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Isomorphisms, Linear Operators, Eigenvalues and Eigenvectors

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Martin, Burda

Lecture 7 notes by Y. Burda 1 Isomorphism of vector spaces We feel that the vector space of quadratic polynomials is the same as the 2 vector space of triples of thei cecients: instead of a polinomial a+bx+cx we can record the coecients b ; if we want to add two polynomials, we c can instead add the triples of their coecient and if we want to multiply a polynomial by a number, we can instead multiply its coecients by this number. As far as vector space structure goes, this is enough in an abstract vector space all we can do is add vectors and multiply them by scalars. The technical word for having vector spaces, which are the same is having isomorphic vector spaces. Lets try to formulate precisely when two vector spaces V and W are isomorphic: we want to associate to any vector of V a vector of W in such a way that the vector associated to v +1v wi2l be the sum of the vectors associated to v 1nd v .2We also want that the vector associated to v will be the product of and the vector associated to v. Moreover, we want exactly one vector of V to be associated to any vector in W. Thus we arrive at the following denition: Denition. Vectors spaces V and W over the same eld K are called iso- morphic if there exists an invertible linear transformation T : V W. Such a transformation is called an isomorphism of V and W. Example: P (n) is isomorphic to K n+1 with the isomorphism T(a + 0 a0 a x + ... + a x ) = . . 1 n a n We can generalize this example and prove the following theorem: Theorem. If dimV = n, then V is isomorphic to K . n n Proof. Choose a basis A = (v ,..1,v ) on V and let T : V K be the transformation that sends each vector to its coordinates relative to A: T(v) = [v]A Ths transformation is linear (exercise) and is invertible: its inverse is 1 T 1 . = 1 1+ ... + n n n 1
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