7
$\begingroup$

Let $p(\lambda)$ be the characteristic polynomial of an $n\times n$ matrix $A$. We know that the roots of $p(\lambda)$ are the eigenvalues of $A$, hence the sum of the roots of the polynomial (taking into account multiplicity) equals $\mathrm{tr}(A)$ and the product of the roots equals $|A|\equiv\mathrm{det}(A)$.

Since $p(\lambda)=\prod_{i=1}^n(\lambda-\lambda_i)$, we have $p'(\lambda_1)=\prod_{i=2}^n(\lambda_1-\lambda_i)$ (arbitrary numbering of eigenvalues). Is there anyway that we can connect this value, i.e. the derivative of the characteristic polynomial at a root/eigenvalue, to other special quantities connected with $A$, like determinants and trace?

I am sorry if the question is a little vague.

Many thanks to all the responders in advance!

$\endgroup$
  • 1
    $\begingroup$ Formula for $p'(\lambda)$? $\endgroup$ – Jon Warneke Jun 29 '16 at 15:49
  • $\begingroup$ Well, it would be nice to have a formula for $p'(\lambda_1)$ in terms of "easily" accesible or "macroscopic" quantities of the matrix, like the determinant or the trace of $A$ or of a function of $A$. $\endgroup$ – Bryson of Heraclea Jun 29 '16 at 15:57
  • 2
    $\begingroup$ Based on this formula, we have $$ p'(\lambda_1) = \operatorname{trace}(\operatorname{adj}(A- \lambda_1 I)) $$ $\endgroup$ – Omnomnomnom Jun 29 '16 at 16:31
1
$\begingroup$

The linear algebraic origins of the question are something of a red herring: This may be taken as a question about the derivative of a complex polynomial $p$ at a root $a$. Write $$ p(x) = (x - a)^{k} q(x),\qquad q(a) \neq 0. $$ If $k > 1$, then $p'(a) = 0$, regardless of the other roots. If $k = 1$, i.e., $a$ is a simple root of $p$, then $$ p'(x) = (x - a) q'(x) + q(x); $$ if $p$ factors completely, with roots $a_{i}$, then $p'(a) = q(a) = \prod_{i} (a - a_{i})$, as you say. This can be expanded as a polynomial in $a$ whose coefficients are the elementary symmetric polynomials in the $a_{i}$.


That said, the value $q(a) = \prod_{i} (a - a_{i})$ does have a linear-algebraic interpretation: If $T:V \to V$ has characteristic polynomial $p$, if $a$ is an eigenvalue, and if $E_{a}$ is a one-dimensional space of $a$-eigenvectors, the operator $aI - T$ induces an operator on the quotient space $V/E_{a}$ whose eigenvalues are the $a - a_{i}$, and whose determinant is therefore $q(a)$.

$\endgroup$
  • $\begingroup$ Thank you very much! I like the simplicity and elegance of your linear-algebra-theoretic answer. From a numerical point of view, how could one deduce a matrix representation of the operator that is induced by $aI-T$ on the quotient space $V/E_a$? $\endgroup$ – Bryson of Heraclea Jun 30 '16 at 7:30
  • 1
    $\begingroup$ You're very welcome. :) To get a matrix for the operator induced by $aI - T$, you can express $T$ in Jordan canonical form, say; the eigenspace $E_{a}$ corresponds to a column containing $(n-1)$ zeros and a single entry $a$. A matrix for the induced operator may be obtained by crossing out the row and column of that "$a$". $\endgroup$ – Andrew D. Hwang Jun 30 '16 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.