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Dml.cz ANNIHILATORS IN NORMAL AUTOMETRIZED ALGEBRAS Ivan Chajda and Ji Rach nek Olomouc u Received November 14 1997 Abstract. The concepts of an annihilator and a relative annihilator in an autometrized l-algebra are introduced. It is shown that every relative annihilator in a normal autometrized l-algebra A is an ideal of A and every principal ideal of A is an annihilator of A. The concepts of an annihilator and a relative annihilator in an autometrized l-algebra are introduced. It is shown...

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Hello IN×39’m Sheldon Axle the author of linear algebra done right this video discusses part two of the section the book titled duality the focus of this video is annihilators and the matrix of dual map if T is a linear map then we would like to be able to describe the null space and the range of the dual map Prime to do that we need the concept of annihilator of a subset of V the supposed is a subset of our vector space V annihilator of U denoted U with superscript 0 is defined to be the Seton linear functional son V such as fee of U equals 0 for all vectors in U reuse the superscript 0 to remind us of the 0 that appears in the definition note that the Annihilator of U is subspace of the dual of V thus annihilator review depends upon Va swell as upon you however we do not put VIN the notation because our choice of Will always be obvious from the contextlet'’s look at some examples firs first example the Annihilator of the subspace consisting of just 0 is the hole vector space V Prime this is because for every fee in the dual of V of 0 is equal to0 our next example states that annihilator of the entire vector space Vi's just the linear functional 0 that×39;because the only linear functional that0 on all of V is 0 linear functionality×39’s look at a more complicated example now for a vector space V we×39’ll use P of which is the vector space consisting of all polynomials with real coefficients for subspace you let#39’s take all polynomial multiples of x squared let V be the linear functional when Po fare defined by V of a polynomial P is the derivative at 0 of The claim is that fee than is annihilator of you why is that we have to show that if we take a polynomial in you then we apply fee to it, we get zero well that follows from the product rule thinking about fee of X applied to squared times the polynomial that#39’s the derivative of x squared times the polynomial and by the product rule whence evaluate that at zero we will get zero be sure you verify this example the next result states that a few is subset of V than the Annihilator of U ISA subspace of the dual of an is very easy to make sure you understand why it×39;true our next theorem gives us a way to compute the dimension of the Annihilator of a subspace u this theorem says that if these finite dimensional and u is subspace of V then the dimension of Plus the dimension of the Annihilator view is equal to the dimension of V we could solve that equation for the dimension of the Annihilator view getting a formula for the dimension of the Annihilator view in terms of the dimension of V and the dimension of you please be sure to read the proof of this result in the book now we come to uninteresting result but allows us to characterize the null space and the range of the dual of a linear map interns of annihilators here's what the result says suppose V and W are finite dimensional vector spaces and T is linear map from V to W then the null space of the dual of T is equal to annihilator of...

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