text: user8601: In Horvath's Weighted network analysis, the author mentions that M^b,  where b is a real number and M is a matrix, denotes the element-wise power. This seems odd to me. The way I was taught, M^b would mean multiplying the matrix M, b times. Should element-wise power not be M^(ob) where where o denotes the Hadamard power? 
  
  user7441: You CAN calculate M^b for b being float number, complex number or even another matrix. 
  As well as you can calculate exp(M), log(M), sin(M) whatever... That is very standard for math people. 
  Let me give two ways how you can think of it.
  1) M^b = exp(b log(M)), so you need to define what is exp and log for matrices.
  That can be done using power series: exp(M) = 1 + M + M^2/2! + M^3/3! + ... 
  log(1+M) = M - M^2/2 + M^3/3 - ... 
  2) More general way to calculate any function of matrix f(M) , looks like that:
  if your matrix is diagonal, then it is true that: f(diag(d_i)) = diag( f(d_i)  ) . 
  So the question how to reduce any matrix to diagonal one.
  That can be done for almost all matrices by conjugation M = C D C^{-1}.
  Then f(M) = C f(D) C^{-1}. 
  
original_text: 8601: <p>In Horvath's <em>Weighted network analysis</em>, the author mentions that M^b,  where b is a real number and M is a matrix, denotes the element-wise power. This seems odd to me. The way I was taught, M^b would mean multiplying the matrix M, b times. Should element-wise power not be M^(ob) where where o denotes the Hadamard power? </p>
   7441: <p>You CAN calculate M^b for b being float number, complex number or even another matrix. 
  As well as you can calculate exp(M), log(M), sin(M) whatever... That is very standard for math people. </p>
  
  <p>Let me give two ways how you can think of it.</p>
  
  <p>1) M^b = exp(b log(M)), so you need to define what is exp and log for matrices.
  That can be done using power series: exp(M) = 1 + M + M^2/2! + M^3/3! + ... 
  log(1+M) = M - M^2/2 + M^3/3 - ... </p>
  
  <p>2) More general way to calculate any function of matrix f(M) , looks like that:
  if your matrix is diagonal, then it is true that: f(diag(d_i)) = diag( f(d_i)  ) . 
  So the question how to reduce any matrix to diagonal one.
  That <a href="https://en.wikipedia.org/wiki/Jordan_normal_form" rel="nofollow noreferrer">can be done for almost all</a> matrices by conjugation M = C D C^{-1}.
  Then f(M) = C f(D) C^{-1}. </p>
  