If we consider $A$ to be a symmetric $n\times n$ matrix, then how can we show that there exists a unitary matrix $G$ such that : $$ A=GTG^{*} $$ where $T$ is a tridiagonal symmetric matrix.
I am not seeking a full answer but I am having a trouble understanding what this question is trying to tell me and how to start.
I know that for any matrix $A$, it can be decomposed via the Schur decomposition to two unitary matrices and an upper triangular matrix. I am not sure how I can see that symmetry here leads to the modification of the Schur decomposition so that $A$ becomes similar to a tridiagonal matrix.
No, actually, the Schur decomposition tells you that for any matrix $A$, there exists a unitary matrix $Q$ and an upper triangular matrix $T$ such that
$$A=QAQ^{-1}$$
Since $Q$ is unitary, $Q^{-1}=Q^*$, so your question is literally just a rephrasing of the Schur decomposition.