Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 2007 — Problem Set 6 Solutions *1. Let P be the linear space of all polynomials (of any degree). Show that no finite set of polynomials { p 1 ( t ) , p 2 ( t ) , . . ., p m ( t ) } can span all of P . (Hint: since there are a finite number of polynomials in my set, there is one of largest degree.) Let N be the largest of the degrees of the polynomials p 1 ( t ) , p 2 ( t ) , . . ., p m ( t ) . Then no linear combination of these polynomials can have degree greater than N . Since P does contain polynomials of degree greater that N (e.g., it contains t N + 1 ), the set { p 1 ( t ) , p 2 ( t ) , . . ., p m ( t ) } does not span all of P . *2. Suppose we are given a set of vectors { v 1 , v 2 , . . ., v k } in R n . Then we can create an n × k matrix by using the vectors v i , in order, as the columns. Call that matrix A . a. What property of A will guarantee that { v 1 , v 2 , . . ., v k } is linearly indepen- dent? Independence tells us that the equation x 1 v 1 + x 2 v 2 + · · · + x k v k = 0 has a unique solution, namely setting all the x i to zero. Now, given how we defined the matrix A , this is the same as saying that the equa- tion Ax = 0 has a unique solution (namely, x = 0 ). That will happen if and only if rref ( A ) has a pivot in every column, so that there are no free variables.
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