Let T be a linear operator on the finite-dimensional vector space V over the field F.

Let p be the minimal polynomial for T, Af, where the Af, are distinct irreducible monic polynomials over F and the Af are positive integers.

Let T be a linear operator on the finite-dimensional vector space V over the field F.

* Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i. e. S ⊆ Q ⊆ T. If there is a unique number c such that a ( S ) ≤ c ≤ a ( T ) for all such step regions S and T, then a ( Q )

Using terms from formal language theory, the precise mathematical definition of this concept is as follows: Let S and T be two finite sets, called the source and target alphabets, respectively.

Let M be a smooth manifold and let x be a point in M. Let T < sub > x </ sub > M be the tangent space at x.

Let the line of symmetry intersect the parabola at point Q, and denote the focus as point F and its distance from point Q as f. Let the perpendicular to the line of symmetry, through the focus, intersect the parabola at a point T. Then ( 1 ) the distance from F to T is 2f, and ( 2 ) a tangent to the parabola at point T intersects the line of symmetry at a 45 ° angle.

Let us denote the time at which it is decided that the compromise occurred as T.

Let T be a linear operator represented by the matrix

Let V and W be vector spaces ( or more generally modules ) and let T be a linear map from V to W. If 0 < sub > W </ sub > is the zero vector of W, then the kernel of T is the preimage of the zero subspace

Let T: X → X be a contraction mapping on X, i. e.: there is a nonnegative real number q < 1 such that

Let T be a set consisting of all infinite sequences of 0s and 1s.

Let K be a closed subset of a compact set T in R < sup > n </ sup > and let C < sub > K </ sub > be an open cover of K. Then

Let the curve be a unit speed curve and let t = u × T so that T, u, t form an orthonormal basis: the Darboux frame.

Let ( m, n ) be a pair of amicable numbers with m < n, and write m = gM and n = gN where g is the greatest common divisor of m and n. If M and N are both coprime to g and square free then the pair ( m, n ) is said to be regular, otherwise it is called irregular or exotic.

Let denote the Bézier curve determined by the points P < sub > 0 </ sub >, P < sub > 1 </ sub >, ..., P < sub > n </ sub >.

Let P < sub > F </ sub > be the domain of a prefix-free universal computable function F. The constant Ω < sub > F </ sub > is then defined as

Let M be a smooth manifold and let x be a point in M. Let I < sub > x </ sub > be the ideal of all functions in C < sup >∞</ sup >( M ) vanishing at x, and let I < sub > x </ sub >< sup > 2 </ sup > be the set of functions of the form, where f < sub > i </ sub >, g < sub > i </ sub > ∈ I < sub > x </ sub >.

Let M be a smooth manifold and let f ∈ C < sup >∞</ sup >( M ) be a smooth function.

Let e be the error in b. Assuming that A is a square matrix, the error in the solution A < sup >− 1 </ sup > b is A < sup >− 1 </ sup > e.

Let us for simplicity take, then < math > 0 < c =- 2a </ math > and.

Moving to groups in general, let H be a subgroup of some group G. Let ~ be an equivalence relation on G, such that a ~ b ↔ ( ab < sup >− 1 </ sup > ∈ H ).

Let r be a non zero real number and let the r < sup > th </ sup > power mean ( M < sup > r </ sup > ) of a series of real variables ( a < sub > 1 </ sub >, a < sub > 2 </ sub >, a < sub > 3 </ sub >, ... ) be defined as

Let t and s ( t > s ) be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then s < sup > 2 </ sup > equals half the harmonic mean of c < sup > 2 </ sup > and t < sup > 2 </ sup >.

Genesis 1: 9 " And God said, Let the waters be collected ". Letters in black, < font color ="# CC0000 "> niqqud in red </ font >, < font color ="# 0000CC "> cantillation in blue </ font >

* Let D < sub > 1 </ sub > and D < sub > 2 </ sub > be directed sets.

Let the open enemy to it be regarded as a Pandora with her box opened ; ;

Let every policeman and park guard keep his eye on John and Jane Doe, lest one piece of bread be placed undetected and one bird survive.

`` Let him be now ''!!

Let us assume that it would be possible for an enemy to create an aerosol of the causative agent of epidemic typhus ( Rickettsia prowazwki ) over City A and that a large number of cases of typhus fever resulted therefrom.

Let Af be the null space of Af.

Let N be a linear operator on the vector space V.

Let V be a finite-dimensional vector space over an algebraically closed field F, e.g., the field of complex numbers.

Let N be a positive integer and let V be the space of all N times continuously differentiable functions F on the real line which satisfy the differential equation Af where Af are some fixed constants.

Let Q be a nonsingular quadric surface bearing reguli Af and Af, and let **zg be a Af curve of order K on Q.

Let us take a set of circumstances in which I happen to be interested on the legislative side and in which I think every one of us might naturally make such a statement.

Let the state of the stream leaving stage R be denoted by a vector Af and the operating variables of stage R by Af.

Let this be denoted by Af.

Let it be granted then that the theological differences in this area between Protestants and Roman Catholics appear to be irreconcilable.

Let not your heart be troubled, neither let it be afraid ''.

The same God who called this world into being when He said: `` Let there be light ''!!

For those who put their trust in Him He still says every day again: `` Let there be light ''!!

Let us therefore put first things first, and make sure of preserving the human race at whatever the temporary price may be ''.

Let her out, let her out -- that would be the solution, wouldn't it??