Solution: By inspection, the standard matrix of T is: A = Where, Ae 1 = 3. , and Ae 3 = 4. , Ae 2 =

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1 This is a typical assignment, but you may not be familiar with the material. You should also be aware that many schools only give two exams, but also collect homework which is usually worth a small part of the grade. Many schools also require the homework to be typed and I believe everyone should be learning to use their computers to prepare and present information. 1. Problem 2, 1.9 T : R R 2, T (e 1 ) (1, ), T (e 2 ) (4, 7), T (e ) ( 5, 4), where e 1, e 2, e are the columns of the identity matrix. Assume that T is a linear transformation, find the standard matrix of T. Solution: By inspection, the standard matrix of T is: A 7 4 Where, Ae 1 1, Ae , and Ae 4 2. Problem 16, 1.9 Fill in the missing entries of the following matrix, assuming that the equation holds for all values of the variables.?? x 1?? x1 2x x 1 +?? 2 x 1 Solution: By inspection: x1 x 1 2x 1 + x 1. Problem 21, 1.9 Let T : R 2 R 2 be a linear transformation such that T (x 1, ) (x 1 +, 4x ). Find x such that T (x) (, 8) Solution: By inspection: x1 x1 + x 2 4x Which leads to this matrix equation: x1 8 Rewriting as an augmented matrix:

2 Now, the reduced row echelon form of this augmented matrix: Hence, the vector x 4 4. Problem 25, 1.9 Let T (x 1,, x, x 4 ) (0, x 1 +, + x, x + x 4 ). Determine if this linear transformation is (a) one-to-one and (b) onto. Justify each answer. Solution: By inspection: Where: A A x 1 x x x x x + x 4 Definition: 1 A mapping T : R n R m is said to be one-to-one if each b in R m is an image of at most one x in R n. Definition: A mapping T : R n R m is said to be onto R m if each b in R m is an image of at least one x in R n. To see if this linear transformation is onto, we need to find at least one x R 4, such that Ax b, for all b R 4. The augmented matrix is: b b b b 4 Here, we can clearly see that b 1 0, and therefore this system does not include all b R 4, i.e. A does not span R 4. This linear transformation is not onto. To see if this linear transformation is one-to-one, solve Ax 0. The augmented matrix in reduced row echelon form is: This definition, and the ones that follow, including theorems can be found in any linear algebra textbook.

3 Therefore, the solution of Ax 0 is not unique. This linear transformation is not oneto-one. 5. Problem 2, 2.1 Compute the matrix sum or product if it is defined. If an expression is undefined, explain why. Let A, B, C, D, E (a) Evaluation of A + 2B: A + 2B (b) Evaluation of C E: C E Not defined, because the size of C and E are different, and addition of matrices is only defined for matrices of the same size. (c) Evaluation of CB: CB (d) Evaluation of EB: EB Not defined, because the column size of E does not equal the row size of B, and multiplication of matrices is only defined when the number of columns of the first matrix factor equals the number of rows of the second matrix factor. 6. Show that in a vector space, the number of linearly independent vectors is less than or equal to the number of vectors that span the space. Specifically, suppose the vectors {x 1,, x,..., x n } span a vector space V, and that the vectors {y 1, y 2, y,..., y m } are in the space and are linearly independent, then m n. 2 Preliminary definitions and theorems: 2 Proof idea: If {x 1,, x,..., x n} span a vector space V, one can write y 1 c 1x 1 + c 2 + c x + + c nx n, not all c i 0, else y 1 0, which can not happen in a linearly independent set (why not?), so some c i 0. Then x 1 is a linear combination of y 1 and the remaining x s, so the set {y 1,, x,..., x n} also spans the vector space V. Continue this to show a contradiction if m > n. Alternatively, complete the argument of page 182, number 27.

4 Definition: Let V be a vector space, and let B {b 1, b 1, b 1,..., b n } be an ordered set of vectors in V. A linear combination of the elements of B is any vector of the form v a 1 b 1 + a 2 b 1 + a b a n b n where the coefficients a 1, a 2, a,..., a n are arbitrary scalars (some of which may be zero). Definition: The span of B, denoted Span (B) is the (unordered) set of all linear combinations of the vectors in B. Definition: The set B is said to be linearly independent if, whenever a 1 b 1 + a 2 b 1 + a b a n b n 0, we must also have a 1 a 2 a n 0. Theorem: Let B {b 1, b 1, b 1,..., b n } be a finite collection of vectors in V. Span (B) is a vector space (a subspace of V ). Then Theorem: A set B {b 1, b 1, b 1,..., b n } with n > 1 is said to be linearly dependent if, and only if, one of the vectors b i can be written as a linear combination of the others. Two proofs are offered, but I believe the first proof given is more elegant and easier to follow. The second proof is given only because it was specified in the hint. 1 First Proof The proof that follows will show that the number of linearly independent vectors is less than or equal to the number of vectors that span the space. Preamble: We re given a finite collection of vectors, X {x 1,..., x n }, and we are told that this finite collection spans a vector space V. We know the Span (X) is an unordered collection of all vectors in V. Let A x 1 x x n, be the m n matrix, where each column is a vector from the set X. The reduced row echelon form of A will have a most m pivots (because we were told that it spans V ). Let p be the number of row pivots, where p m. Case 1: (1) n > p which implies that X is linearly dependent. Proof: The reduced row echelon form of A has p pivots (because we were told that it spans V ), at most one for each row, so if n > p there must be a column without a pivot, hence there is a free variable implying that the columns of A are linearly dependent (i.e. a solution to a 1 x 1 + a 2 x 1 + a x a n x n 0 includes a non-trivial solution). Case 2: (1) n p which implies that X is linearly independent. Proof: The reduced row echelon form of A has p pivots (because we were told that it spans V ), at most one for each row, so if n p each column has a pivot, hence there is a no free variable implying that the

5 columns of A are linearly independent (i.e. the solution to a 1 x 1 +a 2 x 1 +a x 1 + +a n x n 0 is only the trivial solution). It has been shown that the number of vectors that span a space V must be greater than or equal to the number of linearly independent vectors in the space V. Q.E.D. 2 Second Proof The argument of this proof is essentially the same, but I believe it is less precise because it relies on solving equations where the numerical divisors may be zero, and re-arrangements may need to be performed to avoid this. Preamble: Given X {x 1, x 1, x 1,..., x n }, where X spans a vector space V, also given is a set of linearly independent vectors Y {y 1, y 2, y 2,..., y m }, where the y i s are in the vector space V. Since the set Y is linearly independent we know: a 1 y 1 + a 2 y 1 + a y a j y m 0, (1) where we must also have a 1 a 2 a m 0. It is worth pointing out here that this uniqueness of solution guarantees that the reduced row echelon form of y1 y 2 y y m, is equivalent a matrix that has m row pivots. Furthermore, y i 0 for all i, otherwise the above equation (1) would have a non-trivial solution. Since X spans the vector space V, and Y is in this same vector space, we know that each y i can be written as a linear combination of the vectors from the set X. n a i,k x k y i k1 Again, since we know y i 0 for all i, we can further state that at least one numerical coefficient in each equation from n a i,k x k y i must be non-zero. Now solve one equation generated from n k1 k1 a i,k x k y i for x 1, then using one of the m 1 equations remaining to solve for,..., finally the last equation for x m. If n > m, solve for the remaining {x m+1,..., x n }. Now that we have each x i, where i 1, n, in the the span of set that includes one vector from the set Y, and the vectors from X x i. Note: rearrangements may be necessary to avoid division by 0. Let p be the number of pivots in the reduced row echelon form of this matrix of coefficients. Case 1: (1) n > p which implies that X is linearly dependent. Proof: Again, the argument is essentially the same, but this time we are are confronted with an over-determined system of equations. So, the reduced row echelon form of this system will have p pivots (because we were told that it spans V ), at most one for each row, so if n > p there must be a column without a pivot, hence there is a free variable implying that the columns of of this system are linearly dependent.

6 Case 2: (1) n p which implies that X is linearly independent. Proof: Again, the argument is essentially the same, but this time we are confronted with a precisely determined system of quations. So, the reduced row echelon form of this system has p pivots (because we were told that it spans V ), at most one for each row, so if n p each column has a pivot, hence there is a no free variable implying that the columns of this system are linearly independent. It has been shown that the number of vectors that span a space V must be greater than or equal to the number of linearly independent vectors in the space V. Q.E.D.