fix inv warning in matlab

I use this code:

b=inv(A'*A)*A'*y; 

Matlab gives warning. never use inv to solve linear system

How can I fix it?

Answer 1

[Walter Roberson]

b=inv(A'*A)*A'*y; 

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.

[NA]

It means that instead of using

b=inv(A'*A)*A'*y;

I should use

b=A\y

[Walter Roberson]

Yup.

[NA]

so if I have this

di=sum((r*inv(sig)).*r,2);

should be change to

di=sum((r/(sig)).*r,2);

Is it correct?

[Walter Roberson]

That looks strange to me, seeing as it has both a matrix multiply by r ( * operator) and an element-by-element multiplication by r ( .* operator) . But if the * and .* are both correct, then yes, r/sig .* r

[NA]

I have this

RR = WW - HH * inv(HH' * WWInv * HH) * HH';

Can I change it to this one?

RR = WW - HH * (HH' * WWInv * HH)\ HH';

[Walter Roberson]

Yes, but you should be questioning why you are doing that.

Provided that HH and WWInv are square matrices, then

inv(HH' * WWInv * HH)

equals

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

inv(HH)*inv(WWInv)

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

[NA]

Thank you. But

RR = WW - HH * inv(HH' * WWInv * HH) * HH';
RR = WW - HH *inv(HH)*inv(WWInv)
RR = 0

[Walter Roberson]

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