1
$\begingroup$

Last year, I took a course in Computational Physics, where I learnt methods to integrate and differentiate functions, diagonalize matrices, etc. More precisely, I recall we would diagonalize matrices with the QR method, which essentially consisted in performing transformations on the matrix of interest until it was in the desired diagonal form.

However, now that I think of it, I do not understand why we would simply not find the matrix's characteristic polynomial and find its roots, which is a simpler problem, considering matrix computations involve many more arithmetic steps than just calculating a determinant and then reducing the problem to that of finding the roots of some polynomial. What is more, the precission would also be easier to control, and demanding higher precision would only involve more iterations to find a root instead of more matrix products. Why is this not the standard procedure then? Or at least, why were we not taught to do it this way?

Improve this question
$\endgroup$
4
  • $\begingroup$ Key word: Numerical stability. $\endgroup$ Commented Feb 22 at 9:41
  • $\begingroup$ Reminds me of this reddit post: reddit.com/r/math/comments/1p3ya7b/… $\endgroup$ Commented Feb 22 at 10:30
  • $\begingroup$ @hendlim extremely useful comment, just what I was looking for, thanks a bunch! $\endgroup$ Commented Feb 22 at 12:43
  • 1
    $\begingroup$ Then once you find the roots you will need to do a matrix inverse problem for each eigevnalue to get the eigenvectors. $\endgroup$ Commented Feb 22 at 14:16

1 Answer 1

Reset to default
1
$\begingroup$

The first problem with using the characteristic polynomial is not root-finding, stability issues, or computing the eigenvalues.

The immediate problem is calculating the coefficients of the characteristic polynomial. Indeed, the characteristic polynomial of a matrix $A$ is defined as $\det(A-\lambda I)$, where $I$ is the unit matrix. Calculating a determinant by using its definition through permutations is an operation of unacceptable ($n!$ ) computational cost. Clever methods exist, but ... they rely on reducing the original matrix to a special form. I.e., they are equivalent or depend on the QR method.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.