applyting function to all elements in a matrix
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What is1./(1+exp(-z)), (z can be a matrix ) and how is it different from 1/(1+exp(-z))?
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Vladimir Sovkov
2019-12-30
The second version is a little bit strange, although without formal errors. The thing is that exp(-z) is per-element, i.e., it computes the matrix, whose every element is the exponent of the corresponding element of the initial matrix, while the operation / computes the matrix inversion. If you want all the operations to be done in the sense of the matrix algebra, you should use expm(-z). The first version is totally per-element, hence, it looks quite consistent.
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KALYAN ACHARJYA
2019-12-30
First one: 1./(1+exp(-z)), have you noticed {dot} before division sign, it represents the dot array operation (element wise), lets say matrix A & B,
>> a=magic(3);
>> b=rand(3,3);
>> a*b % Matrix operation
ans =
8.1855 8.5246 8.5199
7.8620 6.5847 10.2725
11.6650 9.5398 7.9509
>> a.*b % Array element wise operation
ans =
6.5178 0.9134 1.6710
2.7174 3.1618 3.8282
0.5079 0.8779 1.9150
So the expression 1./(1+exp(-z)), there is no issue considering z may have any matrix.
>> z=rand(4,4) % random z considered
z =
0.9649 0.4854 0.9157 0.0357
0.1576 0.8003 0.7922 0.8491
0.9706 0.1419 0.9595 0.9340
0.9572 0.4218 0.6557 0.6787
>> result=1./(1+exp(-z))
result =
0.7241 0.6190 0.7142 0.5089
0.5393 0.6900 0.6883 0.7004
0.7252 0.5354 0.7230 0.7179
0.7226 0.6039 0.6583 0.6635
Second one: 1/(1+exp(-z))
The denominator 1+exp(-z) have no issue, considering z as a arbitary matrix, but when a scalar divided by matrix (vector), I don't think is there any direct approach. Read this thread, about "What's the meaning of a scalar divided by a vector" because 1 as numerator is scalar on the other hand denominator 1+exp(-z) is a vector.
In Matlab it shows the error related to dimention mismatch
>> result=1/(1+exp(-z))
Error using /
Matrix dimensions must agree.
Hope this helps to clear your doubt.
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Vladimir Sovkov
2019-12-30
Indeed, 1/... does not work as a matrix operation. To get it in the matrix sense, it must be either (1+expm(-z))^(-1) or inv(1+expm(-z)), and these expressions only make sense for square well-conditioned matrices. For some related purposes, the pseudoinversion pinv or the left division \ can be used.
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