Characteristic polynomial of a matrix is monic? Given a $n \times n$ matrix A, I need to show that its characteristic polynomial, defined as $P_A (x) = det (xI-A)$ is monic. I am trying induction. But no clue after induction hypothesis.
 A: You have 
$$c_A(\lambda) = \det(\lambda I - A)
= \sum_{\sigma\in \Sigma_n} {\rm sgn}(\sigma)
\prod_{i=1}^n
(\lambda \delta_{i\sigma(i)} - A_{i\sigma(i)}),$$
where $\Sigma_n$ is the permutation group on $n$ letters.  The only term in the sum giving a $\lambda^n$ term is the identity permutation.  Hence the polynomial is monic.
A: $$A=\begin{pmatrix}a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&a_{22}&\ldots&a_{2n}\\\ldots&\ldots&\ldots&\ldots\\a_{n1}&a_{n2}&\ldots&a_{nn}\end{pmatrix}\implies $$
$$|xI-A|=\begin{vmatrix}x-a_{11}&-a_{12}&\ldots&-a_{1n}\\-a_{21}&x-a_{22}&\ldots&-a_{2n}\\\ldots&\ldots&\ldots&\ldots\\-a_{n1}&-a_{n2}&\ldots&x-a_{nn}\end{vmatrix}=\prod_{i=1}^n(x-a_{ii})+\ldots=x^n+\ldots$$
and this is enough since we know $\;\deg p_A(x)=n\;$ ...
A: If you already know that $P_A$ is a polynomial, and if the matrix has real or complex coefficients, here is a nice trick:
$$ \lim_{|x|\to+\infty} \frac{P_A(x)}{x^n} 
= \lim_{|x|\to+\infty} \det \left( I - \frac{1}{x} A \right) 
= \det I
= 1, $$
because $\det$ is continuous.
From this, you can deduce that $P_A$ is monic of degree $n$.
A: If we use the permutation definition of the determinant, there will be only one term in the sum that is a polynomial of degree $n$, namely the product of all diagonal elements. (The rest of the terms form a polynomial of degree strictly less than $n$.)
This term is of the form $(x - a_{11}) \dots (x - a_{nn})$, which is clearly monic.
A: Let $A=(a_{ij})$ so we have
$$P_A(x)=\left|\begin{array}\\
x-a_{11}&\cdots&\cdots&-a_{1n}\\
-a_{21}&x-a_{22}&\cdots&-a_{2n}\\
\vdots&&\vdots\\
-a_{n1}&\cdots&\cdots&x-a_{nn}
\end{array}\right|$$
We expand the determinant relative to the first line we have
$$P_A(x)=(x-a_{11})\Delta_{11}+\sum_{j=2}^n(-1)^{1+j}a_{1j}\Delta_{1j}$$
where $\Delta_{1j} $ is a polynomial with degree less or equal $n-2$ and then we see that the term on $x^n$ of $P_A(x)$ is in  $(x-a_{11})\Delta_{11}$ and by induction in
$$(x-a_{11})(x-a_{22})\cdots(x-a_{nn})$$
and since the constant term of $P_A(x)$ is $(-1)^n\det A$ hence we have
$$P_A(x)=x^n-\underbrace{\left(\sum_{k=1}^na_{kk}\right)}_{=Tr A}x^{n-1}+\cdots+(-1)^n\det A$$
A: For $n=1$,
\begin{equation}
P_A(x)=det(xI_1-A)=|s-a_{11}|=s-a_{11}.
\end{equation}
So, $P_A(x)$ is monic for $n=1$.
Assume $P_A(x)$ is monic for $n=k$, i.e.
\begin{equation}
P_A(x)=det(xI_k-A)=\begin{vmatrix}
                   (x-a_{11})&-a_{12}&...&-a_{1k}\\
                   -a_{21}&(x-a_{22})&...&-a_{2k}\\
                   \vdots&\vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k2}&...&(x-a_{kk})
                   \end{vmatrix}=x^k+...
\end{equation}
We are now left to show that; $P_A(x)$ is monic for $n=k+1$, so
\begin{equation}
\begin{split}
P_A(x)=&det(xI_{k+1}-A)\\
=&\begin{vmatrix}
                   (x-a_{11})&-a_{12}&...&-a_{1k}&-a_{1,k+1}\\
                   -a_{21}&(x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
                   \vdots&\vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k2}&...&(x-a_{kk})&-a_{1,k+1}\\
                   -a_{k+1,1}&-a_{k+1,2}&...&-a_{k+1,k}&(x-a_{k+1,k+1})
\end{vmatrix}\\
=&\underbrace{(-a_{k+1,1})(-1)^{(k+1)+1}\begin{vmatrix}
                   -a_{12}&...&-a_{1k}&-a_{1,k+1}\\
                   (x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
                   \vdots&\ddots&\vdots\\
                   -a_{k2}&...&(x-a_{kk})&-a_{1,k+1}
                   \end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&\underbrace{(-a_{k+1,2})(-1)^{(k+1)+2}\begin{vmatrix}
                   (x-a_{11})&-a_{13}&...&-a_{1k}&-a_{1,k+1}\\
                   -a_{21}&-a_{23}&...&-a_{2k}&-a_{2,k+1}\\
                   -a_{31}&(x-a_{33})&...&-a_{3k}&-a_{3,k+1}\\
                   \vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k3}&...&(x-a_{kk})&-a_{1,k+1}
                   \end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&...\\
+&(x-a_{k+1,k+1})(-1)^{\overbrace{(k+1)+(k+1)}^{2(k+1)}}det(xI_k-A)\\
=&(x-a_{k+1,k+1})(x^{k}+...)+...\\
=&x^{k+1}+...
\end{split}
\end{equation}
So, $P_A(x)$ is also found monic for $n=k+1$.
$\therefore$ $P_A(x)$ is monic.$\square$
A: The characteristic polynomial is the product of monic degree-1 or degree-2 terms, corresponding to eigenvalues and eigenpairs, respectively. 
The set of monic polynomials is closed under multiplication. That is your induction hypothesis.
done.
