What's the best way to solve a linear system? Backslash is simple, but the factorization it computes internally can't be reused to solve multiple systems (x=A\b then y=A\c). You might be tempted to use inv(A), as S=inv(A) ; x=S*b ; y=S*c, but's that's slow and inaccurate. The best solution is to use a matrix factorization (LU, CHOL, or QR) followed by forward/backsolve, but figuring out the correct method to use is not easy.
In textbooks, linear algebraic expressions often use the inverse. For example, the Schur complement is written as S=A-B*inv(D)*C. It is never meant to be used that way. S=A-B*(D\C) or S=A-(B/D)*C are better, but the syntax doesn't match the formulas in your textbook.
The solution is to use the FACTORIZE object. It overloads the backslash and mtimes operators (among others) so that solving two linear systems A*x=b and A*y=c can be done with this:
An INVERSE method is provided which does not actually compute the inverse itself (unless it is explicitly requested). The statements above can be done equivalently with this:
That looks like a use of INV(A), but what happens is that S is a factorized representation of the inverse of A, and multiplying S times another matrix is implemented with a forward/backsolve.
Wiith the INVERSE function, your Schur complement computation becomes S = A - B*inverse(D)*C which looks just like the equation in your book ... except that it computes a factorization and then does a forward/backsolve, WITHOUT computing the inverse at all.
An extensive demo is included, as well as a simple and easy-to-read version (FACTORIZE1) with fewer features, as an example of how to create objects in MATLAB.
If the system is over-determined, QR factorization is used to solve the least-squares problem. If the system is under-determined, the transpose of A is factorized via QR, and the minimum 2-norm solution is found. In MATLAB, x=A\b finds a "basic" solution to an under-determined system.
You can compute a subset of the entries of the inverse just by referencing them. For example, S=inverse(A), S(n,n) computes the (n,n) entry of the inverse, without computing all of them.
In the RARE case that you need the entire inverse (or pseduo-inverse of a full-rank rectangular matrix) you can force the object to become "double", with S = double (inverse (A)). This works for both full and sparse matrices (pinv(A) only works for full matrices).
And remember ... don't ever multiply a matrix by inv(A).
This code is published as "Algorithm 930: FACTORIZE: an object-oriented linear system solver for MATLAB", T. A. Davis, ACM Transactions on Mathematical Software, Vol 39, Issue 4, pp. 28:1 - 28:18, 2013.
Tim Davis (2021). Don't let that INV go past your eyes; to solve that system, FACTORIZE! (https://www.mathworks.com/matlabcentral/fileexchange/24119-don-t-let-that-inv-go-past-your-eyes-to-solve-that-system-factorize), MATLAB Central File Exchange. Retrieved .
I am having an issue with running factorize that maybe others can shed some light on. I have added the Factorize path. However, when I attempt to use factorize to solve a system, I get the following error:
Undefined function or variable 'spqr'.
Error in factorize (line 143)
throw (me) ;
Does anyone know how I can solve this issue? Thankyou in advance!
inv() is certainly slower than \ unless you have multiple right hand side vectors to solve for. But inv() is NOT inaccurate. The link elaborates further : http://arxiv.org/abs/1201.6035
>Several widely-used textbooks lead the >reader to believe that solving a >linear system of equations Ax = b by >multiplying the vector b by a computed >inverse inv(A) is inaccurate. >Virtually all other textbooks on >numerical analysis and numerical >linear algebra advise against using >computed inverses without stating >whether this is accurate or not. In >fact, under reasonable assumptions on >how the inverse is computed, x = >inv(A)*b is as accurate as the >solution computed by the best
Works as documented in the demo: like a charm!
One question: can this package also be used for eigendecomposition? For example:
A = rand(5);
[S,D] = eig(A);
But this still uses EIG, does this package contain a direct way to compute the eigendecomposition?
Comment from the author: I have addressed Ben's comment about uminus in this update (Sept 2011). You can now do -inverse(A)*b, or -2*inverse(A)*b, which you could not do in the previous version.
-inverse(A)*b can also be done as -(inverse(A)*b), which is more natural than inverse(A)*(-b).
Comment from the author: implementing the uminus function would be rather painful. I would need to return an object that is unchanged from the previous one, except with a flag stating that the result of all other functions must negate their result. This would be rather ugly and would needlessly complicate the code, just to add a rather minor feature.
Thanks for the suggestion ... I agree that uminus "ought" to work. It seems simple from the perspective of an end-user. But implementing it is far more trouble than it's worth, in my opinion.
Very good, it is even faster when I solve a single linear system.
What mathworks are waiting to incorporate this package into Matlab?
All in all this package is a great idea and it looks like a first class implementation. As such, I'd like to help improve it by pointing out an issue. I am trying to solve Ax=-b, so I type in x=-inverse(A)*b, but I get the following error:
??? Undefined function or method 'uminus' for input arguments of type 'factorization_dense_lu'.
The work around is simple: x=inverse(A)*(-b). However, these are mathematically equivalent, so both should work.
Nice with an update to modern MATLAB OOP.
This is an excellent piece of software. Makes great use of Matlab's OOP.
Absolutely splendid! Many thanks to Tim for writing this. A great title too.
a truly professional and INValuable package written with a twinkle in the (author's) eye and a must look-at for people dealing with linear systems...
one pedestrian thought: why not create a stand-alone object given ML's new and powerful OOP-strength...
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