Given real numbers $\,x_1,\dots,x_n,\,$ real functions $\,\phi_0,\dots,\phi_{n-1},\,$ and $\,\varphi_0,\dots,\varphi_{n-1},\,$ I have to prove the following lemma:
$$\det[\phi_{j-1}(x_k)]\det[\varphi_{j-1}(x_k)] = \det[\sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)].$$ The RHS is determinant of the sum of rank $1$ matrices.
Can someone help me ?
You ask about the identity $$\det[\phi_{j-1}(x_k)]\det[\varphi_{j-1}(x_k)] = \det[\sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)] .$$ Define the $\,n\times n\,$matrices $$ A := \{ \phi_{j-1}(x_k)\}_{j,k=1}^n,\quad B := \{ \varphi_{j-1}(x_k)\}_{j,k=1}^n,\\ C := \{ \sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)\}_{j,k=1}^n. $$ Since $\, C = A^T B \,$ by definition of matrix multiplication, using properties of determinants we get $$\det[A]\det[B] = \det[A^T]\det[B] = \det[A^T B] = \det[C] $$ which is the identity in question.
