# Tag Info

## Hot answers tagged linear-algebra

9

Nilesh Padhi, Welcome to the DSP Community. The classic definition of separable means the data (2D) given by $X \in \mathbb{R}^{m \times n}$ can be written as: $$X = \sigma u {v}^{T}$$ Where $\sigma \in \mathbb{R}$, $u \in \mathbb{R}^{m}$ and $v \in \mathbb{R}^{n}$. This is called Rank 1 Matrix. How can you get those parameters and vectors ...

8

I think you are constructing the convolution matrix incorrectly. Let's do an easy example with some numbers to see how it is constructed Let's imagine that youhave the following vectors of dimensions $M$ and $N$, And let's for the sake of it, say that $M = 5$ and $N = 4$ $\mathbf{a} = [a_{1},a_{2},a_{3},a_{4}]^{T}$ $\mathbf{b} = [b_{1},b_{2},b_{3},b_{4},... 8 The book's formula is right. Let $$H(w) = 1 - r e^{j(\theta - w)} = [1-r \cos(\theta - w)] + j [-r \sin(\theta - w)]$$ Since the group delay$\tau$is the negative of the derivative of the phase of$H(w)$, we first define the phase as: $$\phi(w) = \tan^{-1}\left( \frac{-r \sin(\theta - w)}{1-r \cos(\theta - w)} \right)$$ Using the derivative rule for the ... 7 Slope from all samples obtained To summarize the question's problem, you want to calculate the slope based on all samples obtained thus far, and as new samples are obtained, update the slope without going through all the samples again. On the page you cite is the equation for calculation of the slope$m_n$that together with$b_nminimizes the sum of ... 6 I would like to recommend a great book to you (see below), but before that I would like to point out that there is some confusion in your question. If you see a convolution integral in a textbook, then this refers to the convolution of two continuous time signals, which can't be represented by a matrix/vector multiplication. You must be referring to the ... 6 I created a function to create a Matrix for Image Filtering (Similar ideas to MATLAB's imfilter()): function [ mK ] = CreateImageFilterMtx( mH, numRows, numCols, operationMode, boundaryMode ) % ----------------------------------------------------------------------------------------------- % % [ mK ] = CreateImageFilterMtx( mH, numRows, numCols, ... 6 Hint: According to Euler's formula we have $$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$ 6 This is a slightly tedious but nevertheless straightforward exercise in computing the derivative of a function: \begin{align}\tau(\omega)&=-\frac{d\phi(\omega)}{d\omega}=-\frac{d}{d\omega}\arctan(f(\omega))\tag{1}\end{align} with $$f(\omega)=\frac{r\sin(\omega-\theta)}{1-r\cos(\omega-\theta)}\tag{2}$$ From(1)$we have $$\tau(\omega)=-\frac{f'(\... 6 The easiest approach would be writing each case using Matrix Form of the convolution. In this answer we assume the discrete convolution is applied only on valid support (Matching MATLAB's valid parameter for the convolution). Namely, given x \in \mathbb{R}^{m \times n} and h \in \mathbb{R}^{k \times l} then h \ast x \in \mathbb{R}^{ \left( m - k + ... 6 There are few options: Stephen Boyd, Lieven Vandenberghe - Convex Optimization. This is the classic in this field. Very well written book. Also have a look on other papers of Boyd on similar subjects such as the The Alternating Direction Method of Multipliers (ADMM). They also have a great MOOC Course Stanford Online CVX 101 - Convex Optimization. Amir Beck ... 6 You can employ Compressed Sensing / Sparse Representation for Super Resolution in Frequency Domain. One way to do so is solving the problem:$$ \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| F \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \lambda {\left\| \boldsymbol{x} \right\|}_{1} $$Where the {L}_{1} norm is sparsity inducing regularization ... 6 The decomposition of a separable filter is not unique, since u v' = (u a) (v a^{-1})' The solution you are expecting is found by using a -\sqrt{2} factor, namely -U(:,1) .* S(1,1) ./ sqrt(2) and -sqrt(2)*V(:,1). 5 Since \epsilon is a parameter you need to set why not trade it with another parameter you need to set to create an easily solvable problem (Relaxation of the Problem)? You can transform the problem into the following form ( {L}_{1} Regularized Least Squares):$$ \arg \min_{x} \frac{1}{2} \left\| A x - z \right\|^{2} + \lambda \left\| x \right\|_{1}$...

5

If you want to solve for single value of $\lambda$ in the model: $$\arg \min_{x} \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{1}$$ Then you can use Coordinate Descent method which is the fastest and simplest and doesn't require any matrix inversion. I have a MATLAB code for in my ${L}_{1}$ Regularized Least Squares ...

5

Here's the way I think about a discrete Wiener Filter Consider a sequence of observations $\mathbf{y} \in \Re^n$ Form a matrix from the input $\mathbf{x} \in \Re^{n+r-1}$ by shifting columns one sample each: $$X= \begin{bmatrix} x_1 & x_2 & ... & x_r \\ x_2 & x_3 & & x_{r+1} \\ x_3 & x_4 & & x_{r+2} \\ ... & &... 5 Remember that e^z has a very different meaning than e^x (taking z\in\mathbb{C} and x\in\mathbb{R}). If the exponent was real, then, as you state in your question:$$e^x = 1 \iff x=0$$However, when the exponent is complex, this function acquires a very different meaning. Let z=x+jy, where x,y\in\mathbb{R} and j is the imaginary unit. Then$$...

5

There are really great answers. I will try to give the Sequential Least Squares approach which generalizes to any Linear Model. Sequential Least Squares Model We're after solving the Linear Least Squares model: $$\arg \min_{\boldsymbol{\theta}} {\left\| H \boldsymbol{\theta} - \boldsymbol{x} \right\|}_{2}^{2}$$ Now imagine that we have new measurement ...

5

If you have a function $f \left[ m, n \right] \in \mathbb{R}^{M \times N}$ then the DFT of those functions is an orthogonal basis of functions in $\mathbb{C}^{M \times N}$. So if we have: $$F \left[ k , l \right] = \frac{1}{\sqrt{M N}} \sum_{m = 0}^{M - 1} \sum_{n = 0}^{N - 1} f \left[ m, n \right] {e}^{-j 2 \pi \left( \frac{k}{M} m + \frac{l}{N} n \... 5 You may solve it by 3 steps: Show yourself that a Cyclic Convolution with a vector  \boldsymbol{e}_{i}^{N}  is a Cyclic Shift Operator  {T}_{i - 1} \left( \cdot \right) . Where  \boldsymbol{e}_{i}^{N}  is defined as a vector of length  N  which all its elements is zero but the  i - th element which is 1. Decompose  \phi  into  3  vectors of ... 5 Usually Tikhonov Regularization is applied in the following form:$$ \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| \boldsymbol{x} \right\|}_{2}^{2} $$This formulation can be seen as: MAP Estimator with the prior of  \boldsymbol{x} \sim N \left( 0, {\sigma}_{x}^{2} \right) .... 5 The way to build the matrix is playing with indices of the signal data and the convolution kernel. For example: function [ mK ] = CreateConvMtx1D( vK, numElements, convShape ) % ----------------------------------------------------------------------------------------------- % % [ mK ] = CreateConvMtx1D( vK, numElements, convShape ) % Generates a Convolution ... 4 I will try giving you some intuition. The SVD says each matrix can be decomposed into 3 operations - Rotation, Stretching (Scaling) and the another Rotation. What matters is which directions are scaled and how. Directions are vectors (Pointing some direction). The SVD has many uses in Linear Algebra. One its most known use is Low Rank Approximation of a ... 4 This answer comes a little late, but I think that it's necessary to clear up some of the confusion about what the eigenstructure of a Laplacian is and how it is calculated. First of all, it's important to stress that this is not about properties of the local kernel used for calculating discrete derivatives. Instead you have to understand the Laplacian as ... 4 I'll give you a hint: You need the probability distribution of the alphabet (English and Turkish). For English see the Wikipedia "Letter Frequency" page. Now it's possible to compute the Entropy of this distribution. Note that this is a rough approximation (and an upper bound) since in practice, the letters of a word (and a sentence) are correlated. So, for ... 4 Most of the Denoisers in Image Processing make a simple assumption - The data has small number of freedom degrees while noise has high number. Hence if we try to represent the given noisy data with small number of parameters we probably match the data while exclude most of the noise. In the case above we try to limit the number of Singular Values of the ... 4 It's not valid. If any matrix W is invertible (such as the DFT matrix is), then there's the inverse W^{-1} with$$\begin{align} W^{-1}W &=I\\ &\implies\\ X &= Wx \\ &\iff\\ W^{-1} X &= W^{-1}W x\\ &= Ix\\ &=x\text{ .} \end{align}$$Now, the Discrete Fourier transform can be defined to be unitary, so that its inverse$$W^{-1}...

4

It is not clear what are you asking but I will try answer both things. Deriving the Matrix Inversion Lemma The Matrix Inversion Lemma goes as: $${\left( A + U C V \right)}^{-1} = {A}^{-1} - {A}^{-1} U {\left( {C}^{-1} + V {A}^{-1} U \right)}^{-1} V {A}^{-1}$$ Deriving it is by utilizing these useful identities: \begin{align} U + U C V {A}^{-1} U &...

4

The main advantage of OMP is that the residual is orthogonal to the current solution. Let's say you select all $k$ columns from $A$ (also called atoms) at once and let us also presume that $A$ is an overcomplete basis (this is more or less the standard in OMP literature). Now, with your method, if the atom that correlates the most with your measurements $... 4 Did you ever wonder about where$\pi $came from? Watch out... Let us first draw this weird function complex exponential$e^{-2j\pi t}$for several discrete values of$t\in[0,10]$(the little blue circles joined by line segments): On one axis, the variable$t$, on the others the real and imaginary parts, respectively. It looks like an infinite spring. ... 4 Least Squares solution is always well defined for Linear System of Equations. In your case, which is under determined it means there are many solutions to the Linear Equations. The Least Squares solution has nice property, it also minimizes the$ {L}_{2} \$ norm of the solution (Least Norm Solution) hence it is well defined. In practice, in order to solve ...

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