Characteristic roots of symmetrical transformation in euclidean space Need help in understanding a key point in the proof:
In euclidean space, a symmetrical transformation's characteristic roots are reals.
Let $f$ be a symmetrical transformation in a euclidean space $E$, $\dim (E)=n$, let $e:= \{e_1,\ldots ,e_n\}$ be an orthonormed basis in $E$ and let $A:= A_f^e$ be matrix of $f$ with respect to basis $e$. Transformation $f$ has the same characteristic roots as $A$, which are the roots of $\det (A-\lambda I)$. We have an $n-$degree polynomial with real coefficients, which has $n$ roots in $\mathbb{C}$. Objective is to show all roots are real.
Matrix $A$ is symmetrical. ##Now comes the problematic part for me##:
Let $\lambda _0\in\mathbb{C}$ be a root of $\det (A-\lambda I)$ then we have a system of equations with complex coefficients:
$$\begin{cases} (a_{11}-\lambda_0)x_1 +\ldots +a_{1n}x_n = 0\\
a_{21}x_1+(a_{22}-\lambda_0)x_2 +\ldots +a_{2n}x_n= 0\\\ldots\\
a_{n1}x_1 + a_{n2}x_2 +\ldots + (a_{nn}-\lambda_0)x_n=0 \end{cases}$$
The former system has a non-zero solution $(k_1,\ldots ,k_n)\in\mathbb{C}^n$, therefore for every $i\in\{1,\ldots ,n\}$
$$\sum_{j=1}^n a_{ij}k_j = \lambda_0k_i $$
and so on..
Are we not supposed to calculate the determinant, I don't understand where the system of equations came from or how/why it is relevant. Secondly, why does the system of equations have a non-zero solution and thirdly why is the last equality satisfied for all $i$?
 A: Let $A$ be an $n\times n$ real matrix.  We say $A$ is symmetric iff $A=A^T$ is equal to its own transpose.
The compact expression $\det(A-\lambda I) = 0$ is the characteristic polynomial equation in $\lambda$, whose roots are the "characteristic roots" or eigenvalues of $A$.  In other words, those $\lambda$ which give $A-\lambda I$ as singular matrices will result in nontrivial solutions to the homogeneous linear equations:
$$ (A - \lambda I) \vec x = \vec 0 $$
Writing out the separate linear equations results in the system shown in the Question.
The vector $x$, if it is such a nonzero solution, then satisfies $A\vec x = \lambda \vec x$.  Then we say $x$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$.
Now the related facts about a symmetric real $A$ are remarkable.  Not only will the characteristic roots of $A$ be real, there will be a "full set" of these in the sense that there will be a basis of the eigenvectors of $A$ for all of its eigenvalues.  Representing $A$ with respect to such a basis results in a diagonal matrix.  Hence we know $A$ is (real) diagonalizable, i.e. similar to a (real) diagonal matrix, when $A$ is real and symmetric.
