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Furthermore, the product of an anti-diagonal matrix with a diagonal matrix is anti-diagonal, as is the product of a diagonal matrix with an anti-diagonal matrix. An anti-diagonal matrix is invertible if and only if the entries on the diagonal from the lower left corner to the upper right corner are nonzero. The inverse of any invertible anti ...
A circle corresponds to a multiple of the identity matrix. A near-circle corresponds to a near-multiple of the identity matrix whose eigenvalues are nearly equal to the diagonal entries of the matrix. Therefore, the problem of approximately finding the eigenvalues is shown to be easy in that case.
The matrix of the linear map mapping the vector of the entries of a matrix to the vector of a part of the entries (for example the vector of the entries that are not below the main diagonal) See vectorization: Exchange matrix: The binary matrix with ones on the anti-diagonal, and zeroes everywhere else. a ij = δ n+1−i,j: A permutation matrix.
The sum of the entries along the main diagonal (the trace), plus one, equals 4 − 4(x 2 + y 2 + z 2), which is 4w 2. Thus we can write the trace itself as 2w 2 + 2w 2 − 1; and from the previous version of the matrix we see that the diagonal entries themselves have the same form: 2x 2 + 2w 2 − 1, 2y 2 + 2w 2 − 1, and 2z 2 + 2w 2 − 1. So ...
Any great circle s (blue) passing through the poles is secondary to g. In mathematics, two points of a sphere (or n-sphere, including a circle) are called antipodal or diametrically opposite if they are the endpoints of a diameter, a straight line segment between two points on a sphere and passing through its center. [1]
Proof: Let D be the diagonal matrix with entries equal to the diagonal entries of A and let B ( t ) = ( 1 − t ) D + t A . {\displaystyle B(t)=(1-t)D+tA.} We will use the fact that the eigenvalues are continuous in t {\displaystyle t} , and show that if any eigenvalue moves from one of the unions to the other, then it must be outside all the ...
More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint AD.
Thus = and = where is a diagonal matrix whose diagonal values are in general complex. The left and right singular vectors in the singular value decomposition of a normal matrix A = U D V ∗ {\displaystyle A=UDV^{*}} differ only in complex phase from each other and from the corresponding eigenvectors, since the phase must be factored out of the ...