2
$\begingroup$

Let $A\in M_{n}(R)$ and $f(x)$ be the characterestic polynomial of $A$. Is it true that $f'(x)=\sum_{i=1}^{^{n}}\sum_{j=1}^{n}\det(xI-A(i\mid j))$ which $A(i\mid j)$ is a submatrix of $A$ obtained by the cancelation the $i$th row and $j$th colomn?

  • 0
    No. If $A$ is $2 \times 2$, the leading coefficient on the LHS is $2x$ and on the RHS is $4x$. They're not proportional either: Take A = \left( \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix} \right), the LHS is $d(x^2)/dx=2x$ and the RHS is $3x+(x-1) = 4x-1$. I assume there is a reason you asked this; check your examples and try again?2012-12-12

1 Answers 1

3

The good formula is $f'(x)=\sum_{j=1}^n\det(xI-A(j\mid j)),$ which can be established using the definition of the determinant and the derivative of a product of $n$ functions.

The formula in the OP doesn't hold as for $n=2$ it involves extra-diagonal terms.