I once made the claim that det should be modified. With EVERY use of the function, a dialog box should pop up, asking "Do you REALLY want or need to use the determinant here?" The dialog box would have one button only on it: "NO".
Or maybe the function should be completely disabled from ever running, UNLESS the user has passed a test certifying they know and understand enough about numerical analysis and numerical methods on a computer. Of course, the person who would pass the test as I would write it would also know enough to avoid using det in the first place.
Ok, so both of those alternatives might be a tad extreme. :)
As students, we were taught to use the determinant by our teachers. I was! They learned about it from their teachers, etc. And nobody ever told them not to use it, so why should they do any different? Worse, your textbook tells you that a singular matrix will have a determinant of zero. Of course it was written by the same people who tell you to use that tool. We end up with this entire cadre of people writing books, teaching classes, writing papers, creating new people who will do the same. All of those people not understanding that a determinant is almost always the wrong thing to compute.
The first issue is people try to compute the determinant the wrong way, using a recursive method. This is amazingly inefficient, so that even determinants of 10x10 matrices will take a huge amount of time to compute. In fact, there are schemes using an LU decomposition that do allow at least efficient computation of the determinant, or you could use the product of eigenvalues. The MATLAB det function uses the LU scheme, so it is at least well written as you should expect.
Next, det is quite sensitive to scaling. For example,
A = eye(1000);
Clearly the determinant of A is 1, and it is non-singular.
A = eye(1000);
det(A)
ans =
1
B = A*.5;
det(B)
ans =
9.33263618503219e-302
B = A*.4;
det(B)
ans =
0
So simply scaling A by a constant is sufficient to make the determinant underflow to zero, despite the fact that all of these matrices were non-singular and VERY well behaved. Had I wanted to see an overflow to inf, that would have been as simple to do.
B = A*2.5;
det(B)
ans =
Inf
The point is, det is a poor measure of singularity. For any tolerance you will provide, I can trivially provide a clearly non-singular matrix that will cause det to conclude wrongly about the singularity status of that matrix. For example...
A = rand(6,5);
A = [A,A*rand(5,1)];
det(A)
ans =
4.66661953785783e-17
B = A*1000;
det(B)
ans =
19.1144736270657
The matrix A has rank 5. That should be abundantly clear from how it was constructed. det(A) suggests that A might be singular, although it was not zero. Numerical errors in the floating point arithmetic will cause it to be very rarely exactly zero, even when it might have been so occasionally.
But simply by scaling A by 1000, we might conclude that B is not singular. Or is it?
B = A*900;
det(B)
ans =
0
Simply by arbitrary chance, I chose a different value to scale by, and see how unstable the determinant can be!
As you can see, det is just bad to the bone. Avoid it like you would avoid the plague.
And there are good alternatives to det as a test for singularity. The simplest one is rank. It uses the singular values of your matrix to estimate the numerical rank of the matrix. If rank(A) is less than the smaller of the number of rows or columns of A, then it is singular.
rank(A)
ans =
5
rank(B)
ans =
5
Some prefer a different measure, like cond. If cond is large, on the order of 1/eps, then the matrix is numerically singular.
cond(A)
ans =
2.05284970911659e+16
cond(B)
ans =
2.80663046168285e+16
As you can see, cond rather stably predicts both matrices are singular, although numerical trash changed the exact value slightly.
You might also use the singular values of the matrix itself. Of course then you need to understand what they tell you. More information provided, more understanding needed.
