Timeline for Why were matrix determinants once such a big deal?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Sep 22, 2010 at 12:46 | comment | added | Suvrit | Minor nitpick: Gaussian Elimination (in its textbook implementations) takes $\frac{2}{3}n^3$ operations, not $n^2$. | |
| Aug 23, 2010 at 13:41 | comment | added | Jiahao Chen | @Thierry: I suppose it depends on what you mean as "theoretical" and "application". The use of the Slater determinant is key to deriving the actual Hartree-Fock equations, although in the practical calculations of Hartree-Fock theory the determinant is almost never constructed explicitly. | |
| Aug 19, 2010 at 2:06 | comment | added | Victor Protsak | Daniel, if the matrix is $\textit{known to be}$ lower (or upper) triangular, checking that isn't necessary. That was my point. | |
| Aug 18, 2010 at 23:13 | comment | added | Daniel Litt | @Victor Protsak: Computing the determinant of an upper triangular matrix is $O(n)$, but checking that it's upper triangular is $O(n^2)$. | |
| Aug 18, 2010 at 22:21 | comment | added | J. M. isn't a mathematician | ...and since L is unit lower triangular (has all 1's on its diagonal) its determinant is 1, and thus the determinant of A is computed by multiplying together U's diagonal elements (or summing logarithms of the diagonal elements as the case may be). Now, all that neglected to take pivoting into account, which only serves to change the sign of the determinant. | |
| Aug 18, 2010 at 22:17 | comment | added | J. M. isn't a mathematician | Well I had always thought that the objection of "it takes so much work to do a determinant!" applied only to general dense matrices; to add another example to Victor's list, taking the determinant of a tridiagonal matrix via cofactor expansion is in fact equivalent to evaluating a certain three-term recurrence relation (which now links the theory of orthogonal polynomials and "Jacobi matrices" very tightly). Lastly... codes do exploit $\det(AB)=\det(A)\det(B)$; the point of Gaussian elimination is to factor your matrix A as $A=LU$; L unit lower triangular and U upper triangular... | |
| Aug 18, 2010 at 20:57 | comment | added | Thierry Zell | @Jiahao: your example sounds still very theoretical, but it's a good point nonetheless, and I am sure that others could come up with very practical applications of determinants. Still, most of the algebra that ends up being used in real life is determinant-free. | |
| Aug 18, 2010 at 20:47 | comment | added | Thierry Zell | @Victor: I never advocated not teaching people how to think. Your lower-triangular example is a strawman example (take the transpose). But the more important question is: why would people try to compute such a determinant? There are many people out there whose matrix needs can be perfectly filled by Gaussian elimination, because they don't care about the values of their determinants in the first place. As for symplectic matrices, the only people I know who use them are theorists, so need a strong math background anyway, so should know determinants. | |
| Aug 18, 2010 at 19:39 | comment | added | Victor Protsak | I am also befuddled by pronouncements of a definition of a theoretical concept, such as determinant, to be "bad" when what is meant is "computationally awful". Clearly, computational complexity is a serious and relatively novel issue that complements, rather than invalidates, theoretical basics. For example, should we stop teaching the order of an element of a finite group simply because it's computationally infeasible to find the multiplicative order of a random $g\mod p$? To define the determinant by the Gaussian elimination algorithm would be worse than useless (namely, counterproductive). | |
| Aug 18, 2010 at 19:29 | comment | added | Victor Protsak | "The most sensible thing to do to compute a determinant is to use Gaussian elimination" - I beg to differ! To give a trivial example, the determinant of a lower triangular $n\times n$ matrix is the product of its diagonal entries, which can be computed in $O(n)$ operations, while the Gaussian elimination requires $O(n^2)$. A less trivial example: a symplectic matrix always has determinant 1. My point is that various properties, depending on the situation, may be profitably exploited to evaluate a determinant, e.g. $\det(QR)=\det(Q)\det(R)$ is standard. | |
| Aug 18, 2010 at 19:22 | comment | added | Jiahao Chen | It should be pointed out, though, that some scientists do still use N x N determinants. I am most familiar with its use in quantum chemistry, where Slater determinants are used in Hartree-Fock theory to enforce the fermionic antisymmetry of many-electron wavefunctions by forming an explicitly antisymmetric product of one-electron wavefunctions. Presumably this use of determinants could be eliminated by judicious use of exterior products, but I don't know of a pedagogical presentation that doesn't use the determinant form. | |
| Aug 18, 2010 at 19:03 | history | answered | Thierry Zell | CC BY-SA 2.5 |