How to prove that the number of diagonals of a $M times N$ matrix is $M+N-1$?

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Given a matrix $M times N$, how to prove that the number of diagonals that can be drawn, like in the figure below, is equal to $M + N - 1$?



enter image description here



My idea would be to prove it with induction and consider the 3 possible cases: $M=N$, $M>N$ and $M<N$. But I don't know how to do it.




matrices




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  • Induction on which variable?
    – greedoid
    Jul 29 at 10:39











  • @Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
    – reuseman
    Jul 29 at 10:43










  • This is easy with induction. Fix $M$ and induct on $N$.
    – greedoid
    Jul 29 at 10:45










  • So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
    – reuseman
    Jul 29 at 10:46










  • No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
    – greedoid
    Jul 29 at 10:47















up vote
0
down vote

favorite












Given a matrix $M times N$, how to prove that the number of diagonals that can be drawn, like in the figure below, is equal to $M + N - 1$?



enter image description here



My idea would be to prove it with induction and consider the 3 possible cases: $M=N$, $M>N$ and $M<N$. But I don't know how to do it.




matrices




share|cite|improve this question



















  • Induction on which variable?
    – greedoid
    Jul 29 at 10:39











  • @Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
    – reuseman
    Jul 29 at 10:43










  • This is easy with induction. Fix $M$ and induct on $N$.
    – greedoid
    Jul 29 at 10:45










  • So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
    – reuseman
    Jul 29 at 10:46










  • No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
    – greedoid
    Jul 29 at 10:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given a matrix $M times N$, how to prove that the number of diagonals that can be drawn, like in the figure below, is equal to $M + N - 1$?



enter image description here



My idea would be to prove it with induction and consider the 3 possible cases: $M=N$, $M>N$ and $M<N$. But I don't know how to do it.




matrices




share|cite|improve this question











Given a matrix $M times N$, how to prove that the number of diagonals that can be drawn, like in the figure below, is equal to $M + N - 1$?



enter image description here



My idea would be to prove it with induction and consider the 3 possible cases: $M=N$, $M>N$ and $M<N$. But I don't know how to do it.





matrices





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 10:27









reuseman

145




145











  • Induction on which variable?
    – greedoid
    Jul 29 at 10:39











  • @Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
    – reuseman
    Jul 29 at 10:43










  • This is easy with induction. Fix $M$ and induct on $N$.
    – greedoid
    Jul 29 at 10:45










  • So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
    – reuseman
    Jul 29 at 10:46










  • No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
    – greedoid
    Jul 29 at 10:47

















  • Induction on which variable?
    – greedoid
    Jul 29 at 10:39











  • @Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
    – reuseman
    Jul 29 at 10:43










  • This is easy with induction. Fix $M$ and induct on $N$.
    – greedoid
    Jul 29 at 10:45










  • So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
    – reuseman
    Jul 29 at 10:46










  • No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
    – greedoid
    Jul 29 at 10:47
















Induction on which variable?
– greedoid
Jul 29 at 10:39





Induction on which variable?
– greedoid
Jul 29 at 10:39













@Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
– reuseman
Jul 29 at 10:43




@Angle For a moment I've thought it was possible to use the Induction principle both on $M$ and $N$, but I was so wrong
– reuseman
Jul 29 at 10:43












This is easy with induction. Fix $M$ and induct on $N$.
– greedoid
Jul 29 at 10:45




This is easy with induction. Fix $M$ and induct on $N$.
– greedoid
Jul 29 at 10:45












So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
– reuseman
Jul 29 at 10:46




So this means that there are 3 cases to consider $M=N$, $M>N$, $M<N$ with $M$ fixed, and other 3 with $N$ fixed?
– reuseman
Jul 29 at 10:46












No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
– greedoid
Jul 29 at 10:47





No, just go from a matrix $Mtimes N$ to a matrix with one more row, that is a matrix $Mtimes N+1$.
– greedoid
Jul 29 at 10:47











2 Answers
2






active

oldest

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up vote
0
down vote



accepted










This is a counting problem.



Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.



Now go down through the last column and you get $M$ of those.



There is one which is counted twice, so the total is $M+N-1$






share|cite|improve this answer





















  • Is this a valid way to prove it for all the possible values of $M$ and $N$?
    – reuseman
    Jul 29 at 10:44










  • Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
    – Mohammad Riazi-Kermani
    Jul 29 at 10:46


















up vote
-1
down vote













Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry. Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$. Since $iin1,2,ldots,M$ and $jin1,2,ldots,N$, we see that
$$k=i+j-1in1,2,ldots,M+N-1,.$$



Each $kin1,2,ldots,M+N-1$ corresponds to a nonempty diagonal line. For $k=1,2,ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$. For $k=M+1,M+2,ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

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    active

    oldest

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    up vote
    0
    down vote



    accepted










    This is a counting problem.



    Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.



    Now go down through the last column and you get $M$ of those.



    There is one which is counted twice, so the total is $M+N-1$






    share|cite|improve this answer





















    • Is this a valid way to prove it for all the possible values of $M$ and $N$?
      – reuseman
      Jul 29 at 10:44










    • Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
      – Mohammad Riazi-Kermani
      Jul 29 at 10:46















    up vote
    0
    down vote



    accepted










    This is a counting problem.



    Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.



    Now go down through the last column and you get $M$ of those.



    There is one which is counted twice, so the total is $M+N-1$






    share|cite|improve this answer





















    • Is this a valid way to prove it for all the possible values of $M$ and $N$?
      – reuseman
      Jul 29 at 10:44










    • Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
      – Mohammad Riazi-Kermani
      Jul 29 at 10:46













    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    This is a counting problem.



    Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.



    Now go down through the last column and you get $M$ of those.



    There is one which is counted twice, so the total is $M+N-1$






    share|cite|improve this answer













    This is a counting problem.



    Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.



    Now go down through the last column and you get $M$ of those.



    There is one which is counted twice, so the total is $M+N-1$







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 29 at 10:36









    Mohammad Riazi-Kermani

    27.3k41851




    27.3k41851











    • Is this a valid way to prove it for all the possible values of $M$ and $N$?
      – reuseman
      Jul 29 at 10:44










    • Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
      – Mohammad Riazi-Kermani
      Jul 29 at 10:46

















    • Is this a valid way to prove it for all the possible values of $M$ and $N$?
      – reuseman
      Jul 29 at 10:44










    • Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
      – Mohammad Riazi-Kermani
      Jul 29 at 10:46
















    Is this a valid way to prove it for all the possible values of $M$ and $N$?
    – reuseman
    Jul 29 at 10:44




    Is this a valid way to prove it for all the possible values of $M$ and $N$?
    – reuseman
    Jul 29 at 10:44












    Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
    – Mohammad Riazi-Kermani
    Jul 29 at 10:46





    Yes, it works for all values of $M $and $N$. Go through some examples and you figure it out.
    – Mohammad Riazi-Kermani
    Jul 29 at 10:46











    up vote
    -1
    down vote













    Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry. Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$. Since $iin1,2,ldots,M$ and $jin1,2,ldots,N$, we see that
    $$k=i+j-1in1,2,ldots,M+N-1,.$$



    Each $kin1,2,ldots,M+N-1$ corresponds to a nonempty diagonal line. For $k=1,2,ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$. For $k=M+1,M+2,ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.






    share|cite|improve this answer

























      up vote
      -1
      down vote













      Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry. Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$. Since $iin1,2,ldots,M$ and $jin1,2,ldots,N$, we see that
      $$k=i+j-1in1,2,ldots,M+N-1,.$$



      Each $kin1,2,ldots,M+N-1$ corresponds to a nonempty diagonal line. For $k=1,2,ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$. For $k=M+1,M+2,ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.






      share|cite|improve this answer























        up vote
        -1
        down vote










        up vote
        -1
        down vote









        Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry. Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$. Since $iin1,2,ldots,M$ and $jin1,2,ldots,N$, we see that
        $$k=i+j-1in1,2,ldots,M+N-1,.$$



        Each $kin1,2,ldots,M+N-1$ corresponds to a nonempty diagonal line. For $k=1,2,ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$. For $k=M+1,M+2,ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.






        share|cite|improve this answer













        Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry. Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$. Since $iin1,2,ldots,M$ and $jin1,2,ldots,N$, we see that
        $$k=i+j-1in1,2,ldots,M+N-1,.$$



        Each $kin1,2,ldots,M+N-1$ corresponds to a nonempty diagonal line. For $k=1,2,ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$. For $k=M+1,M+2,ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 29 at 11:35









        Batominovski

        22.9k22777




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