On the Determinant and the Trace of a matrix

Suppose you have a squared matrix $A$ with $n$ rows and columns. Is there a relationship between its eigenvalues and its determinant? And what about its trace?

Determinant of A.

Consider the determinant of $\mid A- \lambda I \mid$ , which is a polynomial of degree $n$ . The values of $\lambda$ that solve the equation $\mid A- \lambda I \mid = 0$ , are the eigenvalues of $A$ , and are $n$ as its degree.
We can write such a polynomial using its roots (its eigenvalues) as follows:
$\mid A- \lambda I \mid = (\lambda - \lambda_1)(\lambda - \lambda_2)\cdots(\lambda - \lambda_n)$

Its constant term is: $(-1)^n \lambda_1\cdots\lambda_n$ ,
but it’s also the value of $\mid A- \lambda I \mid$ with $\lambda = 0$ , which obviously is $\mid A \mid$ .

So $\mid A \mid = (-1)^n \lambda_1\cdots\lambda_n$ .

Trace of A.

If we develop the term of degree $n-1$ we obtain:
$-(\lambda_1+\lambda_2+\cdots\+lambda_n)$ .

For simplicity we consider the case of an $3x3$ matrix:

$\mid A- \lambda I \mid = \begin{bmatrix}a_{11}-\lambda & a_{12} & a_{13} \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}$

It can be decomposed as follows:

$\mid A- \lambda I \mid = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}$

Now we can apply the decomposition on the second row of the second matrix:

$\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}=\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}$

And finally we can apply the decomposition on the third row of the last matrix:

$\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}=\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33} \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\0 & 0 & -\lambda \end{bmatrix}$