Derivative of coefficients of characteristic polynomial

Question

Let $ X(t) $ be a square matrix of dimension $ n $. The Jacobi formula expresses the derivative of the determinant of $X(t)$ in terms of the derivative of the matrix itself. Is there an analogous formula for the other coefficients of the characteristic polynomial?

I am actually interested in the $ n = 3 $ case, so the only coefficient I do really care about is the one of $ t $.

Thank you very much!

Answer 1

The Newton Identities can be used to write the characteristic polynomial of $X\in{\mathbb C}^{3\times 3}$ in terms of the traces of the powers of $X$ $$\eqalign{ p(X) &= X^3 +c_2X^2 +c_1X +c_0I \\ &= X^3 - X^2\operatorname{tr}(X) + X\left(\frac{\operatorname{tr}^2(X)-\operatorname{tr}(X^2)}{2!}\right) - I\det(X) \\ }$$ Calculate the differentials of the trace functions $$\eqalign{ \operatorname{tr}(X) &= I:X \quad&\implies d\operatorname{tr}(X) = I:dX \\ & &\implies d\operatorname{tr}^2(X) = 2\operatorname{tr}(X)\,d\operatorname{tr}(X) = 2\operatorname{tr}(X)I:dX\\ \operatorname{tr}(X^2) &= I:X^2 \quad&\implies d\operatorname{tr}(X^2) = I:(X\,dX+dX\,X) = 2X^T:dX \\ }$$ where the colon is a convenient product notation for the trace, i.e. $\;A:B = \operatorname{tr}(A^TB)$

Then calculate the differential, derivative (wrt $t$) and gradient (wrt $X$) of the $c_1$ coefficient $$\eqalign{ dc_1 &= \tfrac{1}{2}(2\operatorname{tr}(X)I-2X^T) : dX \\ \dot{c}_1 &= \big(\operatorname{tr}(X)I-X^T\big) : \dot{X} \\ \frac{\partial c_1}{\partial X} &= \operatorname{tr}(X)I-X^T \\ }$$ As a bonus, do the same for the $c_2$ coefficient $$\eqalign{ dc_2 &= -I:dX \\ \dot{c}_2 &= -I:\dot{X} \\ \frac{\partial c_2}{\partial X} &= -I \\ }$$ Even the determinant (the $c_0$ coefficient) can be written in terms of traces $$\det(X) = \frac{\operatorname{tr}^3(X) -3\operatorname{tr}(X)\operatorname{tr}(X^2) +2\operatorname{tr}(X^3)}{3!} $$

Answer 2

The 1st order coefficient in the (cubic) characteristic polynomial is given by $f : \mathbb R^{3 \times 3} \to \mathbb R$, defined by

$$f ({\rm X}) := \frac12 \mbox{tr}^2 \left( {\rm X} \right) - \frac12 \mbox{tr} \left( {\rm X}^2 \right)$$

as Greg mentioned. Hence,

$$\begin{aligned} f ({\rm X} + h {\rm V}) = \cdots &= f ({\rm X}) + h \left( \mbox{tr} \left( {\rm X} \right) \mbox{tr} \left( {\rm V} \right) - \mbox{tr} \left( {\rm V} {\rm X} \right) \right) + O (h^2)\\ &= f ({\rm X}) + h \left\langle \color{blue}{\mbox{tr} \left( {\rm X} \right) {\rm I}_3 - {\rm X}^\top}, {\rm V} \right\rangle + O (h^2)\end{aligned}$$

where the Frobenius inner product was used. Thus, the gradient of $f$ with respect to $\rm X$ is

$$\boxed{\nabla f ({\rm X}) = \mbox{tr} \left( {\rm X} \right) {\rm I}_3 - {\rm X}^\top}$$

which is the same Greg obtained.

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matrix-calculus scalar-fields gradient