Linear Algebra Solutions

SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA

All the Page numbers, Theorem numbers, etc. without stating the reference are from the textbook [1]. Other references will be specified.

1.9 The Matrix of a Linear Transformation 24. (a) False. In fact, the linear transformation x 7→ Ax can never map R3 to R4 . The range of the this linear transformation, by definition, is the span of the set of column vectors of A. Since there are at most three pivots in the matrix A, in the echelon form, the last row cannot have a pivot. Thus this span cannot be the whole R4 . (b) True. Page 73, Theorem 10. For every linear transformation T from Rn to Rm , there is a unique m × n matrix A such that T (x) = Ax. (Note that in the end of Page 67, it is said that every matrix transformation is a linear transformation, yet there are examples of linear transformations that are not matrix transformations. This refers to general linear transformation on general vector space where there is no preferred basis.) (c) True. Again, Page 73, Theorem 10. (Although technically, it should be stated that the given linear transformation from Rn to Rm is T .) (d) False. One-to-one, (or injectivity), means every vector in the codomain, has at most one preimage. For T to be a mapping, it is automatically, each vector in Rn maps to only one vector in Rm . (e) False. See Table 3 on Page 76. 26. As we have already seen, the standard matrix of T is 1 −2 3 A= . 4 9 −8 By subtracting 4 times first row from the second row, we get the echelon form 1 −2 3 . 0 17 −20 1

2

MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA

There are two pivots, two rows and three columns, thus the columns of A span R2 but are not linear independent. By Theorem 12 on Page 79, T is onto but not one-to-one. 28. As we see in the figure, the column vectors of the standard matrix A, a1 , a2 are not multiple to each other. Thus {a1 , a2 } span R2 and is linear independent. (See Page 60 for linear independence. For span R2 , one can count the pivots in the 2 × 2 matrix A. Since column vectors are linear independent, there is a pivot in each column, thus totally there are two pivots. This implies each row must have a pivot.) Again, by Theorem 12 on Page 79, T is onto and one-to-one. 35. (1) If a linear transformation T : Rn → Rm is onto, then m 6 n. Assume A is the standard matrix of T , then the columns of A span Rm , or equivalently, in the echelon form of A, there is a pivot in each row. Since A has m rows and n columns, this means A has exactly m pivots and (since there can be at most one pivot in each column) m 6 n. (2) If a linear transformation T : Rn → Rm is one-to-one, then m > n. (The argument is similar to above, just switch the roles of rows and columns.) Assume A is the standard matrix of T , then the columns of A are linear independent, or equivalently, in the echelon form of A, there is a pivot in each column. Since A has m rows and n columns, this means A has exactly n pivots and (since there can be at most one pivot in each row) m > n. 36. The reason is that this question is equivalent to (by definition) “Is there a preimage for every vector in the codomain?” or (in terms of matrix equation, A is the standard matrix of T ) “Does Ax = b have a solution for every b?”. In both forms, it is clear that this is an existence question.

2.1 Matrix Operations 2. Let 2 0 −1 7 −5 1 1 2 3 5 −5 A= ,B = ,C = ,D = ,E = 4 −5 2 1 −4 −3 −2 1 −1 4 3 (1) A + 3B =

23 −15 2 7 −17 −7

.

SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013

3

(2) 2C − 3E is not defined since C and E have different size. (3) 26 −35 −12 DB = . −3 −11 −13 (4) EC is not defined, since E has only one column and C has two rows. 4. 

   0 −1 3 25 −5 15 A − 5I3 =  −4 −2 −6  , (5I3 )A =  −20 15 −30  . −3 1 −3 −15 5 10

6. 

 4 −3 1 4   A = −3 5 , B= . 3 −2 0 1 So b1 =

1 3

, b2 =

4 −2

.

and we have 

   −5 22 Ab1 =  12  , Ab2 =  −22  . 3 −2 From either of the two ways of calculation, (which do  −5  AB = [Ab1 Ab2 ] = 12 3

not have much difference) we get  22 −22  . −2

8. B has the same number of rows as BC, which is 5. 10. It is easy to calculate that AB = AC =

−21 −21 7 7

.

12. Let B = [b1 b2 ], then since AB = 0, b1 and b2 are solutions to the matrix equation Ax = 0. By the elementary row operation that adds 23 times of the first row to the second

4


row, we have

3 −6 A∼ . 0 0 Therefore the general solutions of Ax = 0 is 2 x = x2 . 1 For example, we can choose B=

2 4 1 2

.

16. (a) True. This follows directly from the definition. (b) False. In fact, AB = [Ab1 Ab2 Ab3 ]. (c) True. In (AB)T = B T AT , we take B = A, then we get (A2 )T = (AT )2 . (d) False. In fact (ABC)T = C T B T AT not equal to C T AT B T in general. A counter example is 0 1 0 0 1 0 A= ,B = ,C = . 0 0 1 0 0 1 Then 1 0 0 0 T (ABC) = 6= = C T AT B T . 0 0 0 1 (e) True. Since (A + B)T = AT + B T , by simple induction on n, we have the transpose of a sum of n matrices equals the sum of their transposes. 18. The third column of AB is also all zeros. Since if B = [b1 b2 · · · bn ], then AB = [Ab1 Ab2 · · · Abn ]. The third column is Ab3 = A0 = 0. 20. The first two columns of AB are also equal. Same as Problem 18. If b1 = b2 , then Ab1 = Ab2 . 22. Again, assuming B = [b1 b2 · · · bn ] and since the columns of B are linear dependent, we can find c1 , . . . , cn not all zero such that c1 b1 + · · · + cn bn = 0. Therefore applying A to the left we have c1 Ab1 + · · · + cn Abn = A(c1 b1 + · · · + cn bn ) = A0 = 0. This shows that the columns of AB, Ab1 , . . . , Abn are linear dependent.


5

24. Since the columns of A span R3 . We can find the solutions x = xj to Ax = ej , j = 1, 2, 3 where e1 , e2 , e3 are standard unit vectors in R2 . Then D = [x1 x2 x3 ] satisfies AD = I3 . 27. uT v = vT u = −3a + 2b − 5c (think this as a 1 × 1 matrix).     −3a −3b −3c −3a 2a −5a uvT =  2a 2b 2c  , vuT =  −3b 2b −5b  . −5a −5b −5c −3c 2c −5c 28. By the formula (AB)T = B T AT and (AT )T = A, we have uT v = (vT u)T = vT u since this is a 1 × 1 matrix, its transpose is the same as itself. Also we have uvT = (vuT )T . 32. For j = 1, . . . , n, the j-th column of AIn is Aej where e1 , . . . , en are standard unit vector in Rn . Assume A = [a1 · · · an ], then ! X Aej = 0ai + 1aj = aj i6=j

Therefore AIn has exactly the same columns as A, or AIn = A. 34. Use the formula (AB)T = B T AT repeatedly, (ABx)T = (B x )T AT = xT B T AT .

2.2 The Inverse of a Matrix 2. Use Theorem 4 on Page 105, since determinant of the matrix is 3 · 5 − 2 · 8 = −1 6= 0, −1 3 2 5 −2 −5 2 −1 = (−1) = . 8 5 −8 3 8 −3 4. Again, use Theorem 4 on Page 105, since determinant of the matrix is 2 · (−6) − 4 · (−4) = 4 6= 0, −1 3 1 −6 4 2 −4 −2 1 = = . 4 −6 −1 12 4 −4 2

6


6. Since

7 3 −6 −3

−1

1 = 7(−3) − 3(−6)

−3 −3 6 7

=

1 1 −2 − 73

,

The solution to the given system, which is equivalent to 7 3 x1 −9 = −6 −3 x2 4 is

x1 x2

=

7 3 −6 −3

−1

−9 4

=

1 1 −2 − 73

−9 4

=

7. (a) Since det A = 2 6= 0, we have 1 12 −2 6 −1 −1 A = = . − 52 12 2 −5 1 Therefore the solutions to Ax = bj , j = 1, 2, 3, 4 are −9 −1 x1 = A b1 = 4 11 −1 x2 = A b2 = −5 6 −1 x3 = A b3 = −2 13 −1 x4 = A b4 = −5

, , , .

(b) The augmented matrix is [A b1 b2 b3 b4 ] =

1 2 −1 1 2 3 5 12 3 −5 6 5

By subtracting 5 times the first row from the second row, we get 1 2 −1 1 2 3 . 0 2 8 −10 −4 −10 By subtracting the second row from the first row, we get 1 0 −9 11 6 13 . 0 2 8 −10 −4 −10

.

−5 26 3

.


By multiplying

1 2

7

to the second row, we get the reduced echelon form 1 0 −9 11 6 13 0 1 4 −5 −2 −5

which equals [I A−1 b1 A−1 b2 A−1 b3 A−1 b4 ]. The last four columns are exactly the solutions to the four equations. 8. By multiplying P −1 to the left-hand side and P to the right-hand side of the equation A = P BP −1 , we have P −1 AP = P −1 P BP −1 P = IBI = B since P −1 P = I. 10. (a) False. The elementary row operations that reduce A to the identity In are equivalent to multiply A−1 from the left. Therefore it will change A−1 to A−1 A−1 which does not necessarily equal to In . (b) True. Since by definition, A−1 A = AA−1 = I, this shows that A−1 is also invertible and (A−1 )−1 = A. (c) False. If A, B are invertible n × n matrices, then by Theorem 6 part (b) on page 107, we have (AB)−1 = B −1 A−1 which does not necessarily equal to A−1 B −1 . (d) True. Let xj be the solution of Ax = ej . Then for any b = (b1 , · · · , bn ) ∈ Rn , since b = b1 e1 + · · · + bn en , we get that x = b1 x1 + · · · + bn xn satisfies the equation Ax = b. In other words, the columns of A span Rn . Therefore in the reduced echelon form of A, there is a pivot in every row. So there are exactly n pivots, i.e. the reduced echelon form is In . In other words, A is row equivalent to In . Thus A is invertible. (Also see Theorem 8 on page 114.) (e) True. See Theorem 7 on page 109. 12. Since A is invertible, we have A−1 A = I, therefore D = ID = (A−1 A)D = A−1 (AD) = A−1 I = A−1 . 14. Since D is invertible, we can multiply D−1 to the right-hand side of the equation (B − C)D = 0 to get 0 = (B − C)DD−1 = (B − C)I = B − C.

8


Therefore B = C. 16. Since B is invertible, we can write A = AI = A(BB −1 ) = (AB)B −1 . Now both AB and B −1 are invertible matrices, so the product A is also invertible. 18. Since B is invertible, as in Problem 16, A = (AB)B −1 = (BC)B −1 = BCB −1 . 20. (a) Since (A − AX)−1 = X −1 B, we have B = IB = (XX −1 )B = X(X −1 B) = X(A − AX)−1 . Now both X and (A − AX)−1 are invertible, we have that B is also invertible. (b) Since B = X(A − AX)−1, we have B(A − AX) = X, so BA − BAX = X and thus BA = X + BAX = (I + BA)X. We need to invert I + BA, but since X is invertible, we have I + BA = BAX −1 where all of B, A, X −1 are invertible. Therefore I + BA is invertible and we can solve that X = (I + BA)−1 BA. 24. The same explanation as Problem 10 part d. The columns of A span Rn . Therefore in the reduced echelon form of A, there is a pivot in every row. So there are exactly n pivots, i.e. the reduced echelon form is In . In other words, A is row equivalent to In . Thus A is invertible. (Also see Theorem 8 on page 114.) 25. We consider the following two separated cases: Case 1: a = b = 0. Then the matrix A is row equivalent to c d 0 0 which can have at most one pivot. Therefore there is a row without a pivot, we know the system Ax = 0 has infinite solutions. −b Case 2: a and b are not both zero, then let x0 = , we can check a −ab + ba Ax0 = = 0. −cd + da Therefore Ax = 0 has more than one solution: at least 0 and x0 .


30.

3 6 1 0 ∼ 4 7 0 1 1 2 13 ∼ 0 1 43

Therefore

3 6 4 7

9

1 1 2 13 0 1 2 0 3 ∼ 4 7 0 1 0 −1 − 43 1 0 1 0 − 73 2 ∼ . −1 0 1 43 −1 −1

=

− 73 4 3

2 −1

.

32. Since 

1 2  −4 −7 −2 −6 we know the original matrix not invertible.

     −1 1 2 −1 1 2 −1 3  ∼  0 1 −1  ∼  0 1 −1  , 4 0 −2 2 0 0 0 is not row equivalent to the identity matrix. Therefore it is

33. Since       1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0  1 1 0 0 1 0  ∼  0 1 0 −1 1 0  ∼  0 1 0 −1 1 0  0 0 1 0 −1 1 1 1 1 0 0 1 0 1 1 −1 0 1 we have

−1   1 0 0 1 0 0  1 1 0  =  −1 1 0  . 1 1 1 0 −1 1 

Similarly, 

1  1   1 1 

1  0 ∼  0 0 Therefore

0 1 0 0

0 1 1 1

0 0 1 1

0 0 1 1

0 1 0 0 0 −1 1 0 0 0 −1 1 1 0 −1 0 

0 0 0 1

1  1   1 1

1 0 0 0

0 1 1 1

0 1 0 0

0 0 1 1

0 0 1 0

  0 1 0  0  ∼ 0 1   0 1 0 1 0 1   0 1 0   0   0 1 ∼ 0   0 0 0 0 1

0 0 1 1 0 0 1 0

0 1 0 0 0 −1 1 0 0 −1 0 1 1 −1 0 0

 0 0   0  1

0 1 0 0 0 −1 1 0 0 0 −1 1 1 0 0 −1

−1  0 1 0 0   0  −1 1 0 =  0 −1 1 0  1 0 0 −1

 0 0  . 0  1

 0 0  . 0  1

10


In general, the corresponding n × n matrix  1 0  1 1   1 1  A =  .. ..  . .   1 1 1 1

is  ··· 0 0 ··· 0 0   ··· 0 0   . . . .. ..  . . .   1 ··· 1 0  1 ··· 1 1 0 0 1 .. .

We can write A = [aij ]16i,j6n where aij = From the case of n = 3, 4 we can guess  1  −1   0  B =  ..  .   0 0

1 0

if i > j if i < j

the inverse of A is 0 1 −1 .. . 0 0

0 0 1 .. .

··· ··· ··· .. .

0 ··· 0 ···

0 0 0 .. .

0 0 0 .. .



    .   1 0  −1 1

We can also write B = [bij ]16i,j6n where  if i = j  1 bij = −1 if i = j + 1  0 if i < j or i > j + 1 We can check directly that the product C = AB = [cij ]16i,j6n is actually the identity matrix. Since n X cij = aik bkj , k=1

we first use the formula for aik to get cij =

i X

bkj = b1j + · · · + bij .

k=1

Case 1: If i > j, then there are two nonzero terms in the above sum. cij = bjj + bj+1,j = 1 − 1 = 0. Case 2: If i = j, then there are only one nonzero term in the above sum. cij = bjj = 1.


11

Case 3: If i < j, then every term in the above sum is zero, so cij = 0. This shows that 1 if i = j cij = δij = 0 if i 6= j (Kronecker delta symbol, see [2]) or AB = I. There is also an algebraic way to prove AB  0 0 0  1 0 0   0 1 0  J =  .. .. ..  . . .   0 0 0

= I is as follows, let  ··· 0 0 ··· 0 0   ··· 0 0   . . . .. ..  . . .   ··· 0 0  0 0 0 ··· 1 0

Then J is a nilpotent matrix: J n = 0. (See [3]) In fact, J k has entries 1 on the sub-diagonal: i − j = k for k = 1, . . . , n − 1. Now we can write A = I + J + J 2 + · · · + J n−1 , B = I − J and we have AB = (I + J + J 2 + · · · + J n−1 )(I − J) = I − J + J − J 2 + · · · + J n−1 − J n = I − J n = I. Remark: Actually this is a standard property for all nilpotent elements in a commutative ring with a unit (See e.g. [4]). The third way to prove this is by induction. We write An , Bn for the n × n matrices. We need to prove An Bn = In for every n. It is obvious that this is true for n = 1 since A1 = B1 = 1 (as either numbers or 1 × 1 matrices.) Now suppose An−1 = Bn−1 = In−1 , since An−1 0 Bn−1 0 An = , Bn = , T T vn−1 1 wn−1 1 where 

vn−1

  1 0  ..   ..    =  .  , wn−1 =  .  1   0 1 −1

   . 

Therefore by multiplication of partitioned matrices (see Section 2.4) An−1 Bn−1 0 In−1 0 An Bn = = . T T T vn−1 Bn−1 + wn−1 1 (Bn−1 vn−1 + wn−1 )T 1

12


T T It remains to check that Bn−1 vn−1 + wn−1 = 0 which is straightforward. In fact, Bn−1 vn−1 n−1 T is the sum of all the columns of Bn−1 , which is en−1 ∈ R , i.e. the negative wn−1 .

34. Again we start with  1  2 3

3 × 3 matrix    0 0 1 0 0 1 0 0 1 0 0 2 0 0 1 0  ∼  1 1 0 0 21 0  3 3 0 0 1 1 1 1 0 0 13



   1 0 0 1 0 0 1 0 0 1 0 0 ∼  0 1 0 −1 12 0  ∼  0 1 0 −1 21 0  0 1 1 −1 0 13 0 0 1 0 − 21 13 Therefore −1   1 0 0 1 0 0  2 2 0  =  −1 1 0  . 2 3 3 3 0 − 12 13 

For 4 × 4 matrix, we have  1 0 0  2 2 0   3 3 3 4 4 4 

1  0 ∼  0 0

0 1 1 1

0 0 1 1

0 0 0 4

1 0 0 0

0 1 0 0

0 0 1 0

  0 1 0  0  ∼ 1 1   1 1 0 1 1 1

  0 1 0 0 0 1 1   0 −1 2 0 0   0 ∼ 0 −1 0 31 0   0 1 −1 0 0 14 0 

1  0 ∼  0 0

0 1 0 0

0 0 1 0

0 1 0 0

 1 0 0 0 0 21 0 0   0 0 13 0  0 0 0 14

0 0 1 1

0 0 0 1

0 0 1 1

 0 1 0 0 0 0 −1 12 0 0   0 0 − 12 31 0  1 0 − 12 0 41

 0 1 0 0 0 0 −1 21 0 0  . 1 0 0 − 2 13 0  1 0 0 − 13 14

Therefore 

1  2   3 4

0 2 3 4

0 0 3 4

−1   0 1 0 0 0 1  0  0 0   =  −1 2 . 1 1  0 −  0 0  2 3 1 1 4 0 0 −3 4


In general, the corresponding n × n matrix is  1 0 0  2 2 0   3 3 3  A =  .. . .. ..  . .   n−1 n−1 n−1 n n n

··· ··· ··· ...

0 0 0 .. .

0 0 0 .. .

13



    .   n−1 0  n n

··· ···

We can write A = [aij ]16i,j6n where aij = From the case of n = 3, 4 we can guess the  1 0  −1 1 2   0 −1  2 B =  .. ..  . .   0 0 0 0

i 0

if i > j if i < j

inverse of A is 0 ··· 0 ··· 1 ··· 3 .. . . . .

0 0 0 .. .

1 0 · · · n−1 1 0 · · · − n−1

0 0 0 .. .



    .   0  1 n

We can also write B = [bij ]16i,j6n where  1  if i = j  j 1 bij = −j if i = j + 1   0 if i < j or i > j + 1 Again, we can check directly that the product C = AB = [cij ]16i,j6n is actually the identity matrix. Since n X cij = aik bkj , k=1

we first use the formula for aik to get cij =

i X

ibkj = i(b1j + · · · + bij ).

k=1

Case 1: If i > j, then there are two nonzero terms in the above sum. cij = i(bjj + bj+1,j ) = i( 1j − 1j ) = 0. Case 2: If i = j, then there are only one nonzero term in the above sum. cij = ibjj = ji = 1.

14


Case 3: If i < j, then every term in the above sum is zero, so cij = 0. This shows that 1 if i = j cij = δij = 0 if i 6= j or AB = I. The algebraic way to prove this is to write A = D(I + J + J 2 + · · · + J n−1 ) and B = (I − J)D−1 where D is the diagonal matrix with diagonal entries 1, 2, . . . , n:   1 0 0 ··· 0 0  0 2 0 ··· 0 0     0 0 3 ··· 0 0    D =  .. .. .. . . . . . . .  . . .  . . .    0 0 0 ··· n − 1 0  0 0 0 ··· 0 n To see A = D(I +J +J 2 +· · ·+J n−1 ), it is enough to use the expression of I +J +· · ·+J n−1 in the previous problem and the fact that multiply D on the left is equivalent to the row operations that multiply the i-th row by i, i = 1, . . . , n. To see B = (I − J)D−1 , it is enough to see that multiply J on the left is equivalent to remove the last row, then move every other row down for one level and finally put all zeroes in the first row. Now it is easy to see AB = DD−1 = I using the previous problem. We can also prove this by induction. We write An , Bn for the n × n matrices. We need to prove An Bn = In for every n. It is obvious that this is true for n = 1 since A1 = B1 = 1 (as either numbers or 1 × 1 matrices.) Now suppose An−1 = Bn−1 = In−1 , since An−1 0 Bn−1 0 An = , Bn = , 1 T T vn−1 n wn−1 n where 

vn−1

  0 n  ..   ..    =  .  , wn−1 =  .  n   0 1 − n−1 n

   . 

Therefore by multiplication of partitioned matrices (see Section 2.4) An−1 Bn−1 0 In−1 0 An Bn = = . T T T Bn−1 + nwn−1 1 (Bn−1 vn−1 + nwn−1 )T 1 vn−1 T It remains to check that Bn−1 vn−1 + nwn−1 = 0 which is straightforward.


15

37. We assume C=

c11 c12 c13 c21 c22 c23

.

Then the equation CA = I2 is equivalent to the system c11 + c12 + c13 = 1 2c11 + 3c12 + 5c13 = 0 . c21 + c22 + c23 = 0 2c21 + 3c22 + 5c23 = 1 We need to find solutions cij ∈ {−1, 0, 1}. It is easy to see that the solutions are c11 = 1, c12 = 1, c13 = −1. c21 = −1, c22 = 1, c23 = 0. Therefore C=

1 1 −1 −1 1 0

.

Now we can compute AC: 

 −1 3 −1 AC =  −2 4 −1  = 6 I3 . −4 6 −1 (In fact, the product of a 3 × 2 matrix and a 2 × 3 matrix can have at most rank 2, but the identity matrix has rank 3. So they can not equal to each other. See Section 2.7.) 38. We can choose 

1  1 D=  1 0

 0 1  . 1  1

It is not possible that CA = I2 for some matrix C. Assume C = [c1 c2 ], then we have CA = [c1 − c1 + c2 c1 − c2 c2 ]. Therefore the span of columns of CA: Span{c1 , −c1 +c2 , c2 −c2 , c2 } equals to Span{c1 , c2 }, which cannot be R4 . Thus CA 6= I4 .

16


2.3. Characterizations of Invertible Matrices 2. Since

−4 2 6 −3

∼

−4 2 0 0

The matrix is not invertible. 4. Since there is a row of all zero, there are at most two pivots, so the matrix is not invertible. 6. Since 

     1 −3 −6 1 −3 −6 1 −3 −6  0 4 3 ∼ 0 4 3 ∼ 0 4 3 , −3 6 0 0 −3 −18 0 0 − 63 4 we know the matrix has three pivots, thus invertible. 8. Since this is an upper triangular matrix, which is already in the echelon form. There are exactly four pivots, thus the matrix is invertible. 12. (a) True. Since AD = I where both A and D are n × n matrices, both A and D are invertible (Invertible Matrix Theorem (a)⇔(j)(k)), also A = D−1 . Therefore DA = DD−1 = I. (b) False. A can be any n × n matrix, for example, the zero matrix. Then the row reduced echelon form of A is not I. (c) True. Since the columns of A are linear independent, A is invertible, thus the columns of A span Rn . (Invertible Matrix Theorem (e)⇒(a)⇒(h).) (d) False. Since the equation Ax = b has at least one solution for each b, the columns of A span Rn . Therefore A is invertible and x 7→ Ax is one-to-one. (Invertible Matrix Theorem (h)⇒(a)⇒(f).) (e) False. We can take b = 0 and A is not invertible, so the equation Ax = 0 has infinite solutions. 14. A square lower triangular matrix is invertible if and only if all the diagonal entries are nonzero. Let A be a lower triangular matrix, then A is invertible if and only if AT is invertible. AT is a square upper triangular matrix. If all the diagonal entries of A are non-zero, then AT is in echelon form and it has exact n pivots which are the diagonal


17

entries. So A is invertible. Conversely, if A is invertible, then the echelon form of AT must have n pivots. This means that the diagonal entries must be the pivots, thus nonzero. 16. If A is invertible, then so is AT , thus the columns of AT must be linearly independent. (Invertible Matrix Theorem (a)⇒(l) and (a)⇒(e).) 18. No. By an elementary row operation that add the negative of one of the two rows to another, we get a row with all zero. Therefore the matrix cannot have n pivot positions, thus not invertible. (Assume the matrix is n × n.) 20. No. Because the equation Ax = b is consistent for every b in R5 , the matrix A is invertible. Therefore Ax = b has at most one solution for every b. (Invertible Matrix Theorem (g)⇒(a)⇒(f), then use definition of one-to-one.) 22. Since EF = I, by Invertible Matrix Theorem (j)(k))⇒(a), we know E and F are both invertible. Also E −1 = F . Therefore F E = E −1 E = I = EF , i.e. E and F commute. 26. Since the columns of A are linear independent, A is invertible. (Invertible Matrix Theorem (e)⇒(a).) So A2 is also invertible and as a consequence, the columns of A2 span Rn . (Invertible Matrix Theorem (a)⇒(h).) 28. Since AB is invertible, let C = (AB)−1 A, then CB = (AB)−1 AB = I. Therefore B is invertible. (Invertible Matrix Theorem (j)⇒(a)). 30. By definition, if x 7→ Ax is one-to-one, then each b can have at most one preimage. Ax = b cannot have more than one solutions for any b. Therefore If Ax = b has more than one solutions for some b, then the transformation x 7→ Ax is not one-to-one. We can also deduce that this transformation is not invertible. (By Invertible Matrix Theorem.) 32. Since the equation Ax = 0 has only the trivial solution, A has a pivot in every column. Since A is an n × n matrix, A has exactly n pivots. Therefore A has a pivot in every row. This shows that Ax = b has a solution for each b.

18


34. The standard matrix of T is A=

2 −8 −2 7

.

Since det A = −2 6= 0, its inverse is −1

A

1 =− 2

7 8 2 2

=

− 72 −4 −1 −1

.

Therefore T is invertible and T −1 (x1 , x2 ) = (− 72 x1 − 4x2 , −x1 − x2 ).

3.1 Introduction to Determinants 2. Cofactor expansion across the first row: 0 5 1 4 −3 0 =(−1)1+1 0 −3 0 + (−1)1+2 5 4 0 4 1 2 1 2 4 1

+ (−1)1+3 1 4 −3 2 4

=0 − 20 + 22 = 2. Cofactor expansion down the second column: 0 5 1 4 −3 0 =(−1)1+2 5 4 0 + (−1)2+2 (−3) 0 1 2 1 2 1 2 4 1

+ (−1)3+2 4 0 1 4 0

= − 20 + 6 + 16 = 2. 6. Cofactor expansion across the first row: 5 −2 4 3 −5 0 −5 0 3 −5 =(−1)1+1 5 + (−1)1+2 (−2) + (−1)1+3 4 0 3 −4 7 2 7 2 −4 2 −4 7 =5 + 20 − 24 = 1. 10. First we use the cofactor expansion across the second row: 1 −2 5 2 1 −2 2 0 0 3 0 2+3 , = (−1) 3 2 −6 5 2 −6 −7 5 5 0 4 5 0 4 4


19

then we use the cofactor expansion across the second column to get 1 −2 2 2 −6 5 5 0 4

1+2 = (−1) (−2) 2 5 5 4

2+2 + (−1) (−6) 1 2 5 4

= 2(−17) − 6(−6) = 2.

Therefore

1 −2 5 2 0 0 3 0 2 −6 −7 5 5 0 4 4

= −6.

12. First, we use the cofactor expansion across the first row:

4 0 0 0 7 −1 0 0 2 6 3 0 5 −8 4 −3

−1 0 0 = (−1)1+1 4 6 3 0 −8 4 −3

Then we use the cofactor expansion across the first row again: −1 0 0 6 3 0 −8 4 −3

1+1 = (−1) (−1) 3 0 4 −3

= (−1)3(−3).

Therefore

4 0 0 0 7 −1 0 0 2 6 3 0 5 −8 4 −3

= 4(−1)3(−3) = 36.

(Actually, this is a lower triangular matrix. The determinant is the product of the diagonal entries.)

20


14. First we use the cofactor expansion across the fourth row, then the cofactor expansion down the last column, finally the cofactor expansion down the first column: 6 3 2 4 0 3 2 4 0 3 2 4 9 0 −4 1 0 8 −5 6 7 1 =(−1)4+1 3 0 −4 1 0 = (−1)4+1 3(−1)3+4 1 0 −4 1 −5 6 7 1 2 3 2 3 0 0 0 0 2 3 2 0 4 2 3 2 0 4 4+1 3+4 1+1 −4 1 3+1 2 =(−1) 3(−1) 1 (−1) 3 + (−1) 2 3 2 −4 1 =3[−33 + 36] = 9. 18. 1 3 5 2 1 1 = (1 · 1 · 2) + (3 · 1 · 3) + (5 · 2 · 4) − (1 · 1 · 4) − (3 · 2 · 2) − (5 · 1 · 3) = 20. 3 4 2 20. The elementary row operation is to multiply k to the second row. The determinant is also multiplied by k. 22. The elementary row operation is to add k times the second row to the first row. The determinant remains the same. 24. The elementary row operation is to switch the first row and the second row. The determinant becomes the negative of the original determinant. 26. 

 1 0 0 det  0 1 0  = 1. k 0 1 28.  1 0 0 det  0 k 0  = k. 0 0 1 

30. 

 0 0 1 det  0 1 0  = −1. 1 0 0


21

32. The determinant of an elementary scaling matrix with k on the diagonal is k. 36. Since E=

1 0 k 1

,A =

a b c d

,

we have EA =

a b ka + c kb + d

.

We can calculate that det(E) = 1, det(A) = ad − bc and det(EA) = a(kb + d) − b(ka + c) = ad − bc, which shows that det(EA) = det(E) det(A). 38. Since A=

a b c d

,

we have kA =

ka kb kc kd

.

So we can calculate that det(kA) = (ka)(kd) − (kb)(kc) = k 2 (ad − bc). Therefore det(kA) = k 2 det(A). 40. (a) False. Cofactor expansions across rows or down columns are always the same. (b) False. The determinant of a triangular matrix (either upper or lower) is the product of the entries on the main diagonal. 43. In general, det(A + B) 6= det A + det B. For example, 1 0 1 1 2 1 ,B = ,A + B = A= 0 1 1 1 1 2 then det(A + B) = 3, det(A) = 1, det(B) = 0.

22


References [1] David C. Lay, R. Kent Nagle, Edward B. Saff, Arthur David Snider, Linear Algebra and Differential Equations, Second Custom Edition for University of California, Berkeley. [2] http://en.wikipedia.org/wiki/Kronecker delta [3] http://en.wikipedia.org/wiki/Nilpotent [4] Michael Atiyah, Ian G. Macdonald, Introduction to Commutative Algebra.

Linear Algebra Solutions

Recommend Documents