Consider the error function \[ E(\mathbf {w})=\sum _{p=1}^{P}(y_p-\mathbf {w}\cdot \mathbf {x}_p)^2 \] where \(\mathbf {w}\) and \(\mathbf {x}_p\) are \(N\)-component vectors, \(N\ge P\), and \(y_p,\mathbf {x}_p\) are
prescribed.
Show that if \(\mathbf {w}_0\) is a minimizer of \(E\) then \[ [X^TX]\mathbf {w}_0=X^T\mathbf {y} \] where \(X_{pj}=[\mathbf {x}_p]_j\), \(j=1,\ldots ,N\), \(X^T\) is the transpose of the
matrix \(X\), and \(\mathbf {y}=(y_1,\ldots ,y_P)^T\).
Suppose that the weight vector \(\mathbf {w}\) is updated at each time step \(m\) using the
gradient decent algorithm \[ w_j(m+1)=w_j(m)-\eta \frac {\partial E(\mathbf {w}(m))}{\partial w_j(m)}. \] Show that \(\mathbf {w}(m)\to \mathbf {w}_0\) as \(m\to \infty \) provided that the rate parameter
\(\eta \) satisfies \[ |1-\eta \lambda _j|<1,\qquad j=1,\ldots ,N \] where \(\lambda _j\) are the eigenvalues of the symmetric \(N\times N\) matrix \(X^TX\). [Hint:
Explicitly calculate \(\partial E/\partial w_j\), use part (a) to rewrite the difference equation in
terms of \(\Delta \mathbf {w}(m)=\mathbf {w}(m)-\mathbf {w}_0\), and then diagonalize. Assume that \(X^TX\) is invertible so that \(\mathbf {w}_0\) is
unique].