Let $A \in \mathbb{R}^{n\times n}$ be a positive definite matrix, and $d_1, ..., d_m$ be $A$-conjugate directions. Show that $d_1, ..., d_m$ are linearly independent.
This means that \begin{equation} \forall \textbf{h}\in \mathbb{R}^n\backslash \{0\}, \textbf{h}^\top \textbf{A}\textbf{h}>0 \qquad \textbf{(1)} \end{equation} And \begin{equation} \forall i \in \{1, ..., m\}, d_i \neq 0, \forall j \neq i, \textbf{d}_i^\top \textbf{A}\textbf{d}_j = 0 \qquad \textbf{(2)} \end{equation} Let us show that $\textbf{d}_i, ..., \textbf{d}_m$ is linearly independent by contradiction Let us suppose that: \begin{equation} \exists i^\star \in \{1,..., m\}, (\lambda_i)\in \mathbb{R}^m, d_{i^\star} = \sum_{j\neq i^\star}\lambda_j\textbf{d}_j \qquad \textbf{(3)} \end{equation} We can suppose that $i^\star = m$ without loss of generality and then: \begin{align*} \forall i \in \{1,..., m-1\}\quad \textbf{d}_i\textbf{A}\textbf{d}_m &= 0 \quad \text{by (2)} \qquad \textbf{(4)}\\ \text{However}: \textbf{d}_i\textbf{A}\textbf{d}_m &= \textbf{d}_i^\top\textbf{A}\sum_{j = 1}^{m-1}\lambda_j\textbf{d}_j \quad \text{by (3)} \qquad \textbf{(5)}\\ &= \sum_{j = 1}^{m-1}\lambda_j \textbf{d}_i^\top\textbf{A}\textbf{d}_j \qquad \textbf{(6)}\\ &= \lambda_i \underbrace{\textbf{d}_i^\top\textbf{A}\textbf{d}_i}_{>0 \text{ by (1)}} \qquad \textbf{(7)} \end{align*} Therefore $$ \forall i \in \{1, ..., m-1\}, \lambda = 0 $$ This means that $\textbf{d}_m = \sum 0\textbf{d}_i = 0$, contradiction
My question how can we go from (6) to (7). How can we convert $\lambda_j\textbf{d}_j$ to $\lambda_i\textbf{d}_i$?
It uses the $A$-conjugacy: $d_i^TAd_j=0$ for all $i\ne j$, so $$ \sum_j \lambda_j d_i^TAd_j =\lambda_id_i^TAd_i, $$ on other words: all but one summand are zero.