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fix inv warning in matlab

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I use this code:
Matlab gives warning. never use inv to solve linear system
How can I fix it?

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Walter Roberson
Walter Roberson 28 Jul 2019
Multiply both sides on the left by A'*A :
(A'*A) * b = (A'*A) * inv(A'*A) * A' * y
and for any invertable matrix, X * inv(X) is the identity matrix so (A'*A) * inv(A'*A) cancels out to identity, so
(A'*A) * b = A' * y
Multiply both sides on the left by inv(A'):
inv(A') * A' * A * b = inv(A') * A' * y
and inv(A') * A' cancels to the identity on both sides, so
A * b = y
Multiply by inv(A) on the left on both sides:
inv(A) * A * b = inv(A) * y
inverse cancels to identity, so
b = inv(A) * y
Now use the \ operator:
b = A \ y;
The above mathematics might not strictly apply if A is not a square matrix.

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Walter Roberson
Walter Roberson 2 Aug 2019
Yes, but you should be questioning why you are doing that.
Provided that HH and WWInv are square matrices, then
inv(HH' * WWInv * HH)
inv(HH)*inv(WWInv) * inv(HH')
You then right-multiply by HH' so
inv(HH)*inv(WWInv) * inv(HH') * HH'
and you group the last two together and the inv(HH') will cancel the HH, leaving
The name suggests that inv(WWInv) would probably be an existing variable named WW, so you could probably instead be using
RR = WW - HH \ WW
Naime 2 Aug 2019
Thank you. But
RR = WW - HH * inv(HH' * WWInv * HH) * HH';
RR = WW - HH *inv(HH)*inv(WWInv)
RR = 0
Walter Roberson
Walter Roberson 2 Aug 2019
Yes, I missed the HH pre-multiplier, but Yes, that logic appears correct. If you try it with actual random matrices, you will find that RR is 0 to within round-off error (e.g., for rand(15,15) all of the entries come out with absolute value less than 1E-13

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