Diagonalizable linear operators and invariant subspaces.

Question

I am looking through this proof provided here:

https://math.stackexchange.com/a/78090/1039033

And I don't completely understand the induction process being used in the top voted answer.

Lemma. If $v_1 + v_2 + \cdots + v_k \in W$ and each of the $v_i$ are eigenvectors of $A$ corresponding to distinct eigenvalues, then each of the $v_i$ lie in $W$.

Proof. Proceed by induction. If $k = 1$ there is nothing to prove. Otherwise, let $w = v_1 + \cdots + v_k$, and $\lambda_i$ be the eigenvalue corresponding to $v_i$. Then:

$$Aw - \lambda_1w = (\lambda_2 - \lambda_1)v_2 + \cdots + (\lambda_k - \lambda_1)v_k \in W.$$

By induction hypothesis, $(\lambda_i - \lambda_1)v_i \in W$, and since the eigenvalues $\lambda_i$ are distinct, $v_i \in W$ for $2 \leq i \leq k$, then we also have $v_1 \in W$. $\quad \square$

Answer 1

Here is a different proof without using induction. However, it requires some knowledge of polynomials. Since $W$ is $A$-invariant, it is also $f(A)$-invariant for any $f\in F[x]$, where $F$ is the base field. In particular, for any $f\in F[x]$, we have $$f(A)v_1+f(A)v_2+\dots +f(A)v_k\in W.$$ Since $Av_i=\lambda_iv_i$, we have $f(A)v_i=f(\lambda_i)v_i$. Thus the above equation says that $$(*)\qquad f(\lambda_1)v_1+f(\lambda_2)v_2+\dots+f(\lambda_k)v_k\in W.$$ Fix an index $i$, we can construct the polynomial $$f(x)=(x-\lambda_1)(x-\lambda_2)\cdots \widehat{(x-\lambda_i)}\cdots(x-\lambda_k)=\prod_{1\le j\le k, j\ne i}(x-\lambda_j).$$ Here $\widehat{(x-\lambda_i)}$ means the term $x-\lambda_i$ is omitted from the product. Notice that $f(\lambda_i)\ne 0$ (since $\lambda_1,\dots,\lambda_k$ are distinct), and $f(\lambda_j)=0$ for any $j\ne i$. Apply this polynomial to the above equation $(*)$, we know that $f(\lambda_i)v_i\in W$. Since $f(\lambda_i)\ne 0$, we get that $v_i\in W$.