Matrix, Dimensions.
n^[1] by 1, it ends up being n^[1] by m, if m is decisive training set. The dimensions of w^[1] stays the same so is the n^[1] by n^[0] and x instead of being n^[0] by 1 is now all your training examples stamped horizontally, so it’s now n^[0] by m. You notice that when you take a, n^[1] by n^[0]