I am trying to fix this proof of a lemma that I did't correctly write:
If $A\in \mathbb{R}^{n\times n}$ symmetric positive definite with $m$ different Eigenvalues $\lambda_j$, then the Conjugate gradient method for $A$ converges in a maximum of $m$ iterations.
Let $U_m=K_m(r_0,A)=\langle r_0,Ar_0,\dots,A^{m-1}r_{0}\rangle$ the m'th Kyrolv space. We want to show $U_m=U_{m+1}$ (why)
For $v_j$ the Eigenvektors to the $m$ different Eigenvalues, we can write $r_0$ as: $$r_0=\sum_j \gamma_j v_j \in V_m= \langle v_1,\dots,v_m\rangle$$ Let $z\in V_m$, so $z=\sum_j \alpha_j v_j$. Then we have $$Az=\sum_j \alpha_j A v_j=\sum_j \alpha_j \lambda_jv_j=\sum_j \beta_j v_j \in V_m.$$
I don't understand why this gives the desired result.
In extension of the last equation you get $$ A^iz=\sum_jα_jA^iv_j=\sum_jα_jλ_j^ivj\in V_m \implies K_i(z,A)\subset V_m. $$ So as the chain of Krylov spaces has to increase in dimension until it becomes stationary at $U_k=U_{k+1}=...$, one has necessarily $k\le m$, as all these spaces are subspaces of the $m$-dimensional subspace $V_m$. At most $k=m$, so that in every case $U_{m+1}=U_m$.