The relay node with linear relaying transmits the linear combination of its past received signals.The optimization of two-hop relay channel with linear relaying is discussed in this paper.The capacity for the two-hop Gaussian relay channel with linear relaying is derived,which can be formulated as an optimization problem over the relaying matrix and the covariance matrix of the signals transmitted at the source.It is proved that the solution to this optimization problem is equivalent to a "single-letter" optimization problem.We also show that the solution to this "single-letter" optimization problem has the same form as the expression of the rate achieved by Time-Sharing Amplify and Forward (TSAF).In order to solve this equivalent problem,we proposed an iterative algorithm.Simulation results show that if channel gain of one hop is relatively smaller,the achievable rate with TSAF is closer to the max-flow min-cut capacity bound,but at a lower complexity.