Let f ∈ C+ 2π with a zero of order 2p at z. Let r>p and m = n/r. Then there exists a constant c > 0 independent of n such that for all nsufficiently large, all eigenvalues of the preconditioned matrix C−1 n (Km,2r ∗ f)Tn(f) are larger than c.
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Let f ∈ C+ 2π with a zero of order 2p at z. Let r>p and m = n/r. Then there exists a constant c > 0 independent of n such that for all nsufficiently large, all eigenvalues of the preconditioned matrix C−1 n (Km,2r ∗ f)Tn(f) are larger than c.
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- If there is a non-singular matrix P such as P-1AP=D, matrix A is called a diagonalizable matrix. A, n x n square matrix is diagonalizable if and only if matrix A has n linearly independent eigenvectors. In this case, the diagonal elements of the diagonal matrix D are the eigenvalues of the matrix A. A=({{1, -1, -1}, {1, 3, 1}, {-3, 1, -1}}) : 1 -1 -1 1 3 1 -3 1 -1 a)Write a program that calculates the eigenvalues and eigenvectors of matrix A using NumPy. b)Write the program that determines whether the D matrix is diagonal by calculating the D matrix, using NumPy. #UsePythonGiven a matrix of dimension m*n where each cell in the matrix can have values 0, 1 or 2 which has the following meaning: 0: Empty cell 1: Cells have fresh oranges 2: Cells have rotten oranges So we have to determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] (up, down, left and right). If it is impossible to rot every orange then simply return -1. Examples: Input: arr[][C] = { {2, 1, 0, 2, 1}, {1, 0, 1, 2, 1}, {1, 0, 0, 2, 1}}; Output: All oranges cannot be rotten. Below is algorithm. 1) Create an empty Q. 2) Find all rotten oranges and enqueue them to Q. Also enqueue a delimiter to indicate beginning of next time frame. 3) While Q is not empty do following 3.a) While delimiter in Q is not reached (i) Dequeue an orange from queue, rot all adjacent oranges. While rotting the adjacents, make sure that time frame is incremented only once. And time frame is…we represent the finite-length signals as vectors in Euclidean space, many operations on signals can be encoded as a matrix-vector multiplication. Consider for example a circular shift in C: a delay by one (i.e. a right shift) transforms the signal x = (xo X1 X2]" into x = [xı xo xz]" and it can be described by the matrix TO D = [0 1 0'0 0 1'1 0 0] so that x = Dx. Determine the matrix F that implements the one step difference operator in C ie the operator that transforms a signal x into [(x - x)(x1 - x0)(x2 - 1)]
- Bu = ƒ and Cu = f might be solvable even though B and C are singular. Show that every vector f = Bu has ƒ1 + ƒ2+ ……. +fn = 0. Physical meaning: the external forces balance. Linear algebra meaning: Bu = ƒ is solvable when ƒ is perpendicular to the all – ones column vector e = (1, 1, 1, 1…) = ones (n, 1).Type in Latex **Problem**. Let $$A = \begin{bmatrix} .5 & .2 & .3 \\ .3 & .8 & .3 \\ .2 & 0 & .4 \end{bmatrix}.$$ This matrix is an example of a **stochastic matrix**: its column sums are all equal to 1. The vectors $$\mathbf{v}_1 = \begin{bmatrix} .3 \\ .6 \\ .1 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}$$ are all eigenvectors of $A$. * Compute $\left[\begin{array}{rrr} 1 & 1 & 1 \end{array}\right]\cdot\mathbf{x}_0$ and deduce that $c_1 = 1$.* Finally, let $\mathbf{x}_k = A^k \mathbf{x}_0$. Show that $\mathbf{x}_k \longrightarrow \mathbf{v}_1$ as $k$ goes to infinity. (The vector $\mathbf{v}_1$ is called a **steady-state vector** for $A.$) **Solution**. To prove that $c_1 = 1$, we first left-multiply both sides of the above equation by $[1 \, 1\, 1]$ and then simplify both sides:$$\begin{aligned}[1 \, 1\, 1]\mathbf{x}_0 &= [1 \, 1\, 1](c_1\mathbf{v}_1 +…Pls Use Python If there is a non-singular matrix P such as P-1AP=D , matrix A is called a diagonalizable matrix. A, n x n square matrix is diagonalizable if and only if matrix A has n linearly independent eigenvectors. In this case, the diagonal elements of the diagonal matrix D are the eigenvalues of the matrix A. A=({{1, -1, -1}, {1, 3, 1}, {-3, 1, -1}}) : 1 -1 -1 1 3 1 -3 1 -1 a)Write a program that calculates the eigenvalues and eigenvectors of matrix A using NumPy. b)Write the program that determines whether the D matrix is diagonal by calculating the D matrix, using NumPy. Ps: Please also explain step by step with " # "
- Using the below insights: obtain a matrix P such that if A is any matrix with 3 columns, AP is a cyclic shift of the columns of A (namely the first column of A is the second column of AP, second column of A is the third column of AP, and the third column of A becomes the first column of AP). # Let A = a-1, a-2, ..., a-n# x = x-1, x-2, ..., x-n# Ax = x-1*a-1 + x-2*a-2 + ... + x-n*a-n# [x1] [x1]# A = [x2] = [a1 a2 ... an]* [x2] = a1*x1 + a2*x2 + ... + an*xn# [...] [...]# [xn] [xn]Simplify the given function using K map f(a,b,c,d)= ∑m (1, 2, 4, 11, 13, 14, 15) + d (0, 5, 7, 8, 10)Consider a vector ('L', 'M', 'M', 'L', 'H', 'M', 'H', 'H'), where H stands for ‘high’, L stands for ‘low’, and M for ‘medium’ in rstudio. a. Convert this vector into an unordered factor. b. Convert this vector into an ordered factor with Low < Medium < High.
- write a C++ program to Given a matrix of dimension m*n where each cell in the matrix can have values 0, 1 or 2 whichhas the following meaning:0: Empty cell1: Cells have fresh oranges2: Cells have rotten orangesSo we have to determine what is the minimum time required so that all the oranges becomerotten. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1],[i,j+1] (up, down, left and right). If it is impossible to rot every orange then simply return -1.Examples:Input: arr[][C] = { {2, 1, 0, 2, 1},{1, 0, 1, 2, 1},{1, 0, 0, 2, 1}};Output:All oranges can become rotten in 2 time frames.Input: arr[][C] = { {2, 1, 0, 2, 1},Tahir Iqbal Department of Computer Sciences. BULC{0, 0, 1, 2, 1},{1, 0, 0, 2, 1}};Output:All oranges cannot be rotten.Below is algorithm.1) Create an empty Q.2) Find all rotten oranges and enqueue them to Q. Also enqueuea delimiter to indicate beginning of next time frame.3) While Q is not empty do following3.a) While delimiter in…Given the complement of a graph G is a graph G' which contains all the vertices of G, but for each unweighted edge that exists in G, it is not in G', and for each possible edge not in G, it is in G'. What logical operation and operand(s) can be applied to the adjacency matrix of G to produce G'? AND G's adjacency matrix with 0 to produce G' XOR G's adjacency matrix with 0 to produce G' XOR G's adjacency matrix with 1 to produce G' AND G's adjacency matrix with 1 to produce G'Let A be an m × n matrix with m > n. (a) What is the maximum number of nonzero singular values that A can have? (b) If rank(A) = k, how many nonzero singular values does A have?