update

doc/src/week35/week35.do.txt

@@ -2244,11 +2244,11 @@ Bessel's correction) we have
 \]
 !et
 
-meaning that every squared non-singular value of $\bm{X}$ divided by$n$,
-the number of samples, are the eigenvalues of the covariance
+meaning that every squared non-singular value of $\bm{X}$ divided by $n$ (
+the number of samples) are the eigenvalues of the covariance
 matrix. Every singular value of $\bm{X}$ is thus a positive square
 root of an eigenvalue of $\bm{X}^T\bm{X}$. If the matrix $\bm{X}$ is
-self-adjoint, the the sinular values of $\bm{X}$ are equal to the
+self-adjoint, the singular values of $\bm{X}$ are equal to the
 absolute value of the eigenvalues of $\bm{X}$.
 
 !split
@@ -2258,26 +2258,26 @@ For $\bm{X}\bm{X}^T$ we found
 
 !bt
 \[
-$\bm{X}\bm{X}^T$=\bm{U}\bm{\Sigma}\bm{V}^T\bm{V}\bm{\Sigma}^T\bm{U}^T=\bm{U}\bm{\Sigma}^T\bm{\Sigma}\bm{U}^T. 
+\bm{X}\bm{X}^T=\bm{U}\bm{\Sigma}\bm{V}^T\bm{V}\bm{\Sigma}^T\bm{U}^T=\bm{U}\bm{\Sigma}^T\bm{\Sigma}\bm{U}^T. 
 \]
 !et
 Since the matrices here have dimension $n\times n$, we have
 !bt
 \[
-\bm{\Sigma}\bm{\Sigma}^T = \begin{bmatrix} \tilde{\bm{\Sigma}} \\ \bm{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\bm{\Sigma}} 0 \bm{0}\\ \end{bmatrix}=\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}, 
+\bm{\Sigma}\bm{\Sigma}^T = \begin{bmatrix} \tilde{\bm{\Sigma}} \\ \bm{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\bm{\Sigma}}  \bm{0}\\ \end{bmatrix}=\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}, 
 \]
 !et
 leading to
 !bt
 \[
-$\bm{X}\bm{X}^T$=\bm{U}\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}\bm{U}^T. 
+\bm{X}\bm{X}^T=\bm{U}\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}\bm{U}^T. 
 \]
 !et
 
 Multiplying with $\bm{U}$ from the right gives us the eigenvalue problem
 !bt
 \[
-$\bm{X}\bm{X}^T$\bm{U}=\bm{U}\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}. 
+(\bm{X}\bm{X}^T)\bm{U}=\bm{U}\begin{bmatrix} \tilde{\bm{\Sigma}} & \bm{0} \\ \bm{0} & \bm{0}\\ \end{bmatrix}. 
 \]
 !et
 
@@ -2404,7 +2404,7 @@ Using our insights about the SVD of the design matrix $\bm{X}$
 We have already analyzed the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix $\bm{U}$ as
 !bt
 \[
-\bm{X}\bm{\beta} = =\bm{U}\bm{U}^T\bm{y}.
+\tilde{\bm{y}}_{\mathrm{OLS}}=\bm{X}\bm{\beta}  =\bm{U}\bm{U}^T\bm{y}.
 \]
 !et
 
@@ -2413,11 +2413,11 @@ For Ridge regression this becomes
 
 !bt
 \[
-\bm{X}\bm{\beta}^{\mathrm{Ridge}} = \bm{U\Sigma V^T}\left(\bm{V}\bm{\Sigma}^2\bm{V}^T+\lambda\bm{I} \right)^{-1}(\bm{U\Sigma V^T})^T\bm{y}=\sum_{j=0}^{p-1}\bm{u}_j\bm{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\bm{y},
+\tilde{\bm{y}}_{\mathrm{Ridge}}=\bm{X}\bm{\beta}_{\mathrm{Ridge}} = \bm{U\Sigma V^T}\left(\bm{V}\bm{\Sigma}^2\bm{V}^T+\lambda\bm{I} \right)^{-1}(\bm{U\Sigma V^T})^T\bm{y}=\sum_{j=0}^{p-1}\bm{u}_j\bm{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\bm{y},
 \]
 !et
 
-with the vectors $\bm{u}_j$ being the columns of $\bm{U}$. 
+with the vectors $\bm{u}_j$ being the columns of $\bm{U}$ from the SVD of the matrix $\bm{X}$. 
 
 !split
 ===== Interpreting the Ridge results =====