Exercise 6.4.13

If $A$ is Gramian, we have $A$ is symmetric since $A^t={(B^{t}B)}^t=B^{t}B=A$. Also, let $\lambda$ be an eigenvalue with unit eigenvector $x$. Then we have $\mathit{Ax}=\mathit{\lambda x}$ and

Conversely, if $A$ is symmetric, we know that $L_A$ is a self-adjoint operator. So we may find an orthonormal basis $\beta$ such that ${[L_A]}_{\beta }$ is diagonal with the $\mathit{ii}$-entry to be $\mathit{\lambda i}$. Denote $D$ to be a diagonal matrix with its $\mathit{ii}$-entry to be $\sqrt{{\lambda }_i}$. So we have $D^2={[L_A]}_{\beta }$ and

where $\alpha$ is the standard basis. Since the basis $\beta$ is orthonormal, we have ${[I]}_{\beta }^{\alpha }={({[I]}_{\alpha }^{\beta })}^t$. So we find a matrix

such that $A=B^{t}B$.