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scalar multiple वाक्य

"scalar multiple" हिंदी मेंscalar multiple in a sentence
उदाहरण वाक्यमोबाइल
  • This is a group because the product of any two scalar multiples of the identity matrix is another scalar multiple of the identity matrix, and the product of any powers of 2 is another power of 2.
  • This is a group because the product of any two scalar multiples of the identity matrix is another scalar multiple of the identity matrix, and the product of any powers of 2 is another power of 2.
  • A positive scalar multiple of a solution is also a solution, so you could convert this in various ways to a question about affine constraints in \ mathbb { R } ^ { m-1 }.
  • The "'classification theorem "'for C 0 contractions states that two multiplicity free C 0 contractions are quasi-similar if and only if they have the same minimal function ( up to a scalar multiple ).
  • This is somewhat elementary, and perhaps you know this already . but in the special case that all your blocks are scalar multiples of each other, then it is a Kronecker product and you can get the eigenvalues of the big matrix.
  • To prove this result one notes first that a representation is irreducible if and only if the commutant of ? ( " A " ), denoted by ? ( " A " )', consists of scalar multiples of the identity.
  • If q _ 0 + iq _ 1 + jq _ 2 + kq _ 3 is not a unit quaternion then the homogeneous form is still a scalar multiple of a rotation matrix, while the inhomogeneous form is in general no longer an orthogonal matrix.
  • This shows that ? is irreducible if and only if any such ? " g " is unitarily equivalent to ?, i . e . " g " is a scalar multiple of " f ", which proves the theorem.
  • A function that is Fr�chet differentiable at a point is necessarily continuous there and sums and scalar multiples of Fr�chet differentiable functions are differentiable so that the space of functions that are Fr�chet differentiable at a point form a subspace of the functions that are continuous at that point.
  • Since the space of modular forms of weight 2 " k " has dimension 1 for 2 " k " = 4, 6, 8, 10, 14, different products of Eisenstein series having those weights have to be equal up to a scalar multiple.
  • The set of harmonic functions on a given open set " U " can be seen as the kernel of the Laplace operator ? and is therefore a vector space over "'R "': sums, differences and scalar multiples of harmonic functions are again harmonic.
  • Recall that " n "-dimensional projective space \ mathbb { P } ^ n is defined to be the set of equivalence classes of non-zero points in \ mathbb { A } ^ { n + 1 } by identifying two points that differ by a scalar multiple in " k ".
  • The set of real scalar multiples of this null vector, called a " null line " through the origin, represents a " line of sight " from an observer at a particular place and time ( an arbitrary event we can identify with the origin of Minkowski spacetime ) to various distant objects, such as stars.
  • When the condition number is exactly one ( which can only happen if " A " is a scalar multiple of a linear isometry ), then a solution algorithm can find ( in principle, meaning if the algorithm introduces no errors of its own ) an approximation of the solution whose precision is no worse than that of the data.
  • This holds more generally for any algebra " R " over an uncountable algebraically closed field " k " and for any simple module " M " that is at most countably-dimensional : the only linear transformations of " M " that commute with all transformations coming from " R " are scalar multiples of the identity.
  • Alternatively, it is possible to rewrite the equation of the plane using dot products with \ mathbf { p } in place of the original dot product with \ mathbf { v } ( because these two vectors are scalar multiples of each other ) after which the fact that \ mathbf { p } is the closest point becomes an immediate consequence of the Cauchy Schwarz inequality.
  • Scalar multiples of the identity operator of course commute with everything, and are usually identified with the scalars themselves, so one can seemingly add a scalar to an operator without multiplying it with I first . ( The self-adjoint operators are, however, closed under the operation p, q \ mapsto i ( pq-qp ), which makes them into a Lie algebra.
  • The elements of the polynomial ring k [ x _ 0, \ dots, x _ n ] are not functions on \ mathbb { P } ^ n because any point has many representatives that yield different values in a polynomial; however, for homogeneous polynomials the condition of having zero or nonzero value on any given projective point is well-defined since the scalar multiple factors out of the polynomial.
  • If a vector x maximizes R ( M, x ), then any non-zero scalar multiple kx also maximizes R, so the problem can be reduced to the Lagrange problem of maximizing \ sum _ { i = 1 } ^ n \ alpha _ i ^ 2 \ lambda _ i under the constraint that \ sum _ { i = 1 } ^ n \ alpha _ i ^ 2 = 1.
  • Specializing further, if it happens that " M " has a Riemannian metric on which " G " acts transitively by isometries, and the stabilizer subgroup " G " x of a point acts irreducibly on the tangent space of " M " at " x ", then the Casimir invariant of ? is a scalar multiple of the Laplacian operator coming from the metric.
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