# Show each eigenvector belong to invariant subspace

Let $f\in End(V)$, $n$- dimensional and $W$ be an $f$-invariant subspace of $V$ show that:

if $v_1,..,v_k$ are eigenvectors and $a_1,..,a_k$ are distinct, respective eigenvalues of $v_i$ and if $v_1+…+v_k \in W$ then we have $v_i \in W$ for each $i=1,..k$

my try:

since W is invariant then from $v_1+…+v_k \in W$ we have $f(v_1+…+v_k)=a_1v_1+…+a_kv_k \in W$ but I don't know how to show they singulary belongs to W.

• What do those $\;a_i\;$ have to do with the question? – Timbuc Mar 20 '15 at 20:50
• @Timbuc all that was meant to say is that the eigenvalues are distinct; somehow that's "clearer" if you call them $a_i$. – Omnomnomnom Mar 20 '15 at 20:53

By induction we have $v_1\in W$ and assume the result : "if a linear combination of $v_1,\ldots,v_{k-1}$ belongs to $W$ then each $v_i$ belongs to W" is true. Now we prove the result for $v_1,\ldots, v_k$ so

$$v_1+\ldots+v_k\in W\tag1$$ and then as you did

$$a_1v_1+\ldots+a_kv_k\in W\tag2$$ so $a_k(1)-(2)$ gives $$(a_k-a_1)v_1+\cdots+(a_k-a_{k-1})v_{k-1}\in W$$ and we apply the IH to get that $v_1,\ldots,v_{k-1}\in W$ and then $v_k\in W$.

• could you explain what is HR ? – claire Mar 20 '15 at 21:00
• Sorry it's IH: induction hypothesis. – user63181 Mar 20 '15 at 21:01
• I have question: you conclude that $v_k \in W$ from (1) ? if so how to expalin then or you do $a_1(1)-(2)$ and from this $v_2,…,v_k\in W$ so $v_1,…,v_n \in W$ ? – claire Mar 20 '15 at 21:24
• If we proved that $v_1,\ldots,v_{k-1}\in W$ then $v_1+\ldots+v_{k-1}\in W$ and then from $(1)$ we find that $v_k\in W$. – user63181 Mar 20 '15 at 21:28
• Would you consider voting to undelete/reopen? – Namaste Mar 21 '15 at 16:55

Hint: the vectors $f^m(v_1+v_2+\cdots+v_k)$ for $m=0,\dots,k-1$ are all linearly independent. However, by our assumption, they all lie in $W$.

Here, $f^k(v)$ means $\overbrace{f(f(\cdots f}^k(v) \cdots ))$.