If E is a K-ev finite dimensional, then dim (E) = dim (E *) = dim (E **). I - 2 - 2 Comments In fin

I - 1 - Introduction I - 1 - 1. Prrequis The study of the duality ncessite few pralables knowledge. The K-vector kitchen interior design space structure (Abbrev in K-ev). The notion of subspace (in Abbrev sev). The lees families, free, gnratrices, bases, dimensions, especially the basic body K can be seen as a K-ev dimension 1 whose canonical basis is BK = (1 K) No. 1 is the neutral K K. The notion of multiplicative linear application of E to K-ev-ev K F, its core, its image. In finite dimensions, the formula dim (E) = dim (Ker (f)) + dim (Im (f)) The K-ev L (E, F) of linear maps from E to F. In finite dimensions, the matrix of a linear application. In finite dimensions, the formula: dim [L (E, F)] = dim (E) dim (F) The Kronecker kitchen interior design symbol: ij = 1 if i = j and ij = 0 if i j. I - 1 - 2. polynmes Lagrange connatre It will be interesting to those polynmes kitchen interior design ultrieures some questions. For n> 1, n considrons separate elements of K: A 1, A 2, ..., an. Seek the n polynmes L 1, L 2, ..., L n of degree common n-1, the following dfinis Manire. For each i, 1 in the polynme L ia roots for a 1, ..., a i-1, i + 1, ..., an and takes the value 1 have. Note that these conditions rsument in:
Let a 1, ..., separate scalar ann, n> 2. polynmes is called Lagrange Associs kitchen interior design these scalar n, n polynmes L 1, L 2, ..., L ndfinis by: for every pair (i, j), L i (aj) = ij. So: It is quite simple to prove that the family (L 1, ..., L n) is a basic n-1 K [X]. I - 1 - 3. Examples kitchen interior design preliminary as Ex I - 1. E is the R-ev R 3, a, b, c three donns ral. Considrons kitchen interior design the u: u: ER dfinie kitchen interior design by u ((x, y, z)) = ax + by + cz Ex I - 2. E is the C-ev continuous applications [0,1 ] C. Considrons in the mapping v: v: EC dfinie by v (f) = Ex I - 3. E = R [X] is the R-ev of polynmes or has the ment of R. Considrons fix the implementation w: w: ER dfinie by w (P) = P (k) (a) Ex I - 4. E a dimension of K-ev n> 0, BE = (e 1, ..., in) a basic E. Then all lment x E possde a single tuple of coordinated on BE: (x 1, ..., xn) such that: For any j between 1 and n, considrons the implementation pj pj: EK dfinie by pj (x) = x j = x coordonnenj. pj is called j-me BE projection report. Ex I - 5. U an open R n, has a point of U, f: UR has a diffrentiable application. The diffrentielle of f at a is the daf Application: R n R dfinie by these five examples borrowing very varis fields of mathematics have in common is introduced applications linear from E to K, K as considr as a K-ev. There is thus the omnipresence such applications in mathematics. I - DEFINITIONS 2. Large I - 2 - 1 linear forms, dual space, bidual space
Let E be a K-ev. Called kitchen interior design linear form on E any linear mapping from E into K. The K-ev L (E, K) of these linear forms is called the dual space of E. It is noted E *. As E * = L (E, K) is a K-ev, it is then considrer K-ev L (E * K) of the linear kitchen interior design forms of E * to K. This will be the space of linear forms of E * to K, so the dual space E *. It's called the bidual E and E ** note. Suppose E of finite dimension. Since dim (K) = 1, using the result: dim [L (E, F)] = dim (E) dim (F) is obtained dim (E *) = dim (E **) = dim (E).
If E is a K-ev finite dimensional, then dim (E) = dim (E *) = dim (E **). I - 2 - 2 Comments In finite dimension, the equality between dim (E), dim (E *) and dim (E **) is remarkable. By cons, if dim (E) is not over, the situation is not as simple. Consider an example. Ex I - 6. Let E = R [X]. Each rel has considrons the map f (a): ER dfinie f (a) (P) = P (a). It verifies that for any rel a, f (a) is a linear form E. In fact, for any P and Q in E, and u and v in any R: f (a) ( uP + vQ) = (uP + vQ) (a) = uP (a) + vQ (a) = uf (a) (P) + vf (a) (Q). Let us study the family (f (a)) a R. We know that this family is free iff every finite subfamily is free. So let's take separate nrels: a 1, ..., an and root inclination is studied family (f (a 1), ..., f (an)) in E *. For that we will solve the equation: where ui are unknown nrels and O * toadflax form zero. Then, for any polynme P, one can write: This formula is especially Last verifies for polynmes Lagrangian L 1, ..., L n Associs to have. This gives, for all j such that 1 jn: Given the formula L j (ai) = ij, remains: for all j such that 1 jn, uj = 0. Conclusion: the family of linear forms (f (ai)) 1 <i <n is free. So the family (f (a)) a R is free. This means that the dual space of R [X] contains a free family of cardinal one of R, then R [X] possde "simply" a dnombrable base. I - 2 - 3. Cro