questions about solving a linear equations ( \ vs pinv vs least-square)

I wonder why the answer with using \ is different to solutions with pinv or least-square.

     X = [13.5, 6.75, 6.75;6.75,6.75,0;6.75,0,6.75];
     y = [0.62;0.31;0.31];

     X\y         % [-0.0791, 0.125, 0.125]

     pinv(X) * y % [0.0306, 0.0153, 0.0153]

     pinv(X * X') * X' * y % [0.0306, 0.0153, 0.0153]

I use MatLab R2016a in Windows 7 SP1 64bit.

Answer 1

[Massimo Zanetti]

Your matrix X has full rank, so the difference between the two methods shouldn't be so high. Running it on my pc I indeed obtain:

X = [13.5, 6.75, 6.75;6.75,6.75,0;6.75,0,0]
y =  [0.62;0.31;0.31];

format long
a1=X\y
a2=pinv(X)*y

a1 =

     0.045925925925926
                     0
                     0

a2 =

     0.045925925925926
     0.000000000000000
    -0.000000000000000

They are the same. Differences among the two methods show up when the matrix X has not full rank. Then, if y does not belong to R(X), the range space of X, then pinv method returns the orthogonal projection of y onto R(X). While other methods triggered by \ operator return different approximations, depending on which one is triggered. The collection of methods that \ can trigger (depending on properties of matrix X) is listed here (bottom of the page):

https://it.mathworks.com/help/matlab/ref/mldivide.html

[Chen ChingChuan]

Sorry, I have a typo. The last element of X is 6.75 not 0.

[Massimo Zanetti]

Then your matrix

X = [13.5, 6.75, 6.75;6.75,6.75,0;6.75,0,6.75]

is exactly rank(X)=2, so rank is NOT FULL (any row is a linear combination of the other two). Fore instance, row1=row2+row3.

As a consequence the solutions of the two methods are different, because of what I sayd before.