Linear Algebra Solutions Review Questions

Solutions: Review Questions 1. The solution is:    −4 3  0   1   x=  3 +r 0 0 0    1     + s 0    2  4 2. If AB is invertible, then by the IMT, there is a W such that (AB)W = I. By the rules of matrix multiplication, A(BW ) = I. Now, by the IMT, since A is square and there is a C such that AC = I, A is invertible. 3. If {v 1 , v 2 , v 3 } are linearly dependent, then there are infinitely many (nontrivial) solutions to the equation: c1 v 1 + c2 v 2 + c3 v 3 = 0 Apply the linear transformation T to both sides, and simplify. This shows that any solution to the equation above is also a solution to: c1 T (v 1 ) + c2 T (v 2 ) + c3 T (v 3 ) = 0 which says that {T (v 1 ), T (v 2 ), T (v 3 )} form a linearly dependent set. 4. If Hx = v is consistent for all v, then every row of H has a pivot. Therefore, there are 7 pivots. Since H has only 7 columns, there must be a pivot in every column as well. Therefore, there are no free variables, and therefore, the solution must be unique. 5. We can show the mapping is nonlinear in a couple of ways. For example, T (0, 0) 6= (0, 0) 6. Suppose that {v 1 , v 2 , v 3 } is linearly independent. Let V be a matrix whose columns are the v i ’s. Then: V c = 0 has only the trivial solution. Now the question is whether or not AV c = 0 has only the trivial solution. The answer is yes. Because A is invertible, we can rewrite the previous equation as A−1 AV c = A−1 0. Now this has only the trivial solution, therefore so does AV c = 0. 7. False. The description given is for a function; that is, each x has only one y associated with it. For a function to be 1-1, each y must have come from a unique x, x1 6= x2 ⇒ T (x1 ) 6= T (x2 ) 8. You can ignore the matrix B, although it is one of the matrices C. We are looking for the columns of C, Ac1 = e1 , Ac2 = e2 , Ac3 = e3 To solve these at one time, we will row reduce the augmented matrix [A | I2 ]:     1 0 1 0 1 1 1 −1 1 0 −→ 1 0 1 0 1 0 1 −2 1 −1 The interpretation is (by column): x1 = −x3 , x2 = 1 + 2x3 , x3 = x3 (free) ⇒ c1 = (0, 1, 0)T + x3 (−1, 2, 1)T and x̂1 = 1 − x̂3 , x̂2 = −1 + 2x̂3 , x̂3 = x̂3 ⇒ c2 = (1, −1, 0)T + x̂3 (−1, 2, 1) Now choose some arbitrary numbers for x3 and x̂3 to find three different C’s- Note that C will be 3 × 2 9. For a unique solution, h 6= 6, and k is anything. For infinitely many solutions, h = 6 and k = 2. For no solution, h = 6 and k 6= 2. 10. Let us say that the coefficient matrix is m×n. If the system is underdetermined, then m < n. Therefore, at best, we can have m pivots, which leaves at least n − m free variables. Therefore, if the system is consistent, it must have an infinite number of solutions. 1 11. Note that in each case, E needs to be 3 × 3 for EA to be defined.      1 3 0 1 0 0 1 E =  0 1 0  , E =  0 0 1  , E =  −2 0 0 1 0 0 −3 0  0 0 1 0 , 0 0 12. Asking if the set spans IRn means that we are checking if every vector in IRn can be written as a linear combination of those vectors. If we put the vectors in the matrix A, we are asking if Ax = b has a solution for every b. Once we take the RREF of A, we then check to see if there is a pivot in every row. (Recall that a ”pivot in every column” is for linear independence). 13. This question is basically the same as before, except we are asking for a solution for one value of b. 14. (a) y 1 , y 2 must be in IR3 , and x1 , x2 must be in IR4 . (b) Ax = w has a solution. By the matrix multiplication properties, if w = y 1 + y 2 , then x must be x1 + x2 . 15. The two solutions are (in order):           −3 4 −3 7 4 x = r 1  + s 0 , x =  0  + r 1  + s 0  0 1 0 0 1 The latter solution is a translated version of the former solution. 16. This is true. It says that the two vectors are constant multiplies of each other. 17. Lots of examples. For instance, let v 1 = v 2 = v 4 = (1, 1, 1, 1)T . Let v 3 = (0, 1, 1, 1)T .   a b 18. If A = , then we need to satisfy the following sets of equations: c d a + 2b c + 2d −2a + b −2c + d = −3 =0 =1 =2 We have a variety of ways to solve this. a = −1, b = −1, c = − 45 , d = 2 5 19. If Ax = 0 has only the trivial solution, there must be a pivot in every column. In this case, there are n. But since A is n × n, this means that there must be a pivot in every row as well. Therefore, the columns do span IRn . In general, this is not true- That is, in general, having a pivot in every column does not imply that there is a pivot in every row. 20. See Problem 2, and let B = A. 21.  x1 x2  1 = −3  13 7 6 3  −4 1  22. If Ax = b has a solution for every b, there must be a pivot in every row. Since A is square, this implies that there must be a pivot in every column. Therefore, the columns of A also span IRn . This is not true in general. 23. False. Because [A|I] → [I|A−1 ], the row operations that reduce A to I also “reduce” I to A−1 . 24. The first and second columns of AB are: (−13, 32)T and (20, −49)T . The (3, 1) entry of B T A is −7. We can’t find the inverse of B T B because it is not row reducible to the identity (we get one free variable, x3 ). 2 25. Suppose x is a vector satisfying Ax = 0. We will show that this implies that x = 0, so the only solution to Ax = 0 is the trivial solution. Ax = 0 Given CAx = C0 = 0 Multiply by C In x = 0 Since CA = In x=0 We have shown that, if Ax = 0, then x = 0. Therefore, x = 0 is the only solution to Ax = 0.     1 −1 cos(θ) − sin(θ) 26. The general matrix is , so the matrix is this case is: √12 1 1 sin(θ) cos(θ) 27. T (x) = [3 −4 0 8]x 28. Show that • T1 (T2 (u + v)) = T1 (T2 (u)) + T1 (T2 (v)): T1 (T2 (u + v)) = T1 (T2 (u) + T2 (v)) by the linearity of T2 T1 (T2 (u) + T2 (v)) = T1 (T2 (u)) + T1 (T2 (v)) by the linearity of T1 • T1 (T2 (au)) = aT1 (T2 (u)) T1 (T2 (au) = T1 (aT2 (u)) = aT1 (T2 (u))  29. The matrix is: 1 −2 3 4 9 −8  30. Note that we can answer these questions without row reduction: To see if T is 1-1, there can be at MOST one solution to Ax = b for each b. This means that there cannot be free variables, and this in turn means that there must be a pivot in every column. Since there are only two columns, and they are not constant multiples of each other, we can conclude that there will be a pivot in each column, and therefore, T will be 1-1. Since A is 3 × 2, we can have at most 2 pivots (and indeed, we do have two pivots). Therefore, there cannot be a pivot in every row, and that means that the columns will not span IR3 . Therefore, the function T is not onto IR3 . 31. I−BAB −1 = C ⇒ I−C = BAB −1 ⇒ B −1 I−B −1 C = AB −1 ⇒ B −1 B−B −1 CB = A ⇒ I−B −1 CB = A 32. (a) If the columns of A are linearly independent, then the only solution to: c1 a1 + c2 a2 + . . . + cn an = 0 is c1 = c2 = . . . = cn = 0 (this is the definition of linear independence). (b) Rewrite the above equation in Matrix-Vector form:    c1 a1 + c2 a2 + . . . + cn an ⇔ [a1 a2 . . . an ]   c1 c2 .. . cn (c) From (a) and (b), the only solution to Ac = 0 is c = 0 3     ⇔ Ac 