I have a Weiner process $\{W(t)\}_{t\ge0}$ with $\sigma^2=\text{Var}(W(1))=1$. For a real constant $\epsilon>0$ consider the differential ratio process $\Delta_\epsilon=\{\Delta_\epsilon(t)\}_{r>0}$ given by \begin{equation} \Delta_\epsilon(t) = \frac{W(t+\epsilon)-W(t)}{\epsilon}, \quad \text{fo...

Proposal: Add the tag grad-curl-div; make gradient, curl and divergence aliases thereof. Edit 2: I've revised the proposal statement to Alexander Gruber's (in my view superior) suggestion in the comments. Implementing this proposal poses some advantages over the status quo: It would (productivel...

At the moment, Mathematics SE has tags hessian-matrix and jacobian. However, tag gradient is a synonym for tag vector-analysis. Why? Motivation Since tag gradient does not exist, tags gradient-descent and gradient-flows are often used instead, which is unfortunate. Two tags are being corrupted b...

Suppose you have a $n \times 1$ column vector $$a=\begin{bmatrix}a_1\\{a_2}\\ \vdots\\{a_n}\end{bmatrix}$$ and a $1 \times m$ row vector $$\quad b=\begin{bmatrix}b_1 & b_2 & \ldots & b_m\end{bmatrix}$$ If you then multiply these $$A=ab=\begin{bmatrix} a_{1}b_1 & a_{1}b_2 & \ldots & a_{1}b_m\...

If $A$ is a complex $n \times n$ matrix of rank $1$, then $$\det(I+A) = 1 + \mbox{tr}(A)$$ How to approach this problem? Rank-$1$ matrices have special properties. Also, thinking about the determinant of a matrix as the product of its eigenvalues and the trace of the matrix as the sum of it...

I have seen a lot of papers mentioning that a certain matrix is rank-$1$. What properties does a rank-$1$ matrix have? I know that if a matrix is rank-$1$ then there are no independent columns or rows in that matrix.

Let $A$ be a complex matrix of rank $1$. Show that $$\det (I+A) = 1 + \operatorname{Tr}(A)$$ where $\det(X)$ denotes the determinant of $X$ and $\operatorname{Tr}(X)$ denotes the trace of $X$. Any hint, please. I do not get how to combine the ideas of rank, determinant and trace. Thank you.

Assume I want to minimise this: $$ \min_{x,y} \|A - x y^T\|_F^2$$ then I am finding best rank-1 approximation of A in the squared-error sense and this can be done via the SVD, selecting $x$ and $y$ as left and right singular vectors corresponding to the largest singular value of A. Now instead, ...

$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$. I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?

First let me make two statements to give my question the proper context. Consider $D$ a diagonal matrix and $u v^T$ a rank 1 matrix. From my current knowledge there exist computationally cheap algorithms to diagonalize $D+uv^T$ as discussed here: maximum eigenvalue of a diagonal plus rank-one ...