MATH 51TA Section Notes for Thu 02 Oct 08Jason LoAdmin: homework 1 due: name, student ID, section numberToday’s topics:• matrix-vector multiplication• nullspaceExample 1. Let A =3 1 40 1 2, v =2−21.1. compute Av2. find a vector in R3orthogonal to the vector0123. find a vector in R3orthogonal to the vector201Solution 1. Av =80.2. Since the second component in the product Av is 0, we know straightaway that the transpose of the second row of A is orthogonal to v.3. Since201can be obtained from012by permuting the entries, we canfind a vector orthogonal to the former by permuting the entries of v in thesame manner (think through this: look at what happens when we compute dotproducts). So12−2is such a vector.Example 2. Suppose A is a 2 by 3 matrix over R. Supp ose v1, · · · , vkarelinearly independent vectors in N (A), the nullspace of A. Then:1. how many components does each vector vihave?2. what is the le ast possible value of k?1Solution 1. Each visatisfies the equation Avi= 0. For this to make sense,the number of rows of vishould equal the numbe r of column of A, i.e. 3.2. (cf. p.55 of Levandosky) Let R := rref (A), the reduced row echelon formof A. There can be at most 2 pivot 1’s in R, meaning there is at least one columnwithout a pivot 1. So there is at least one free variable to the system of linearequations Ax = 0 where x =x1x2x3. (Do an example: say R =1 0 00 0 1.)Hence the least possible value of k is 1.Example 3. Let A =3 1 40 1 2be as in Example 1. (From Example 2, weknow that N(A) contains nonzero vectors.) Find N(A).Solution We need to find N (A) = {x ∈ R3: Ax = 0}. Write x =x1x2x3.Since N (A) = N(rref(A)), let us find rref(A) first. Using row operations, wegetrref(A) =1 0230 1 2So x3is a free variable in the s ystem of linear equations Ax = 0, and everysolution to this system is of the formx =x1x2x3=23x32x3x3= x32321That is, N(A) = N(rref(A)) = span2321Example 4. Let B =3 10 1. Note that this is obtained from A in Example3 by omitting the third column. Find N (B). How many pivot 1’s does rref(B)have?Solution Note that exactly the same row operations that take A to rref (A)will take B to rref (B). Hence rref (B) =1 00 1. So rref (B) has two pivot1’s. And by Proposition 8.3, N(B) = {0}. We can also see the last statementdirectly by observing that the system of linear equations rref(B)x1x2= 0 hasthe zero vector as the only solution.2Example 5. (Ex 8.16)Solution Refer to Proposition 8.2 and the picture on p.53.Example 6. (Ex 8.28)Solution Refer to Proposition 8.3 on
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