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I'd like to proof:

The caracteristic polynomial of $A \in M(n\times n, K)$ has the form:

$P_A(\lambda) = (-1)^n \lambda^n + (-1)^{n-1} \operatorname{tr}(A)\lambda^{n-1} +\dots +\det(A)$

My proof idea:

(1) Showing that for $n\times n$ matrix $A$ follows: the characteristic polynomial is $f(\lambda) = (a_{11} - \lambda)(a_{22} - \lambda)...(a_{nn}-\lambda) + g(\lambda)$ with $g(\lambda)$ being a polynomial of degree.. $n-3$ ??? No I'm not sure..

Proof this first:

let $n=2$ then it's obvious. SO let n>2 then do expansion along the first row. The first term is

$T_1 = (a_{11} - \lambda) \cdot \det \begin{pmatrix} a_{22}-\lambda & ... & a_{2n} \\ ... \\ a_{n2} & ... & a_{nn}-\lambda \end{pmatrix}$

however, this determinant is the characteristic polynomial of a (n-1)x(n-1) matrix, so according induction hypothesis we have

$T_1 = (a_{11} - \lambda) \cdot (a_{22}-\lambda)...(a_{nn} -\lambda) + g(\lambda)$. I'm not sure, but is $g(\lambda)$ still polynomial with degree n-3 or n-2 ??

Let's take a look at the further terms:

$T_2 = -a_{12} \cdot \det(...)$, well now I HAVE to show that this matrix also is a polynomial of degree $n-3$ (or $n-2$, depending on what's correct above)$

(2)

Take now a look at $(a_{11} - \lambda)(a_{22} - \lambda)...(a_{nn} - \lambda) = (-\lambda)^n \cdot (-\lambda)^{n-1}(a_{11}+a_{22}+...+a_{nn}) + h(\lambda)$ (h has degree n-3 or n-2 or..??)

Well in this case we have

$f(\lambda) = (-1)^n \lambda^n + (-1)^{n-1} \cdot tr(A)\lambda^{n-1} + a_{n-2}\lambda^{n-2} + ... + a_1 \lambda + a_0$.

Since $f(0) = a_0$ and $f(0) = \det(A)$ we have $a_0 = \det(A)$.

**So the main problems and questions are:

  1. Is there a shorter proof?
  2. What's the correct degree in every step? Why so?
  3. How can I show that all TERMS but the FIRST ONE are polynomials of degree ($n-2$ or $n-3$ or..)**

Kind Regards!

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  • $\begingroup$ Put $\lambda=0$ in the definition $\text{det}(A-\lambda I)$ and you get the last term. Divide all the elements of the matrix $A-\lambda I$ by $\lambda$ and put $\lambda\rightarrow\infty$ and you get the $(-1)^{n}$ of the first term. $\endgroup$ – OR. Mar 28 '14 at 9:49
  • $\begingroup$ TL:DR. Related. $\endgroup$ – Git Gud Mar 28 '14 at 9:52

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