# New PDF release: A Second Semester of Linear Algebra

By S. E. Payne

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Extra resources for A Second Semester of Linear Algebra

Sample text

Let f ∈ F [x], and let c ∈ F . Then f is divisible by x − c if and ony if f (c) = 0. Proof. By the theorem, f = (x − c)q + r where r is a scalar polynomial. 5, f (c) = 0q(c) + r(c) = r(c). Hence r = 0 if and only if f (c) = 0. Definition Let F be a field. An element c ∈ F is said to be a root or a zero of a given polynomial f ∈ F [x] provided f (c) = 0. 4. A polynomial f ∈ F [x] of degree n has at most n roots in F. Proof. The result is obviously true for polynomials of degree 0 or 1. We assume it to be true for polynomials of degree n − 1.

Is a determinant function. Furthermore, the formulas exhibited in Example 2. make it easy to see that the D given in Example 3. is the unique determinant function. 6. Let D be an n-linear function on n × n matrices over K. Suppose D has the property that D(A) = 0 when any two adjacent rows of A are equal. Then D is alternating. Proof. Let B be obtained by interchanging rows i and j of A, where i < j. We can obtain B from A by a succession of interchanges of pairs of adjacent rows. We begin by interchanging row i with row i + 1 and continue until the rows are in the order α1 , .

T (u) = b1 T (w1 ) + · · · br T (wr ). 34 CHAPTER 4. LINEAR TRANSFORMATIONS This last equation implies that (T (w1 ), . . , T (wr )) spans Im(T ), so at least Im(T ) is finite dimensional. To show that (T (w1 ), . . , T (wr )) is linearly independent, suppose that r ci T (wi ) = 0 for some ci ∈ F. i=1 It follows easily that ri=1 ci wi ∈ null(T ), so ri=1 ci wi = ki=1 di ui . Since (u1 , . . , uk , w1 , . . , wr ) is linearly independent, we must have that all the ci ’s and di ’s are zero. Hence (T (w1 ), .