To me is given the following task in connection with the conjugate gradient method with a positive definite and symmetric matrix:
$$ \begin{array}{l} \text { Let } g^{i} \neq 0, i \leq t \in > \mathbb{N} \text { : Show } \\ \operatorname{span}\left\{g^{0}, \ldots, > g^{t}\right\}=\operatorname{span}\left\{A^{0} g^{0}, \ldots, A^{t} > g^{0}\right\} \\ \operatorname{span}\left\{g^{0}, \ldots, > g^{t}\right\}=\operatorname{span}\left\{d^{0}, \ldots, d^{t}\right\} > \\ \left(g^{t}, d^{i}\right)=0, \quad i \leq t-1 \\ \left(d^{t}, A > d^{i}\right)=0, \quad i \leq t-1 \end{array} $$
I would have thought that I could show this, more or less for all statements at the same time, by means of induction and have now started with the beginning of induction and the induction step:
Induction start $(t=0)$: For $t=0$ the following relationships apply: $$ \begin{array}{l} \operatorname{span}\left\{g^{0}\right\}=\operatorname{span}\left\{A^{0} g^{0}\right\}=\operatorname{span}\left\{d^{0}\right\} \\ \left(g^{0}, d^{0}\right)=0 \text { and }\left(d^{0}, A d^{0}\right)=0 \end{array} $$
Induction step (edit): $$ g^{k+1}=g^{k}+\alpha_{k} A d^{k} $$
From this follows: $$ \alpha_{k} A d^{k}=g^{k+1}-g^{k} $$
Shows that $\alpha_{k} A d^{k}$ lies in $\operatorname{span}\left\{g^{k+1}, g^{k}\right\}$.
$A d^{k}$ is orthogonal to $A d^{0}, \ldots, A d^{k-1}$, since the directions $d^{i}$ are orthogonal in pairs.
$\alpha_{k} A d^{k}$ is in the orthogonal complement of $\operatorname{span}\left\{A d^{0}, \ldots, A d^{k-1}\right\}$ with respect to span $\left\{g^{k+1}, g^{k}\right\}$.
This means that $\operatorname{span}\left\{g^{k}, A d^{k}\right\}=\operatorname{span}\left\{A d^{0}, \ldots, A d^{k-1}\right\} \oplus \operatorname{span}\left\{g^{k+1}, g^{k}\right\}$ is a direct orthogonal decomposition.
Is that correct so far?
conjugate gradient method: Start: $x^{0} \in \mathbb{R}^{n}, > d^{0}:=-g^{0}=b-A x^{0}$ Iteration: $$ \begin{aligned} t & =0,1,2, > \ldots \\ x^{t+1} & =x^{t}+\alpha_{t} d^{t}, > \alpha_{t}=-\frac{\left(g^{t}, d^{t}\right)}{\left(d^{t}, A > d^{t}\right)} \\ g^{t+1} & =g^{t}+\alpha_{t} A d^{t}(g^{t+1}=A > x^{t+1}-b=\underbrace{A x^{t}-b}_{g^{t}}+\alpha_{t} A d^{t}) \\ > d^{t+1} & =-g^{t+1}+\beta_{t} d^{t}, \beta_{t}=\frac{\left(g^{t+1}, A > d^{t}\right)}{\left(d^{t}, A d^{t}\right)} \end{aligned} $$