Let Tmn be a BTTB matrix with a generating function f(x, y) ∈ C2π×2π. Let λmin(Tmn) and λmax(Tmn) denote the smallest and largest eigenvalues of Tmn, respectively. Then we have fmin ≤ λmin(Tmn) ≤ λmax(Tmn) ≤ fmax, where fmin and fmax denote the minimum and maximum values of f(x, y), respectively. In particular, if fmin > 0, then Tmn is positive definite
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Let Tmn be a BTTB matrix with a generating function f(x, y) ∈
C2π×2π. Let λmin(Tmn) and λmax(Tmn) denote the smallest and largest eigenvalues
of Tmn, respectively. Then we have
fmin ≤ λmin(Tmn) ≤ λmax(Tmn) ≤ fmax,
where fmin and fmax denote the minimum and maximum values of f(x, y), respectively. In particular, if fmin > 0, then Tmn is positive definite.
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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.Given 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…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]
- 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. #UsePythonwrite 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…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 +…
- Given and nxn matrix filled with either 0 or 1 your task is to turn as few 0 as possible into 1 such that every index meets the requirement that its neighbour (i+1,i-1,j+1,j-1) sum is even. return minimum no of zero you need to convert into 1, return -1 if no soltuion possible constraints : 1<=tc<=10 1<=n<=10 input: tc = 2 n = 3 1 1 1 1 1 1 0 0 0 n = 3 0 0 0 1 0 0 0 0 0 output : for first tc : -1 for second : 3we 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)]Let C be a linear [n, k]-code with alphabet GF(q). Let C+ be any code obtained from C by lengthening, that is, first adding one extra bit to every codeword of C and then adding one extra row to a generating matrix for C while maintaining linear independence. (a) Find the length and dimension of C+. (b) Suppose C is perfect and 1-error-correcting. Prove C+ cannot be 1-errorcorrecting.
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