For a matrix $A\in\mathbb{K}^{n\times n}$ where $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ the characteristic polynomial is defined as
$\chi_A(\lambda) := \text{det}(A-\lambda I_n) = \sum_{k=0}^n c_k \lambda^k$
with the recursive definition of the coefficients
$c_n = (-1)^n,\;\;\;\;c_{n-1}=(-1)^{n-1}\text{tr}A,\;\;\;\;c_0 = \text{det}A$
and the trace of the matrix $A$
$\text{tr}A := \sum_{k=1}^n a_{kk}$
But when i try to compute the polynomial my "hand-computed" solution is not the same as the one Mathematica provides.
$A:=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}$
leads for me to the polynomial
$\chi_A^{Hand}(\lambda) = -\lambda^3+\lambda^2 (a+b+c)-\lambda (a+b+c) +abc$
but calling
CharacteristicPolynomial[{{a, 0, 0}, {0, b, 0}, {0, 0, c}}, \[Lambda]]
in Mathematica leads to the result
$\chi_A^{Mathematica}(\lambda)=a b c-a b \lambda -a c \lambda -b c \lambda +a \lambda ^2+b \lambda ^2+c \lambda ^2-\lambda ^3$
Can anyone explain me, what i am doing wrong?