Find eigenvalues & eigenspaces $Tv=\lambda v \Rightarrow a_1(\lambda - 1)w_1 + a_2(\lambda - 1)w_2 + \sum_{i = 3}^{n}a_i(\lambda + 1)w_i = 0$

Question

I got a formula for $T$ and $v$ base on an orthonormal base $\{w_1,w_2,...,w_n\}$

Therefore, for $Tv=\lambda v$ we get:

$$ \lambda (\sum_i^n a_iw_i) = a_1w_1 + a_1w_2 - \sum_{i=3}^na_i(\lambda + 1)w_i $$

Getting:

$$ a_1(\lambda - 1)w_1 + a_2(\lambda - 1)w_2 + \sum_{i = 3}^{n}a_i(\lambda + 1)w_i = 0 $$

NOW THE PROBLEM

From here, the solution says that:

$$ \lambda = 1,-1 $$

Namely, those are the eigenvalues.

And for the eigenspaces for those $\lambda$'s, the solution gets: $$ V_{-1} = \{w_3,w_4,...,w_n\}, V_1 = \{w_1,w_2\} $$

And i dont understand how they got to those eigenvalues and those eigenspaces.

maybe somehow the polynom represent the charactaristic polynom but i dont understand.

How, for example, $\lambda = 1$ gets this polynom to $0$, as an eigenvalue should do.

Help.

Thank you.