Omnomnomnom
    Jan 1 at 23:28








1




1




$begingroup$
The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
$endgroup$
– Saucy O'Path
Jan 1 at 22:37






$begingroup$
The map cannot be injective because it's linear and the domain has dimension larger than the codomain. The first two properties are redundant, because they are included in $m_{i,j}+m_{j,k}=m_{i,k}$. I would guess that the map is surjective onto the vector space of matrices that satisfy that property, and that it should not be all that challenging to prove. I'd look for a basis, and then look for a continuous function that realizes each of its elements.
$endgroup$
– Saucy O'Path
Jan 1 at 22:37














$begingroup$
Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
$endgroup$
– Omnomnomnom
Jan 1 at 23:06




$begingroup$
Possibly a helpful perspective: relative to the Frobenius inner product, we could say that $M$ lies in the orthogonal complement of ${E_{ij} + E_{jk} - E_{ik} mid 1 leq i,j,k leq n}cup {E_{ij} + E_{ji} mid 1 leq i,j leq n}$.
$endgroup$
– Omnomnomnom
Jan 1 at 23:06












$begingroup$
It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
$endgroup$
– Omnomnomnom
Jan 1 at 23:28




$begingroup$
It is notable that we don't use any information about the function between $0$ and $1$; perhaps $F_n$ ought to be the set of integrable functions on $[1,n]$.
$endgroup$
– Omnomnomnom
Jan 1 at 23:28










1 Answer
1






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oldest

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1












$begingroup$

Let $S$ denote the subspace of matrices satisfying $m_{ij} = -m_{ji}$ and $m_{ij} + m_{jk} = m_{ik}$.



We first note that an element of $S$ is completely specified once we know the values $m_{i,i+1}$ for $i = 1,dots,n-1$. Moreover, we can construct a basis for $S$ by taking $M_i$ to be the matrix for which $m_{i,i+1} = 1$ and $m_{j,j+1} = 0$ for all $j neq i$.



For instance, we would have
$$
M_1 = pmatrix{0&-1&cdots&-1&-1\1&0\ vdots &0&ddots\&vdots&ddots&0\1&0&cdots &0&0}
$$
Thus, $S$ has dimension $n-1$.



Of course, $S$ contains the image of $xi$. To see that the image of $xi$ is all of $S$, it suffices to consider the functions
$$
f_i(x) = begin{cases}
1 & ileq x leq i+1\
0 & text{otherwise}
end{cases}.
$$
We can verify that $xi(f_i) = M_i$. Since the image of $xi$ contains all $M_i$, it contains their span, which is all of $S$.






share|cite|improve this answer









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    active

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    1












    $begingroup$

    Let $S$ denote the subspace of matrices satisfying $m_{ij} = -m_{ji}$ and $m_{ij} + m_{jk} = m_{ik}$.



    We first note that an element of $S$ is completely specified once we know the values $m_{i,i+1}$ for $i = 1,dots,n-1$. Moreover, we can construct a basis for $S$ by taking $M_i$ to be the matrix for which $m_{i,i+1} = 1$ and $m_{j,j+1} = 0$ for all $j neq i$.



    For instance, we would have
    $$
    M_1 = pmatrix{0&-1&cdots&-1&-1\1&0\ vdots &0&ddots\&vdots&ddots&0\1&0&cdots &0&0}
    $$
    Thus, $S$ has dimension $n-1$.



    Of course, $S$ contains the image of $xi$. To see that the image of $xi$ is all of $S$, it suffices to consider the functions
    $$
    f_i(x) = begin{cases}
    1 & ileq x leq i+1\
    0 & text{otherwise}
    end{cases}.
    $$
    We can verify that $xi(f_i) = M_i$. Since the image of $xi$ contains all $M_i$, it contains their span, which is all of $S$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $S$ denote the subspace of matrices satisfying $m_{ij} = -m_{ji}$ and $m_{ij} + m_{jk} = m_{ik}$.



      We first note that an element of $S$ is completely specified once we know the values $m_{i,i+1}$ for $i = 1,dots,n-1$. Moreover, we can construct a basis for $S$ by taking $M_i$ to be the matrix for which $m_{i,i+1} = 1$ and $m_{j,j+1} = 0$ for all $j neq i$.



      For instance, we would have
      $$
      M_1 = pmatrix{0&-1&cdots&-1&-1\1&0\ vdots &0&ddots\&vdots&ddots&0\1&0&cdots &0&0}
      $$
      Thus, $S$ has dimension $n-1$.



      Of course, $S$ contains the image of $xi$. To see that the image of $xi$ is all of $S$, it suffices to consider the functions
      $$
      f_i(x) = begin{cases}
      1 & ileq x leq i+1\
      0 & text{otherwise}
      end{cases}.
      $$
      We can verify that $xi(f_i) = M_i$. Since the image of $xi$ contains all $M_i$, it contains their span, which is all of $S$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $S$ denote the subspace of matrices satisfying $m_{ij} = -m_{ji}$ and $m_{ij} + m_{jk} = m_{ik}$.



        We first note that an element of $S$ is completely specified once we know the values $m_{i,i+1}$ for $i = 1,dots,n-1$. Moreover, we can construct a basis for $S$ by taking $M_i$ to be the matrix for which $m_{i,i+1} = 1$ and $m_{j,j+1} = 0$ for all $j neq i$.



        For instance, we would have
        $$
        M_1 = pmatrix{0&-1&cdots&-1&-1\1&0\ vdots &0&ddots\&vdots&ddots&0\1&0&cdots &0&0}
        $$
        Thus, $S$ has dimension $n-1$.



        Of course, $S$ contains the image of $xi$. To see that the image of $xi$ is all of $S$, it suffices to consider the functions
        $$
        f_i(x) = begin{cases}
        1 & ileq x leq i+1\
        0 & text{otherwise}
        end{cases}.
        $$
        We can verify that $xi(f_i) = M_i$. Since the image of $xi$ contains all $M_i$, it contains their span, which is all of $S$.






        share|cite|improve this answer









        $endgroup$



        Let $S$ denote the subspace of matrices satisfying $m_{ij} = -m_{ji}$ and $m_{ij} + m_{jk} = m_{ik}$.



        We first note that an element of $S$ is completely specified once we know the values $m_{i,i+1}$ for $i = 1,dots,n-1$. Moreover, we can construct a basis for $S$ by taking $M_i$ to be the matrix for which $m_{i,i+1} = 1$ and $m_{j,j+1} = 0$ for all $j neq i$.



        For instance, we would have
        $$
        M_1 = pmatrix{0&-1&cdots&-1&-1\1&0\ vdots &0&ddots\&vdots&ddots&0\1&0&cdots &0&0}
        $$
        Thus, $S$ has dimension $n-1$.



        Of course, $S$ contains the image of $xi$. To see that the image of $xi$ is all of $S$, it suffices to consider the functions
        $$
        f_i(x) = begin{cases}
        1 & ileq x leq i+1\
        0 & text{otherwise}
        end{cases}.
        $$
        We can verify that $xi(f_i) = M_i$. Since the image of $xi$ contains all $M_i$, it contains their span, which is all of $S$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 23:27









        OmnomnomnomOmnomnomnom

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        127k790178






























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