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3 Consider the vector space R3. The vectors (1,0,0), (0, 1,0) generate a subspace of R3, say 8. Show that 8{(0, 0, I)} and 8{(1, 1, I)} are two possible complementary one-dimensional subspaces of 8. Show that, in general, the choice of a complementary subspace 8 c of 8 c V is not unique. 4 Let 8 1 and 8 2 be the subspaces of the vector space R3 spanned by {(I, 0, 0), (0,0, I)} and {(O, 1, 1), (1,2, 3)}, respectively. Find a basis for each of the subspaces 8 1 n 8 2 and 8 1 + 8 2. 5 Let F = {O, 1, 2} with addition and multiplication defined modulo 3.

3) spans all solutions, then it follows that they form a basis for the vector space S, and consequently, dimeS) If b = (f3}, /32, ... ,13m) =m - r =m - dim(Y I ). , b is a linear combination of bl, b2 , ••• ,bm - r . Use the fact that Xl! 2). This completes the proof. 1) is the socalled non-homogeneous equation, for some known vector y =I O. 1} always has a solution, namely, the null vector in Fm, a nonhomogeneous equation may not have a solution. Such an equation is said to be inconsistent. For example, let Xl = (I, 1, 1), X2 = {I, 0, I} and X3 = {2, 1, 2} be three vectors in the vector space R 3 {R).

It is known that every vector space admits a basis. If S consists of only the zero-vector, then S is a zero-dimensional subspace of V. If every vector in S is of the form ax for some fixed non-zero vector x and for some a in F, then S is a one-dimensional subspace of V. If every vector in S is of the form aXl + f3x2 for some 25 Vector Spaces fixed set of linearly independent vectors Xl and X2 and for some 0: and f3 in F, then S is a tw~dimensional subspace of V . The schematic way we have described above is the way one generally obtains subspaces of various dimensions.

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Actes Du Congres International Des Mathematiciens 1970 by Anon.


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