Search Results

The matrix is symmetric, and it also looks like it may be related to upper-lower, triangular? Wikipedia calls it a hollow matrix and Wolfram alpha gave a weird answer and none of the steps. … I get stuck on the determinant step of finding eigenvalues where the 4x4 matrix looks too big to compute the determinant efficiently. Any ideas on how to move forward with this? …

This generated a triangular matrix (M1) where the diagonal is 1 because A1 = A1, A2 = A2, etc. All other comparisons are between 0 (no similarity) and 1 (identical). … I then did the same thing for protein B, comparing the 5 protein B homologs (B1..B5) to generate a second triangular matrix (M2). …

My current approach is as follows: I use the array of numbers to generate a triangular matrix $A$ by inserting the values of the array into the upper triangular part. … The covariance matrix would then be given by $ABA$. However, this matrix is not necessarily positive definite. How can I ensure that the resulting matrix is positive definite? …

I've read books and articles but I don't understand the difference between these two concepts since we use a lower triangular matrix when imposing restrictions in a SVAR (by referring to the theory), and …

\right)=s-\bar{Q}\left( \tau_{a} \mid X, z \right) $$ [3]: Omitting subscripts for simplicity, the earnings-cum-education model presented above can be written in the format of an exactly identified triangular … right)=\gamma_{s} X+\pi z+\xi G_{a}^{-1}\left(\tau_{a}\right) & (9)\end{array} $$ Given the restrictions imposed by (6) and $(7),$ the key parameter of interest $\Pi\left(\tau_{a} \tau_{u}\right)$ is a matrix …

The essential idea behind the algorithm is to compute $P_t = S_t S_t'$, as such a multiplication will always yield a symmetric non negative matrix. … In the paper, it is shown that one can replace the time update equation by constructing an orthogonal matrix $G$ and an upper triangular matrix $M$ such that $$\begin{pmatrix}M \\0 \end{pmatrix} = G …

We have that a lower-triangular matrix $L$ parametrizes the SPD matrices by a function $f$ which exponentiates the diagonal elements of $L$ and then multiplies the result with its transpose. … In the space of symmetric matrices, you can just make $\theta$ the lower triangular entries again, which are unconstrained, and then at each step compute the $\Sigma$ corresponding to the symmetric matrix …

You are looking for the probability of being in state $n$ after time $t$ for the CTMC with rate (or generator) matrix $$ Q = \begin{bmatrix} -\lambda_1 & \lambda_1\\ & -\lambda_2 & \lambda_2\\ & & \ddots … Note also that you can omit the last row and column, since the matrix is upper triangular, and reduce to $$ p = e_n^T\exp\left(t \begin{bmatrix} -\lambda_1 & \lambda_1\\ & -\lambda_2 & \lambda_2\\ & & …

I've made a $7\times7$ partial covariance matrix with $6$ non-zero elements outside the diagonal ($6$ for each triangular part, since it's symmetric). … I haven't found much literature using the computed partial covariance matrix. Most sources just see which elements are non-zero. …

Although the formula is inefficient (it cannot be simplified like the right hand side of the original equation), it's fast enough for many datasets because the sum can be expressed as a matrix product: … putting the $z_i$ in rows of an $n\times p$ matrix $Z$ (where $p$ is the dimension of the coordinates, $p=3$ in your case), the expression for $\sigma^2$ is the mean of the upper triangular part of $ZZ …

I know that if you do a $QR$ factorization of $X$ such that $ X = QR $ where Q is an $m x n$ orthonormal matrix and $R$ is an $n x n$ upper triangular matrix, then you can derive $\beta$ by: $\beta$ = … In numpy this looks like this: beta = np.linalg.inv(R).dot(Q.T.dot(y)) However, my understanding is that, from an optimization standpoint, it's always a bad idea to take the inverse of a matrix. …

First, observe that $R \beta = Q^\top y$ involves a triangular matrix $R$, which is easy to solve for $\beta$ without forming an explicit inverse by using forward substitution or backward substitution. … In python, we can solve this using the specialized triangular system solver: beta = scipy.linalg.solve_triangular(R, Q.T.dot(y)) …

p}$ is an upper-triangular matrix. … Z$ is the matrix consisting of $X$'s remaining $p - 1$ columns. …

Cholesky based proof The main insight for this proof is the use of the cholesky decomposition of the covariance matrix, i.e. $\Sigma = LL^T$ where $L$ is a lower triangular matrix. … Or starting from iid standard normal $X\sim\mathcal{N}(0, \mathbb{I})$ we can obtain our distribution $$Y=L X \sim \mathcal{N}(0,\Sigma).$$ Due to the lower triangular nature, only $X_1$ is used for $Y …

For example, if $X$ is symmetric, then $X(K)$ can be the upper triangular part. … of (structured matrix) wrt (structured matrix). …