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Mathematics

MATH 125

Scott

Fall

Description

MATH 125 (R1) Winter 2017
Additional Practice Problems for Final { Solutions
Notes: Some of the solutions here are incomplete (their main purpose is to enable you
to check your own answers). In particular, row reduction of matrices and determinant
calculations are omitted for the most part. In your exam, however, you should show all your
work, and give suitable justi▯cation for your answers. Note also that there are often multiple
ways to solve a particular problem; in these solutions, we typically just give one approach.
1. a) De▯ne what it means for an n ▯ n matrix to be invertible.
b) Find the inverse (if it exists) of the matrix
2 3
1 0 1
4 5
A = 0 1 0
2 1 1
c) For which value(s) of a is the matrix
2 3
1 0 a
40 1 a 5
2 1 a
invertible? Justify your answer.
Solution: (a) An n ▯ n matrix A is invertible if there exists another n ▯ n matrix B
such that AB =nI = BA, wherenI is the n ▯ n identity matrix.
(b) We use the standard algorithm:
2 3 2 3
1 0 1 1 0 0 1 0 0 ▯1 ▯1 1
4 0 1 0 0 1 0 5 ▯! ▯▯▯ ▯!4 0 1 0 0 1 05
2 1 1 0 0 1 0 0 1 2 1 ▯1
Since the reduced row echelon form 3f A is I (the left side of the partitioned matrix),
A is invertible and 2 3
▯1 ▯1 1
▯1 4 5
A = 0 1 0
2 1 ▯1
(the right side of the partitioned matrix).
(c) Let B be the given matrix. Then: ▯ ▯
▯1 0 a ▯
det(B) =▯0 1 a ▯ (r3▯ 2r1)
▯ ▯
0 1 ▯a
▯ ▯
▯ 1 a ▯
= ▯ 1 ▯a ▯ (expand along 1st column)
= ▯2a:
Hence B is invertible if and only if ▯2a 6= 0, i.e., if and only if a 6= 0.
2 1 ▯3 6 3
2. Let A =4 2 ▯5 115:
▯1 1 ▯3
a) A is invertible. Compute.A
b) Use A▯1to solve the following system of equations:
x 1 ▯ 3x 2 + 6x3 = 1
2x1 ▯ 5x 2 + 11x 3 = ▯3
▯x 1 + x2 ▯ 3x3 = 1
Solution (a) We use the standard algorithm:
2 3 2 3
1 ▯3 6 1 0 0 1 0 0 4 ▯3 ▯3
4 2 ▯5 11 0 1 0 5▯! ▯▯▯ ▯! 4 0 1 0 ▯5 3 1 5
▯1 1 ▯3 0 0 1 0 0 1 ▯3 2 1
Since the reduced row echelon form o3 A is I (the left side of the partitioned matrix),
A is invertible and
2 4 ▯3 ▯3 3
▯1
A = 4 ▯5 3 1 5
▯3 2 1
(the right side of the partitioned matrix).
2 3
1
(b) The matrix equation of the systx =is A~ 5, so by (a), the unique solution
1
of the system is
2 3 2 32 3 2 3
1 4 ▯3 ▯3 1 10
x = A▯14 ▯3 5 = 4 ▯5 3 154 ▯3 5 = 4 ▯13 5:
1 ▯3 2 1 1 ▯8 3. Compute the determinant of each of the following matrices:
2 3 7 0 ▯63 2 3 2 2 0 0 03
6 7 1 2 3 6 7
a)6 7 9 0 37 b)4▯1 ▯2 ▯3 5 c)6▯1 6 0 07
4▯9 ▯3 0 55 4 5 9 1 05
1 5 0 3 4 1 4 7 8 3 ▯3
2 3
2 3 1 2 3 0 4
1 0 1 0 60 1 0 0 17
60 1 0 1 7 6 7
d)41 1 1 2 5 e)61 1 1 1 ▯2 7
1 1 0 1 41 ▯2 1 0 15
0 0 1 0 2
Solution (a) Since the matrix contains a column of zeros, its determinant is 0.
(b) Since the matrix contains a row which is a multiple of the other (r = ▯r ), its
2 1
determinant is 0.
(c) Since the matrix is triangular, its determinant is the product of its diagonal entries,
i.e., (2)(6)(1)(▯3) = ▯36.
(d) The determinant is 1 (standard calculation omitted: use cofactor expansion/row
operations/column operations or a combination of all these methods).
(e) The determinant is 5 (standard calculation omitted: use cofactor expansion/row
operations/column operations or a combination of all these methods).
4. Find the rank and nullity of
2 3
1 6 11 16 21
63 8 13 18 237
65 10 15 20 25 7
4 5
7 12 17 22 27
9 14 19 24 29
(Hint: The rank is 2, and it shouldn’t take many computations to see this). What is
the determinant of the matrix?
Solution: Denote the given matrix by A. Observe that each row of A is obtained from
the previous one by adding 2 to each entry. Thus: 2 3 2 3
1 6 11 16 21 1 6 11 16 21
6 7 6 7
63 8 13 18 23 7 r2▯ r 1 6 2 2 2 2 2 7
A = 65 10 15 20 25 7 ▯! r3▯ r 2 6 2 2 2 2 2 7
4 5 4 5
7 12 17 22 27 r4▯ r 3 2 2 2 2 2
9 14 19 24 29 r5▯ r 4 2 2 2 2 2
2 3 2 3
1 6 11 16 21 1 6 11 16 21
6 2 2 2 2 2 7 r2▯ 2r 1 6 0 ▯10 ▯20 ▯30 ▯40 7
6 7 6 7
▯! r3▯ r2 6 0 0 0 0 0 7 ▯! 6 0 0 0 0 0 7 :
r4▯ r2 4 0 0 0 0 0 5 4 0 0 0 0 0 5
r5▯ r2 0 0 0 0 0 0 0 0 0 0
The last matrix is in row echelon form. Since it has two non-zero rows, A has rank
2. By The Rank Theorem, it follows that the nullity of A is 5 ▯ 2 = 3 (i.e., (no. of
columns) - rank).
Since the nullity of A is not 0, A is not invertible, and so its determinant is equal to 0.
2 x x x 3
1 2 3
5. Let A = 4 y1 y2 y35. Suppose detA = ▯4. Compute the determinant of:
z z z
1 2 3
2 3
3y1 2x1 125 z1+ x 1
63y 2x 2016 z + x 7
6 2 2 2 27
43y3 2x3 42 z3+ x 35
0 0 2 0
Solution :
▯ ▯
▯3y1 2x1 125 z1+ x 1 ▯ ▯ ▯
▯ ▯ ▯3y 1 2x 1 z1+ x 1 ▯
▯3y2 2x2 2016 z 2 x 2 ▯= ▯2 ▯3y 2x z + x ▯ (expand along 4th row)
▯3y3 2x3 42 z3+ x 3 ▯ ▯ 2 2 2 2 ▯
▯ ▯ ▯3y 3 2x 3 z3+ x 3 ▯
0 0 2 0
▯ ▯
▯ y1 x1 z1+ x1 ▯ 1 1
= ▯2(2)(3) ▯ y2 x2 z2+ x2 ▯ (c1▯ ;c ▯2)
▯ ▯ 3 2
y3 x3 z3+ x3
▯ ▯
▯ y x z ▯
▯ 1 1 1 ▯
= ▯12 ▯ y2 x2 z2 ▯ (c3▯ c2)
▯ y3 x3 z3 ▯ ▯ ▯
▯x1 y1 z1 ▯
= 12 ▯x2 y2 z2 ▯ (c1$ c2)
▯ ▯
x3 y3 z3
= 24det(A ) = 12det(A) = 12(▯4) = ▯48:
2 3
2 3 1 k
6 7
6. Let A = 61 0 3 0 7 .
40 0 1 1 5
0 1 1 0
a) Compute det(A).
b) For what value(s) of k is the matrix A invertible? Justify your answer.
c) For what value(s) of k is the set
▯▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯
2 3 1 k ; 1 0 3 0 ; 0 0 1 1 ;0 1 1 0
linearly independent? Justify your answer.
Solution: (a) Using direct cofactor expansion, we ▯nd that det(A) = k + 8.
(b) A is invertible if and only if det(A) 6= 0, which (by part (a)) holds if and only if
k 6= ▯8.
(c) In general, the n rows of an n ▯ n matrix are linearly independent if and only if
that matrix is invertible. Since the set given here consists of the 4 rows of A, it follows
that it is linearly independent if and only if k 6= ▯8 (by part (b)).
7. Is 8 2 3 2 3 2 3 9
< 1 1 ▯1 =
4 5 4 5 4 5
: 1 ; ▯1 ; 1 ;
▯1 1 1
3
a basis of R ? Justify your answer.
Solution: Let 2 3
1 1 ▯1
4 5
A = 1 ▯1 1
▯1 1 1
be the 3▯3 matrix with the three given vectors as its columns. Using cofactor expansion
or elementary row operations, one checks that det(A) = ▯4 6= 0, and so A is invertible.
3
Since A is invertible, its columns form a basis of R , so the answer is yes. 8. Is 82 3 2 3 2 39
< 1 1 ▯1 =
6 17 6 ▯17 6 17
> 4▯15 4 15 4 15>
: ;
0 0 1
4
a basis of R ? Justify your answer.
4 4
Solution: No! Every basis of R contains exactly 4 vectors (R has dimension 4), but
the given set only contains 3 vectors.
4
9. Do the following four vectors form a basis of R ? Justify your answer.
2 3 2 3 2 3 2 3
1 1 1 0
6 1 6 1 6 706 71
6 7 ; 7 6;7 6;7
4 1 4 0 4 514 51
0 1 1 1
Solution Let 2 3
1 1 1 0
6 1 1 0 1 7
A = 4 5
1 0 1 1
0 1 1 1
be the 4 ▯ 4 matrix with the four given vectors as its columns. Using elementary row
operations (or cofactor expansion), one computes that det(A) = ▯3 6= 0, and so A is
4
invertible. Since A is invertible, its columns form a basis of R , so the answer is yes.
10. a) Give the de▯nition of the coordinate vector ox in S with respect to a
basis B of a subspace S of R .
b) Let S be the subspace of R spanned by the three vectors
2 3 2 3 2 3
1 ▯2 1
405;4 0 5;415 :
1 ▯2 5
Find a basis B of S.
c) With S, B as in b), compx] for
B
2 3
0
x =425
8 d) With S, B as in b), compy if~
▯5 ▯
[yB =
▯3
Solution: (a) Let B =1 ~2;:::~kg. Ix 2 S, then the coordinate vecx with ~
respect to the basis B is the k ▯ 1 vector
2 3
c
6 17
6 c27
[xB = 4 . 5 ;
.
ck
where 1 ;2 ;::k;c is the unique sequence of scalars such that
x = 1 1+ c2 2 ▯▯▯ + k k:
(b) Let 2 3
1 ▯2 1
A = 4 0 0 1 5 ;
1 ▯2 5
so that S = col(A). To ▯nd a basis of S, we bring A to reduced row echelon form:
2 3 2 3
1 ▯2 1 1 ▯2 0
A = 4 0 0 1 5▯! ▯▯▯ ▯! 4 0 0 1 5
1 ▯2 5 0 0 0
Since the leading entries of rref(A) lie in columns 1 and 3, columns 1 and 3 of A form
a basis of col(A) = S. In other words,
8 9
2 1 3 2 13
< =
B = 4 0 5 ; 15
: 1 5 ;
is a basis of S.
(c) We need to exprx as a linear combination of the two basis vectors found in (b).
We solve the appropriate linear system:
2 3 2 3
1 1 0 1 0 ▯2
4 5 4 5
0 1 2 ▯! ▯▯▯ ▯! 0 1 2
1 5 8 0 0 0 2 13 2 1 3 ▯ ▯
▯2
Thusx = ▯24 05 + 24 1 5 , and sx]B= 2 .
1 5
▯ ▯
5
(d) IyB = ▯3 , then
2 1 3 2 1 3 2 23
y = 5 0 5 ▯ 34 1 5= 4 ▯3 5 :
1 5 ▯10
Remark : The ordering of the vectors in B matters here!
11. a) Find a basis B of sv;~;x), where
2 3 2 3 2 3
1 ▯1 5
4 5 4 5 4 5
v = 1 ;~ = ▯1 ;x = 5
1 2 ▯1
b) Compute the coordinate vectvB ,~]B, andx]Bwith respect to the basis B
in a).
Solution (a) Set S = spv;~;x), and let
2 3
1 ▯1 5
A = 4 1 ▯1 5 5;
1 2 ▯1
so that S = col(A). To ▯nd a basis of S, we bring A to reduced row echelon form:
2 3 2 3
1 ▯1 5 1 0 3
4 5 4 5
A = 1 ▯1 5 ▯! ▯▯▯ ▯! 0 1 ▯2
1 2 ▯1 0 0 0
Since the leading entries of rref(A) lie in columns 1 and 2, columns 1 and 2 of A form
a basis of col(A) = S. In other words,
8 2 3 2 39
< 1 ▯1 =
B = 4 1 5;4 ▯1 5 = fv;~g
: 1 2 ;
is a basis of S.
(b) Sinv = 1v + ~, and sinc~ = v + 1~, we have
▯ ▯ ▯ ▯
1 0
[vB = and [~]B= :
0 1 Finally, from the computation of rref(A) in part (a), we seex = 3v ▯ 2~, and so
▯ ▯
[x] = 3
B ▯2
(remember that the dependency relations among the columns of rref(A) are the same
as those among the columns of A).
12. Let
▯2 1 ▯
A =
1 1
2
Then A is invertible and therefore its columns form a basis B of R . If we set
▯ ▯ ▯ ▯
u = 2 and v = 1 ;
1 1
2
then B = fu;vg. Give a formula that computes x]Bfor everyx in R .
▯ ▯
a 2
Solution: Suppose ~x = b 2 R . We want formulae for the coe▯cients which
express x as a linear combination ou and v. For this, we solve the relevant linear
system:
▯ ▯ ▯ ▯
2 1 a 1 0 a ▯ b
1 1 b ▯! ▯▯▯ ▯! 0 1 2b ▯ a
Thus x = (a ▯ b)u + (2b ▯ av, and so
▯ a ▯ b ▯
x]B=
2b ▯ a
(the coe▯cient matrix A of our linear system is invertible here, so the above answer
▯ a ▯
may also be obtained as A▯1 ).
b
Remark : The above observation extends as follows: If B =~ ~ ;:::;~ g is a basis
n n 1 2 n
of R , andx 2 R , then
x]B= A ▯1x;
where ▯ ▯
A = v1 ~2 ▯▯▯ vn :
2 3
1 2 6 ▯1 ▯5
13. Let A = 4 ▯1 ▯1 ▯1 0 1 5
▯2 ▯2 ▯2 1 3 n
a) State the de▯nition of the dimension of a subspace of R .
b) Find a basis of row(A).
c) Find a basis of col(A).
d) Find a basis of null(A).
e) For each of the subspaces in b), c), d), determine their dimensions.
f) State the rank and nullity of A.
Solution (a) The dimension of a subspace S of R is the number of vectors in any
basis of S.
To answer the remaining parts, we compute rref(A):
2 3 2 3
1 2 6 ▯1 ▯5 1 0 ▯4 0 2
A = 4▯1 ▯1 ▯1 0 1 5▯! ▯▯▯ ▯! 4 0 1 5 0 ▯3 5
▯2 ▯2 ▯2 1 3 0 0 0 1 1
(b) A basis of row(A) is given by the non-zero rows of rref(A):
▯
[ 1 0 ▯4 0 2 ];[0 1 5 0 ▯3 ];[ 0 0 1 1 ] :
(c) Since the leading entries of rref(A) are in columns 1, 2 and 4, columns 1, 2 and 4
of A form a basis of col(A):
8 2 3 2 3 2 39
< 1 2 ▯1 =
4 ▯1 5 ; ▯1 5;4 0 5
: ;
▯2 ▯2 1
(d) To ▯nd a basis of null(A), we have to solve the homogeneousx = 0.r system A~
But, from the above computation of rref(A):
2 3
1 0 ▯4 0 2 0
~ 4 5
[Aj0] ▯! ▯▯▯ ▯! 0 1 5 0 ▯3 0
0 0 0 1 1 0
We see that there are three leading v1r2ables4(x , x and x ). Assigning the parametric
values s and t to the free var3ables5x and x , respectively, we get that the general
solution of the system is
2 3 2 3
4 ▯2
6 ▯5 7 6 3 7
6 7 6 7
x = 6 17 + t6 0 7 (s;t 2 R):
4 05 4 ▯1 5
0 1 Hence 82 3 2 39
> 4 ▯2 >
<6 ▯5 7 6 37=
6 17 ; 07
>4 5 4 5>
: 0 ▯1 ;
0 1
is a basis of null(A).
(e) The dimension of row(A) is 3, the dimension of col(A) is 3, and the dimension of
null(A) is 2.
(f) rank(A) = 3, nullity(A) = 2.
2 3
0 1 2 3 0 1
14. Let A =0 2 4 5 1 3 5
0 ▯3 ▯6 ▯8 1 ▯1
a) State the de▯nition of a basis of a subspace of R .
b) Find a basis of row(A).
c) Find a basis of col(A).
d) Find a basis of null(A).
e) Find the rank and nullity of A.
n
Solution: (a) Let S be a subspace of R . A subset B of S is a basis of S if B is linearly
independent and span(B) = S.
To answer the remaining parts, we ▯nd rref(A):
2 3 2 3
0 1 2 3 0 1 0 1 2 0 0 ▯1=2
A =4 0 2 4 5 1 35 ▯! ▯▯▯ ▯!4 0 0 0 1 0 1=2 5
0 ▯3 ▯6 ▯8 1 ▯1 0 0 0 0 1 3=2
(b) A basis of row(A) is given by the non-zero rows of rref(A):
▯
[0 1 2 0 0 ▯1=2 ];[ 0 0 1 0 1=2 ];[ 0 0 0 1 3=2 ] :
(c) Since the leading entries of rref(A) are in columns 2, 4 and 5, columns 2, 4 and 5
of A form a basis of col(A):
8 2 3 2 3 2 39
< 1 3 0 =
4 25 ; 55 ; 1 5
: ;
▯3 ▯8 1 ~
(d) To ▯nd a basis of null(A), we have to solve the homogenex = 0.near system A~
But, from the above computation of rref(A):
2 3
0 1 2 0 0 ▯1=2 0
[Aj0] ▯! ▯▯▯ ▯!0 0 0 1 0 1=2 0 5
0 0 0 0 1 3=2 0
We see that there are three leading 2a4iable5 (x , x and x ). Assigning the parametric
values r and s and t to the free 1ariab3es x 6nd x and x , respectively, we get that
the general solution of the system is
2 3 2 3 2 3
1 0 0
6 0 7 6 ▯2 7 6 1=2 7
6 7 6 7 6 7
x = r 0 7+ s6 1 7+ t6 0 7 (r;s;t 2 R):
6 0 7 6 0 7 6 ▯1=2 7
4 0 5 4 0 5 4 ▯3=2 5
0 0 1
Hence 82 3 2 3 2 39
> 1 0 0 >
>6 0 7 6 ▯2 7 6 1=2 7>
<6 7 6 7 6 7=
6 0 7;6 1 7;6 0 7
>6 0 7 6 0 7 6 ▯1=2 7>
>4 0 5 4 0 5 4 ▯3=2 5>
: ;
0 0 1
is a basis of null(A).
(e) rank(A) = 3, nullity(A) = 3.
15. In each of the following cases, determine if the given set is a subspace of R (for the
appropriate n) or not. For each case in which the set is a subspace, provide a proof
that it is subspace of R and ▯nd a basis for the subspace. For each case in which the
set is not a subspace of R , state one of the properties of a subspace that does not
hold and give a counter-example showing that the property fails.
▯▯ ▯ ▯
x ▯
a) 1 = 2 R2▯x = 0
y
▯▯x▯ ▯ ▯
b) 2 = 2 R2▯x + y ▯ 0
y
( ▯ ▯ ▯ )
x 2▯
c) 3 = y 2 R ▯ the point (x;y) is on the x-axis or y-axis (or both) 8 2 3 ▯ 9
< x ▯ =
d) S4= 4y5 2 R 3▯ 2x = 4y and y = 3z
: z ▯ ;
Solution: We recall that a subset S of R is a subspace of R if the following hold:
~
(S1) 0 2 S.
(S2) Ifu;v 2 S, thenu +~v 2 S.
(S3) Ifv 2 S and ▯ 2 R, then v 2 S.
(a) S is a subspace of R . We show this by checking that (S1-S3) all hold:
1
~
(S1) Since the ▯rst entry of the zero vector is certainly equal 1o 0, 0 2 S , so (S1) holds.
(S2) Let u;v 2 S1. Then
▯ ▯ ▯ ▯
0 0
u = y and v = z
for some y;z 2 R. Hence
▯ ▯
u +v = 0 ;
y + z
which again lies in1S (▯rst entry is 0). Thus (S2) holds.
▯ ▯
(S3) Let v 2 S , so thav = 0 for some y 2 R. If ▯ 2 R is a scalar, then
1 y
▯ ▯ ▯ ▯
▯(0) 0
▯v = ▯ = ;
▯y ▯y
which again lies in1S (▯rst entry is 0). Thus (S3) holds.
Remark: More directly, one may simply note that S1is a subspace because it is the
solution set of the homogeneous linear equation x = 0 (i.e., the null space of the matrix
[1 0]).
▯ ▯
Now, since every vector in1S is of the form y for some y 2 R, we have
1
▯▯ ▯▯
0
S1= span :
1
Since a single non-zero vector is (trivially) linearly independent, it follows that
▯▯ ▯▯
0
1 is a basis of1S .
Remark: As per the previous remark, you may also view this as the problem of ▯nding
a basis of the null space of the matrix [1 0].
(b) S2is not a subspace of R . Indeed, while conditions (S1) and (S2) hold here,
condition (S3) fails in general. For example, if
▯ ▯
1
v = ;
0
then v 2 S2, because 1 + 0 ▯ 0. However,
▯ ▯
▯1
▯~v =
0
does not lie in2S , since ▯1 + 0 < 0. Hence (S3) (with ▯ = ▯1) fails.
(c) S is not a subspace of R . Indeed, while conditions (S1) and (S3) hold here, (S2)
3
does not. For example, the vectors
▯ ▯ ▯ ▯
1 0
u = and v =
0 1
both lie in 3 , since (1;0) lies on the x-axis and (0;1) lies on the y-axis. However,
▯ ▯
u +~v = 1
1
does not belong to S , since (1;1) lies neither on the x-axis nor y-axis. Thus, condition
3
(S2) fails in general.
(d) S4is a subspace of R . We show this by verifying that (S1-S3) all hold:
2 3
0
(S1) 0 =4 0 5 2 S4, since 2(0) = 4(0) and 0 = 3(0). Hence (S1) holds.
0
(S2) Letu;v 2 S4. Then
2 3 2 3
a d
u = 4 b 5 and v = 4 e 5 ;
c f
where
2a = 4b; b = 3c; 2d = 4e and e = 3f (i) Now 2 3
a + d
u +~v = 4 b + e5 ;
c + f
and by (i), we have
2(a + d) = 2a + 2d = 4b + 4e = 4(b + e)
and
b + e = 3c + 3f = 3(e + f):
Hence u +~v 2 S4, and so (S2) holds.
2 3
a
(S3) Letv 2 S , so thav = 4 b 5, where 2a = 4b and b = 3c. If ▯ 2 R is a scalar,
4
c
then 2 3
▯a
▯v = 4 ▯b 5 :
▯c
Since 2a = 4b and b = 3c, we have
2(▯a) = ▯(2a) = ▯(4b) = 4(▯b)
and
▯b = ▯(3c) = 3(▯c):
Hence ▯v 2 S , and so (S3) holds.
4
3
Remark : More directly, one can note tha4 S is a subspace of R because it is the
solution set of the homogeneous linear system
2x ▯ 4y = 0
;
y ▯ 3z = 0
i.e., the null space of the matrix
▯ ▯
2 ▯4 0
A = 0 1 ▯3 :
As per the remark, ▯nding a basis of4S amounts to ▯nding a basis of null(A). We
begin by solving the homogeneous linear systex = 0:
▯ ▯ ▯ ▯
2 ▯4 0 0 1 0 ▯6 0
[Aj0] = ▯! ▯▯▯ ▯!
0 1 ▯3 0 0 1 ▯3 0
We see that there are two leading variabl1s 2x ;x ). Assigning the parametric value t
to the remaining variabl3 x , we ▯nd that the general solution of the system is 2 3
6
4 5
x = t 3 (t 2 R);
1
and so 8 2 3 9
< 6 =
4 3 5
: ;
1
is a basis of4S .
n n
16. Let V and W be subspaces of R . Let S be the set of all vectorsx in R such that
x = v + ~, for somev 2 V and some w~ 2 W, i.e.,
S = fx 2 R j x = v + ~; for some ~v 2 V and some w~ 2 Wg:
n
Show that S is a subspace of R .
Solution : We verify that conditions (S1-S3) (see the previous solution) hold:
(S1) Since V and W are subspaces of R , 0 2 V and 0 2 W. Hence
0 = 0 + 0 2 S;
and so (S1) holds.
(S2) Let x and y be vectors in S. Then
x = ~1+ w~1 and y = ~2+ w~2;
where v1 ~22 V and w ~1;~22 W. Since V is a subspace of R ,~1+ v~22 V , and since
n
W is also a subspace of R ,~1+ w~22 W. Thus:
x + y = (v ~1+ w~1) + ~2+ w~2)
= (v ~ + v~ ) + ~ + w~ ) 2 S;
1 2 1 2
(since it is the sum of an element of V and an element of W), and so condition (S2)
holds.
(S3) Let x 2 S, so thax = v + ~ for some v 2 V and w~ 2 W. If ▯ 2 R, then
▯x = ▯(v + ~) = ▯v + ▯~:
n n
But since V is a subspace of R , v 2 V , and since W is a subspace of R ,~ 2 W.
We thus conclude that ▯x 2 S, being the sum of an element of V and an element of
W. Hence condition (S3) holds.
n
We have veri▯ed that (S1-S3) hold, and hence shown that S is a subspace of R . 5 5 5
17. Let T : R ! R be a linear transformation. Let S be the set of all vectors x 2 R
such that T(x) = 2x, i.e.,
5
S = fx 2 R j T(~x) = 2xg:
Show that S is a subspace of R .
Solution: We verify that conditions (S1-S3) (see the solution to Q. 15) hold:
(S1) Since T is a linear transformation, T(0) = 0 = 2(0), and so 0 2 S. Hence (S1)
holds.
(S2) Let u;v 2 S, so that
T(u) = 2u and T(v) = 2v: (ii)
Then
T(u +~v) = T(~ u) + T(v) (since T is a linear transformation)
= 2~u + 2v (by (ii))
= 2(~u +~v);
and so u +~v 2 S. Hence (S2) holds.
(S3) Let v 2 S, so that Tv) = 2v. If ▯ 2 R, then
T(▯~v) = ▯T(~ v) (since T is a linear transformation)
= ▯(2~ v) (by assumption)
= 2(▯~ v);
and so ▯v 2 S. Hence (S3) holds.
5
We have veri▯ed that (S1-S3) all hold, and hence that S is a subspace of R .
18. Determine whether the following are linear transformations from R to R . If the map
is a linear transformation, provide a proof that it is linear transformation. If the map
is a not linear transformation, state one of the properties of a linear transformation
that does not hold and give a counter-example showing that the property fails.
▯▯ x▯ ▯x + y▯
a) T1 =
y x ▯ y
▯▯ ▯▯ ▯ ▯
b) T x = xy
2 y x 2
▯▯ ▯▯ ▯ ▯
x 2x + y
c) T3 y = 0 ▯▯ ▯▯ ▯ ▯
x 2x + y
d) T 4 y = 1
n m
Solution: Recall that a map T : R ! R is a linear transformation if the following
conditions hold:
n
(LT1) T(~u +~v) = T(u) + T(v) for alu;v 2 R .
n
(LT2) T(▯~v) = ▯T(v) for alv 2 R and ▯ 2 R.
(a) T is a linear transformation. We show this by verifying that (LT1) and (LT2) hold:
1
▯ ▯ ▯ ▯
a c
(LT1) Let u = and v = be vectors in R . Then
b d
▯▯ ▯▯ ▯ ▯
a + c (a + c) + (b + d)
T1(u +~v) = T1 b + d = (a + c) ▯ (b + d) ;
and
▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯
a + b c + d a + b + c + d (a + c) + (b + d)
T1(u)+T 1v)) = a ▯ b + c ▯ d = (a ▯ b) + (c ▯ d) = (a + c) ▯ (b + d) :
Hence T1(u +~v) = T1(u) + 1 v), and so (LT1) holds.
▯ ▯
a
(LT2) Let v = 2 R . If ▯ 2 R, then
b
▯▯ ▯▯ ▯ ▯
▯a ▯a + ▯b
T1(▯v) = T1 = ;
▯b ▯a ▯ ▯b
and ▯ ▯ ▯ ▯ ▯ ▯
a + b ▯(a + b) ▯a + ▯b
▯T 1v) = ▯ = = :
a ▯ b ▯(a ▯ b) ▯a ▯ ▯b
Hence T1(▯v) = ▯T 1v), so (LT2) holds.
We have shown that both (LT1) and (LT2) hold, and hence that T1is a linear trans-
formation.
(b) T2is not a linear tra▯sfo▯mation. Indeed, conditions (LT1) and (LT2) both fail
1
here. For example, lev = . Then
0
▯▯ ▯▯ ▯ ▯
2 0
T 22v) = T2 0 = 4 ; but ▯▯ ▯▯ ▯ ▯ ▯ ▯
2 0 0
2T 2v) = 2T 2 0 = 2 1 = 2 :
Hence T (2v) 6= 2T v), and so (LT2) fails in general (This also shows that (LT1) fails:
2 2
T2(v +~v) 6= 2v) + T2(v)).
Rem

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