Originally Posted by

**beberenne** I have experimented with a 4X4 image. The matrix obtain from the image is B.

$\displaystyle

B = \begin{bmatrix} 1 & -3.8858e-016 & 1.0408e-016 & -2.7756e-017 \\ -3.8858e-016 & 1 & -2.3592e-016 & -2.7756e-01 \\ 1.0408e-016 & -2.3592e-016 & 1 & -1.8041e-016 \\ -2.7756e-017 & -2.7756e-016 & -1.8041e-016 & 1\end{bmatrix}

$

Originally Posted by

**beberenne** Sorry about the typo error. The matrix is symmetrical. The matrix is obtained through a test matrix A multiplied by A transposed. The matrix is generated from a test image of 4 X4. It seems that any image i use, i still get a symmetrical matrix with ones in the diagonal.

The matrix B comes from the product of A and A transposed.

$\displaystyle A= \begin{bmatrix} 0.12452 & 0.65243 & 0.56703 & 0.48714 \\ 0.32858 & 0.34674 & 0.25193 & -0.84163 \\ 0.49466 & -0.65422 & 0.56457 & 0.092583 \\ 0.79489 & 0.16158 & -0.5443 & 0.21398\end{bmatrix}

$

$\displaystyle A'= \begin{bmatrix} 0.12452 & 0.32858 & 0.49466 & 0.79489 \\ 0.65243 & 0.34674 & -0.65422 & -0.16158 \\ 0.56703 & 0.25193 & 0.56457 & -0.5443 \\ 0.48714 & -0.84163 & 0.092583 & 0.21398\end{bmatrix}

$

Originally Posted by

**beberenne** JakeD, if i let the roundings to be zero, then i would not get the correct answer. As you can see, matrix B is the product of A and A transpose as shown above. By letting the roundings to be zero, then i would not get back A.

Multiplying AA' yields

Code:

1.0000e+00 -1.4105e-06 -8.6817e-06 3.1304e-06
-1.4105e-06 1.0000e+00 2.6298e-06 -6.2810e-06
-8.6817e-06 2.6298e-06 1.0000e+00 6.8791e-06
3.1304e-06 -6.2810e-06 6.8791e-06 1.0000e+00

which is not the matrix B you gave above. Is this a different matrix, or is it different because the A matrix wasn't printed with enough precision? Anyway, the B matrix still looks strange. Why is it so close to the identity matrix?