# Diagonalizable linear operators and invariant subspaces.

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 $$v_1 \in W$$. $$\quad \square$$

• Please do not use images to convey information; here is an explanation why. Commented May 26, 2022 at 18:04
• If $w$ is an eigenvector and $k\neq 0$, then $kw$ is also an eigenvector. Here, $\lambda_j-\lambda_1\neq 0$, so each of the summands on the right hand side is an eigenvector. Thus, you have a sum of $k-1$ eigenvectors of $A$ which lies in $W$, so by the inductio hypothesis each of the $(\lambda_j-\lambda_1)v_i$ are in $W$. Multiplying by $1/(\lambda_j-\lambda_1)$ gives you that $v_2,\ldots,v_k\in W$, and since $v_1+\cdots+v_k$ is also in $W$, so is $v_1$. Commented May 26, 2022 at 18:06

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 $$v_i\in W$$.