# Derivative of a characteristic polynomial at an eigenvalue

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!

• Formula for $p'(\lambda)$? – Jon Warneke Jun 29 '16 at 15:49
• 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 $A$. – Bryson of Heraclea Jun 29 '16 at 15:57
• Based on this formula, we have $$p'(\lambda_1) = \operatorname{trace}(\operatorname{adj}(A- \lambda_1 I))$$ – Omnomnomnom Jun 29 '16 at 16:31

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 $q(a)$.
• 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 $V/E_a$? – Bryson of Heraclea Jun 30 '16 at 7:30
• 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 "$a$". – Andrew D. Hwang Jun 30 '16 at 10:47