# Please explain this notation equation

I am confused by this equation as I rarely use math in my job but need this for a program that I am working on. What exactly does the full expression mean? Note that $m^*_i{_j}$ refers to a matrix whose values have already been obtained.

Define the transition matrix $M =${$m_i{_j}$} as follows: for $i\not=j$ set $m_i{_j}$ to $m^*_i{_j}/|U|$ and let $m_i{_i} = 1-\Sigma_{j\not=i} m_i{_j}$

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It might be helpful to know what $\lvert U\rvert$ is too. The equation is telling you how to get the off-diagonal entries of the transition matrix first (from the $m^*_{ij}$ and $\lvert U\rvert$), and then the diagonal entries of the transition matrix from the off-diagonal entries in the same row. –  Andrew Jul 11 '12 at 17:00
$|U|$ is the length of a set of elements that changes depending on user input. In my test program it is 24. Also, could you please simplify what you mean by diagonal and off diagonal entries? As I say, Math really is not my strong point at all. –  adohertyd Jul 11 '12 at 17:02
So, starting with the matrix $(m^*_{ij}),$ away from the diagonal, all entries are divided by $24$ and the diagonal entries set to $0.$ Once this is done, any diagonal entry is given by adding up the entries in that row, then subtracting the sum from $1.$ –  Andrew Jul 11 '12 at 17:06
@Andrew Would a diagonal entry refer to m[0,0], m[1,1], m[2,2] etc where [x,y] is a co-ordinate of the matrix?? –  adohertyd Jul 11 '12 at 17:08
Yes, that is exactly what is usually meant by diagonal entry for a matrix. –  Andrew Jul 11 '12 at 17:10

To obtain the transition matrix $M$ from the matrix $M^*=(m^*_{ij})$, the rule gives us two steps. First, for all off-diagonal terms $m^*_{ij}$ where $i\neq j$ we simply divide the existing entry by $\lvert U\rvert$ (in this case $\lvert U\rvert =24$), and we temporarily replace the diagonal entries $m^*_{ii}$ by $0.$ Second, to get the $i^{\rm th}$ diagonal entry $m_{ii}$ of $M$ we sum up all entries in the $i^{\rm th}$ row of this intermediate matrix and subtract the resulting sum from $1,$ giving $1-\sum_{j\neq i}m_{ij}.$

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