Let's say we have an $n\times n$ symmetric matrix $A \in M_n(\Bbb R)$ and function $f(\lambda)=\det(A-\lambda I)$, where $I$ is the identity matrix. If we have some number $\lambda_0$ that is a solution to $f(\lambda)=0$, then $\lambda_0$ is a real number. I know how to solve it using the complex and complex conjugate method. The thing is that my task is to prove it using induction. This is the proof until the point I get stuck:
We will use induction:
$1)$ n=1 $f(\lambda)=a_{11}-\lambda$. We have that $f(\lambda)$ is real and that $a_{11}$ is real and therefore $\lambda$ is real.
$2)$ Let's assume it's true for n, we'll prove it for n+1
We develop $f(\lambda_0)$ by the (n+1)-th row. Therefore $f(\lambda_0)$ = $\sum_{i=0}^n(a_{n+1,i}A_{n+1,i})$ + $(a_{n+1,n+1}-\lambda_0)A_{n+1,n+1}$=$\sum_{i=0}^{n+1}(a_{n+1,i}A_{n+1,i})$-$\lambda_0 A_{n+1,n+1}$
$A_{n+1,n+1}$ is a real number using the induction assumption. We have to prove that $\sum_{i=0}^{n+1}(a_{n+1,i}A_{n+1,i})$ is a real number and therefore we'll have that $\lambda_0 \in \Bbb R$.
P.S I don't know how to make matrices in this site, so I attach a photo. enter image description here
