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?
The derivative of characterestic polynomial?
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linear-algebra
matrices
polynomials
derivatives
determinant
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0No. 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
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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.