New answers tagged

1

Self-answer: The QR algorithm itself produces eigenvectors only if the matrix is normal. In general, it only produces the Schur form. To compute eigenvectors we must do backward substitution, knowing that the span of the Schur vectors is equal to the whole eigenspace. The implementation is done! https://github.com/frozenca/Ndim-Matrix


1

You're underestimating the number of parameters required. In a $\text{VAR}(p)$ with $k$ series and no mean, there are $pk^2$ regression coefficients, $pk$ per variable. The function MTS::VARorder tests up to $p=13$, which for $k=14$ gives 2548 parameters, 182 per variable. This is already more than your 126 observations. You need to reduce the largest $p$ ...


3

NO. Let $B=\pmatrix{1&0\\0&1}$ and $A=\pmatrix{1&1\\1&1}$. Now calculate their determinants. And yes, both are positive semidefinite.


3

Given $\tilde{X} = XP$ for $P$ an invertible matrix, we have $$\begin{align} {\beta}(\tilde{X}) &= (\tilde{X}^T \tilde{X})^{-1}\tilde{X}^T y \\ &= (P^TX^TXP)^{-1}P^TX^T y \\ &= P^{-1}(X^TX)^{-1}P^{-T}P^TX^T y \\ &= P^{-1}(X^TX)^{-1}(PP^{-1})^TX^T y \\ &= P^{-1}(X^TX)^{-1}X^T y \\ &= P^{-1}\beta(X) \end{align}$$ which uses the facts ...


Top 50 recent answers are included