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{{rpl|Identity matrix}}

...''_''i''} such that the [[overlap matrix]] is not equal to the [[identity matrix]], ...e metric tensor is a function of position and cannot be transformed to an identity matrix by a global transformation, i.e., by a single transformation holding on t

where '''E''' is the 3×3 identity matrix and superscript T indicates the transposed matrix. Clearly a rotation matri

where '''E''' is the 3×3 [[identity matrix]].

where we used '''B B'''^−1 = '''E''' (the identity matrix). \mathrm{Tr}( \mathbf{E}) &= n \qquad\hbox{(trace of identity matrix)}\\

...trix|Hermitian matrices]] which form an orthogonal basis (along with the [[identity matrix]]) for the real [[Hilbert space]] of 2&nbsp;&times;&nbsp;2 Hermitian matric ...mbol]], <math>\delta_{ij}</math> is the [[Kronecker delta]], and I is the identity matrix.

...als the tensor is a real number &sigma;₀ times a 3&times;3 [[identity matrix]]. The scalar &sigma;₀ is the [[conductivity coefficient]] and

...'&times;2''n'' matrix ''J'' that has 2&times;2 block structure and Id (the identity matrix) in the lower left block, -Id in the upper right block, and 0 in the rest.

...3&times;3 [[matrix]] obeying '''A'''^T'''A''' = '''E''' (the [[identity matrix]]), then rotation of the coordinate axes by '''A''' gives

...mp; '''t''' then the modified Eq. (1) gives a rotation matrix close to the identity matrix, which is to be expected and not very useful. ...basically due to the fact that the two vectors are connected by minus the identity matrix, '''which is not a rotation matrix'''. If we want to rotate '''f''' to '''

where <math>\mathbf{E}\,</math> is the [[identity matrix]].

...and ''a fortiori'' linearly independent, the matrix '''S''' becomes the [[identity matrix]]. Such a matrix has unit determinant. In that case ''K'' = (N!)<sup>1/2</s

where '''E''' is the 3&times;3 [[identity matrix]].

Hi Richard, I'm glad to see that you accepted my opinion that unit matrix and identity matrix are synonyms. At WP I had to fight to have "unit matrix" renamed to "matrix

...i.e., to "track") the desired output <math>y_k^d</math>. If ''C'' is the [[identity matrix]] and <math>y_k^d=x_k^d \in \mathbb{R}^n</math> then the tracking problem c

is in general not (a constant times) the [[identity matrix]], the two vectors, '''L''' and '''&omega;''', are in general not parallel, ...n of a position vector with respect to time, followed by insertion of the identity matrix gives,

Obviously, the identity matrix

The Cartesian metric tensor is the [[identity matrix]] and hence in Cartesian coordinates, ...e [[orthogonal matrix|orthogonal]] (matrix times its transpose gives the [[identity matrix]]), so that

If the AOs were orthonormal, '''S''' would be the [[identity matrix]]. '''S''' contains overlap integrals, which are of the form

where the last matrix on the right is the [[identity matrix]] '''I'''. Now