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 R 3 to R 4 . 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 R 4 . (b) True. Page 73, Theorem 10. For every linear transformation T from R n to R m , 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 R n to R m 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 R n maps to only one vector in R m . (e) False. See Table 3 on Page 76. 26. As we have already seen, the standard matrix of T is A = 1 -2 3 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
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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
A =
[1 −2 3
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 eche-
lon 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
A =
[2 0 −1
4 −5 2
], B =
[7 −5 1
1 −4 −3
], C =
[1 2
−2 1
], D =
[3 5
−1 4
], E =
[−5
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)
DB =
[26 −35 −12
−3 −11 −13
].
(4) EC is not defined, since E has only one column and C has two rows.
4.
A− 5I3 =
0 −1 3
−4 −2 −6
−3 1 −3
, (5I3)A =
25 −5 15
−20 15 −30
−15 5 10
.
6.
A =
4 −3
−3 5
0 1
, B =
[1 4
3 −2
].
So
b1 =
[1
3
],b2 =
[4
−2
].
and we have
Ab1 =
−5
12
3
, Ab2 =
22
−22
−2
.From either of the two ways of calculation, (which do not have much difference) we get
AB = [Ab1 Ab2] =
−5 22
12 −22
3 −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 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
row, we have
A ∼[
3 −6
0 0
].
Therefore the general solutions of Ax = 0 is
x = x2
[2
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 = BTAT , we take B = A, then we get (A2)T = (AT )2.
(d) False. In fact (ABC)T = CTBTAT not equal to CTATBT in general. A counter example
is
A =
[0 1
0 0
], B =
[0 0
1 0
], C =
[1 0
0 1
].
Then
(ABC)T =
[1 0
0 0
]6=[
0 0
0 1
]= CTATBT .
(e) True. Since (A+ B)T = AT + BT , 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,