Search results
Results from the WOW.Com Content Network
If you want to transpose a matrix like A = np.array([[1,2],[3,4]]), then you can simply use A.T, but for a ...
The transformation can as you said be written as a linear transformation, but in the vector representation of the matrix: $$ T(\mathrm{vec}(\mathbf(A))) = \mathbf{P}\mathrm{vec}(\mathbf(A)) $$ where $\mathbf{P}\in\mathbb{R}^{n^2\times n^2}$. The matrix $\mathbf{P}$ is a permutation matrix known as a stride permutation or a perfect shuffle matrix.
4. Lets consider the matrix A characterizing the values of some variables aij, j = 1...m with values at different times i = 1...n, as in the OP example, but transposed. If the variables are normalized in mean, the matrix 1 mATA is the estimator of the covariances sj1j2 = E(a ⋅ j1a ⋅ j2) ≈ 1 m ∑ aj1aj2 for the set of random variables a ...
Sorted by: 50. Most people would just claim this is obvious and omit the proof, but if you don't want to do that then perhaps you could first prove that. [M N]T =[MT NT] [M N] T = [M T N T] and. [M N]T =[MT NT]. [M N] T = [M T N T]. Then. [A C B D]T =⎡⎣⎢⎢⎢⎢[A C]T [B D]T⎤⎦⎥⎥⎥⎥ =[AT BT CT DT].
I can follow the definition of the transpose algebraically, i.e. as a reflection of a matrix across its diagonal, or in terms of dual spaces, but I lack any sort of geometric understanding of the transpose, or even symmetric matrices. For example, if I have a linear transformation, say on the plane, my intuition is to visualize it as some ...
Lets have invertible matrix A, so you can write following equation (definition of inverse matrix): AA − 1 = I. Lets transpose both sides of equation. (using IT = I , (XY)T = YTXT) (AA − 1)T = IT. (A − 1)TAT = I. From the last equation we can say (based on the definition of inverse matrix) that AT is inverse of (A − 1)T.
Especially the following formula over there leaves no doubt that a matrix multiplied with its transpose IS something special: B = [B(BTB) − 1 / 2][(BTB)1 / 2] Least Squares methods (employing a matrix multiplied with its transpose) are also very useful with Automated Balancing of Chemical Equations. Share. Cite.
38. So the obvious way to transpose a matrix is to use : for( int i = 0; i < n; i++ ) for( int j = 0; j < n; j++ ) destination[j+i*n] = source[i+j*n]; but I want something that will take advantage of locality and cache blocking. I was looking it up and can't find code that would do this, but I'm told it should be a very simple modification to ...
Show activity on this post. I'm trying to transpose a 2D matrix (10x10) in place: for (b = 0; b < 10; b++) {. tmp = matrix[a][b]; matrix[b][a] = matrix[a][b]; matrix[a][b] = tmp; If I can increase the starting value 'b' of the inner for statement by 1, it works fine. However, when one loop is turned, the value of the variable is set to 0.
For two operand instructions using SIMD you can show that the number of operations necessary to transpose a nxn matrix is n*log_2(n) whereas using scalar operations it's O(n^2). In fact, later I'll show that the number of read and write operations using the scalar registers is 2*n*(n-1). Below is a table showing the number of operations to ...