We want to solve the linear system $Ax=b$, where $A$ is SPD. We use a method similar to steepest decent method. For the first searching direction $d^1, d^2, \cdots d^n$ are chosen to be the standard unit vector $e_1, e_2,\cdots, e_n$. Then the next $n$ search directions$ d^{n+1}, d^{n+2}\cdots, d^{2n}$ are again $e_1, e_2,\cdots, e_n$. We have $x^{(k+1)}=x^{(k)}+\alpha_k d^k$, where $\alpha_k = \frac{d^k\cdot (b-Ax^{(k)})}{Ad^k\cdot d^k}$. The question is prove that each group of $n$ step is one iteration of Gauss Seidel method.
I tried to express $x^{(k+1)}$ and $x^{(k)}$ as vector, but it differs to much to the formula of Gauss- Seidel method.
Let $A=(a_{ij})=L_*+U$, where $L_*$ is the lower triangular part of $A$. Let $x$ and $x^+$ be the initial and final approximation of the Gauss-Seidel method so that $$ x^+=x+L_*^{-1}r, \quad r := b-Ax. $$ Now consider solving $L_*c=r$, $c:=[\gamma_1,\ldots,\gamma_n]^T$. Since $L_*$ is lower triangular, we can successively solve for $\gamma_1,\ldots,\gamma_n$ and write $$ x^+=x+\gamma_1e_1+\ldots+\gamma_ne_n. $$ Let $x_0:=x$. We can define a recursion $$ x_i=x_{i-1}+\gamma_ie_i, \quad i=1,\ldots,n, $$ in which we successively update the components of the solution vector to obtain finally $x^+=x_n$. We just need to show now that $\gamma_i=e_i^T(b-Ax_{i-1})/a_{ii}$.
The coefficient $\gamma_i$ is the $i$th component of the solution of $L_*c=r$. By considering a partitioning of the leading $i\times i$ part of this system, we have $$ \begin{bmatrix}\tilde{L}_* & 0 \\ a_i^T & a_{ii}\end{bmatrix} \begin{bmatrix}\tilde{c}_{i-1}\\\gamma_i\end{bmatrix}=\begin{bmatrix}\ast \\ e_i^Tr\end{bmatrix}, $$ where $a_i:=[a_{i1},\ldots,a_{i,i-1}]^T$ and $\tilde{c}_{i-1}:=[\gamma_1,\ldots,\gamma_{i-1}]^T$, so $$ \gamma_i=\frac{1}{a_{ii}}\left(e_i^Tr-a_i^T\tilde{c}_{i-1}\right). $$ Note that the components of $\tilde{c}_{i-1}$ are the first $i-1$ components of $x_{i-1}-x$ and the remaining components of the latter are zero. Hence $a_i^T\tilde{c}_{i-1}=e_i^TA(x_{i-1}-x)$ and we have $$ \gamma_i = \frac{1}{a_{ii}}\left[ e_i^Tb-e_i^TAx-e_i^TA(x_{i-1}-x) \right]=\frac{1}{a_{ii}}e_i^T(b-Ax_{i-1}). $$