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Example text

Of vectors al, (Ye,. . , (Y,,,in S is clearly in W. Thus W contains the set L of all linear combinations of vectors in S. The set L, on the other hand, contains S and is non-empty. If (Y, /3 belong to L then CYis a linear combination, CY= Xlffl + x2ayz+ * f * + XmQ, of vectors (pi in S, and ,B is a linear combination, P = of vectors YlPl + Y2P2 + * * * + Y&I @j in S. For each scalar c, Cff + P = 5 (CXi)ai +jgl yjPj* i=l Hence ca! + ,Obelongs to L. Thus L is a subspace of V. Now we have shown that L is a subspace of V which contains S, and also that any subspace which contains S contains L.

Find a basis {AI, AZ, Aa, 44) 9. Let V be a vector space over a subfield F of the complex numbers. Suppose a, fi, and y are linearly independent vectors in V. Prove that (a + /3), (0 + y), and (y + (Y) are linearly independent. 10. Let V be a vector space over the field F. Suppose there are a finite of vectors al, . . , LYEin V which span V. Prove that V is finite-dimensional. number 11. Let V be the set of all 2 X 2 matrices A with complex entries which satisfy AlI $ Azz = 0. (a) Show that V is a vector space over the field of real numbers, with the usual operations of matrix addition and multiplication of a matrix by a scalar.

K,. For each j in J, let Ei be the solution obtained by setting xj = 1 and xi = 0 for all other i in J. We assert that the (n - r) vectors Ej, j in J, form a basis for the solution space. Since the column matrix Ej has a 1 in row j and zeros in the rows indexed by other elements of J, the reasoning of Example 13 shows us that the set of these vectors is linearly independent. That set spans the solution space, for this reason. If the column matrix T, with entries t1, . . , t,, is in the solution space, the matrix N = 2; tjEj J is also in the solution space and is a solution such that xi = tj for each j in J.

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