Linear Algebra II { Homework 1 Solution
Math 341 Section 51: Tuesday, September 11, 2007
Due: Tuesday, September 18, 2007
1. (Each 5 points) Show that each of the following is a linear transformation:
(1) T : R2 ! R2 dened by T (x; y ) = (2x y; x).
(2) T : R3 ! R2 dened by T (x; y; z ) = (z; x + y ).
(3) T : R ! R2 dened by T (x) = (2x; x).
(4) T : R2 ! R3 dened by T (x; y ) = (x + y; y; x).
Proof. Since R, R2 , and R3 are vector spaces, all T 's above are transformations. We check
the linearity, i.e.,
T (av + bw) = aT (v ) + bT (w);
where v and w are arbitrary vectors and a, b are arbitrary scalars, i.e., real numbers. First
we simplify the given transformations to save time in computation and then using the
simplied form, we check the linearity.
(1) T (x; y ) = (2x y; x) = (2x; x) + ( y; 0) = x(2; 1) + y ( 1; 0) = xA + yB;
where A = (2; 1) and B = ( 1; 0). For v = (x1 ; y1 ), w = (x2 ; y2 )
numbers a; b, we have
2 R2 and arbitrary
av + bw = (ax1 ; ay1 ) + (bx2 ; by2 ) = (ax1 + bx2 ; ay1 + by2 );
T (av + bw) = T (ax1 + bx2 ; ay1 + by2 )
= (ax1 + bx2 )A + (ay1 + by2 )B
(A and B are dened above.)
= a (x1 A + y1 B ) + b (x2 A + y2 B )
= aT (x1 ; y1 ) + bT (x2 ; y2 ) = aT (v ) + bT (w):
(2) T (x; y; z ) = (z; x+y ) = (z; 0)+(0; x)+(0; y ) = z (1; 0)+x(0; 1)+y (0; 1) = xA+yB +zC;
where A = (0; 1) = B and C = (1; 0). For v = (x1 ; y1 ; z3 ), w = (x2 ; y2 ; z3 )
arbitrary numbers a; b, we have
2 R3 and
av + bw = (ax1 ; ay1 ; az1 ) + (bx2 ; by2 ; bz2 ) = (ax1 + bx2 ; ay1 + by2 ; az1 + bz2 );
T (av + bw) = T (ax1 + bx2 ; ay1 + by2 ; az1 + bz2 )
= (ax1 + bx2 )A + (ay1 + by2 )B + (az1 + bz2 )C (A; B; C are dened above.)
1
= a(x1 A + y1 B + z1 C ) + b(x2 A + y2 B + z2 C )
= aT (x1 ; y1 ; z1 ) + bT (x2 ; y2 ; z2 ) = aT (v ) + bT (w):
(3) T (x) = (2x; x) = x(2; 1) = xA;
where A = (2; 1). For v, w 2 R and arbitrary numbers a, b, we have
T (av + bw) = (av + bw)A = av A + bwA = aT (v ) + bT (w):
(4) T (x; y ) = (x + y; y; x) = (x; 0; x) + (y; y; 0) = x(1; 0; 1) + y (1; 1; 0) = xA + yB;
where A = (1; 0; 1) and B = (1; 1; 0). For v = (x1 ; y1 ), w = (x2 ; y2 ) in R2 and arbitrary
numbers a, b, we have
av + bw = a(x1 ; y1 ) + b(x2 ; y2 ) = (ax1 + bx2 ; ay1 + by2 );
T (av + bw) = (ax1 + bx2 )A + (ay1 + by2 )B
= a(x1 A + y1 B ) + b(x2 A + y2 B )
= aT (x1 ; y1 ) + bT (x2 ; y2 ) = aT (v) + bT (w):
2. (Each 5 points) Show that each of the following is not a linear transformation:
(1) T : R2 ! R2 dened by T (x; y ) = (x2 ; y 2 ).
(2) T : R3 ! R2 dened by T (x; y; z ) = (x + y + z; 1).
(3) T : R ! R2 dened by T (x) = (1; 1).
(4) T : R2 ! R3 dened by T (x; y ) = (xy; y; x).
Proof. We use the same argument as used in the solution of the previous problem. First
we simplify the given transformations and then check the linearity.
(1) T (x; y ) = (x2 ; y 2 ) = x2 (1; 0) + y 2 (0; 1) = x2 A + y 2 B;
where A = (1; 0), B = (0; 1). For v = (x1 ; y1 ), w = (x2 ; y2 ) 2 R2 , and any a, b, we have
av + bw = a(x1 ; y1 ) + b(x2 ; y2 ) = (ax1 + bx2 ; ay1 + by2 );
T (av + bw) = (ax1 + bx2 )2 A + (ay1 + by2 )2 B
= (ax21 + 2abx1 x2 + b2 x22 )A + (ay12 + 2aby1 y2 + b2 y22 )B
6= a(x21A + y12B ) + b(x22A + y22B )
= aT (x1 ; y1 ) + bT (x2 ; y2 ) = aT (v ) + bT (w):
2
(2) T (x; y; z ) = (x + y + z; 1) = x(1; 0) + y (1; 0) + z (1; 0) + (0; 1) = xA + yB + zC + D;
where A = (1; 0) = B = C and D = (0; 1). For v = (x1 ; y1 ; z1 ), w = (x2 ; y2 ; z2 ) 2 R3 , and
any a, b, we have
av + bw = a(x1 ; y1 ; z1 ) + b(x2 ; y2 ; z2 ) = (ax1 + bx2 ; ay1 + by2 ; az1 + bz2 );
T (av + bw) = (ax1 + bx2 )A + (ay1 + by2 )B + (az1 + bz2 )C + D
= a(x1 A + y1 B + z1 C ) + b(x2 A + y2 B + z2 C ) + D
6= a(x1A + y1B + z1C + D) + b(x2A + y2B + z2C + D)
= aT (x1 ; y1 ; z1 ) + bT (x2 ; y2 ; z2 ) = aT (v) + bT (w):
(3) T (x) = (1; 1) = A:
For v , w 2 R, and any a, b, we have
T (av + bw) = A 6= aA + bA = aT (v ) + bT (w):
(4) T (x; y ) = (xy; y; x) = xy (1; 0; 0) + y (0; 1; 0) + x(0; 0; 1) = xA + yB + xyC;
where A = (0; 0; 1), B = (0; 1; 0), and C = (1; 0; 0). For v = (x1 ; y1 ), w = (x2 ; y2 ) 2 R2 ,
and any a, b, we have
av + bw = a(x1 ; y1 ) + b(x2 ; y2 ) = (ax1 + bx2 ; ay1 + by2 );
T (av + bw) = (ax1 + bx2 )A + (ay1 + by2 )B + (ax1 + bx2 )(ay1 + by2 )C
6= a(x1A + y1B + x1y1C ) + b(x2A + y2B + x2y2C )
= aT (x1 ; y1 ) + bT (x2 ; y2 ) = aT (v) + bT (w):
(on Problem 1 & 2). In order to make our argument to be simple and to
understand clearly and easily, let us consider the functions f , g and h on real numbers.
The graph of f (x) = ax is a straight line passing through the origin. So without using
the equation for linearity, f (x) is a linear function. All transformations in the problem
1 have the same form as f (x): (1) T (x; y ) = xA + yB (just extension), (2) T (x; y; z ) =
xA + yB + zC (just extension), (3) T (x) = xA, and (4) T (x; y ) = xA + yB .
However, for a non{zero constants a, b, the graphs of g (x) = ax + b and h(x) = ax2
are not straight lines passing through the origin. Thus g (x) and h(x) are not linear.
In the problem 2 above, (1) T (x; y ) = x2 A + y 2 B has the same form as h(x) and (2)
T (x; y; z ) = xA + yB + zC + D has the same form as g (x). (If D = 0, then T is linear.)
(3) T (x) = A represents a non{zero constant function, which is clearly a straight line, but
Remark 1
3
not passing through the origin. So it is not linear. In (4) T (x; y ) = xA + yB + xyC , the
term xyC yields the non{linearity. (If C = 0, then T is linear by the same reason as in
(1) in the problem 1.)
So what? The point is as follows: before we check the linearity by computing the linearity
equation, simplifying the transformation as above and comparing with the linear function
dened on real numbers will save great amount of time.
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3. (10 points) Let V be the vector space of all n n matrices, and let B be a xed n n
matrix. If T (A) = AB BA, show that T is a linear transformation from V to V .
Proof. Before we prove, the Remark above implies T is linear because it looks like f (x) =
xb
bx dened on real numbers, where b is some constant, and f (x) is linear.
Let A, A0 be arbitrarily chosen in V and any numbers a, a0 . Then we have
T (aA + a0 A0 ) = (aA + a0 A0 )B B (aA + a0 A0 )
= a(AB BA) + a0 (A0 B BA0 ) = aT (A) + a0 T (A0 ):
Remark 2.
What is the kernel space of T ? We observe
ker(T ) = fA 2 V : 0 = T (A) = AB
BA; i:e:; AB = BAg :
That is, the kernel space of T is the set of all matrices in V which commute with B with
respect to the multiplication. So ker(T ) has the identity matrix, the zero matrix and the
inverse matrix of B (if exists). From my past experience, the transformation AB BA for
a xed B is useful and powerful. I remember this operator was studied in the Theory of
Ordinary Dierential Equations and the Matrix Analysis.
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4. (Each 5 points) (1) Let T : R2 ! R2 dened by T (x; y ) = ( y; x). Show that
T 4 = I , where I is the identity transformation, i.e., I (x; y ) = (x; y ), and T 4 (x; y ) =
T (T (T (T (x; y )))).
Proof. Let us just compute T 4 :
T (x; y ) = ( y; x);
T 2 (x; y ) = T (T (x; y )) = T ( y; x) = ( x; y );
T 3 (x; y ) = T (T 2 (x; y )) = T ( x; y ) = (y; x);
T 4 (x; y ) = T (T 3 (x; y )) = T (y; x) = (x; y ):
Thus T 4 = I , identity transformation.
4
(2) Let S : R2 ! R2 dened by S (x; y ) = (x cos(=n) y sin(=n); x sin(=n) + y cos(=n)),
where n is a xed natural number. Show that S 2n = I . (Hint: Use the idea of (1) above.)
Proof. If we compute S 2 , S 3 , and S 4 just as we did in (1), we will get into a disaster which
is horrible as much as an earthquake. It is time to use our brain. We simplify S as follows:
letting x = (x; y ) and
0
1
cos(
=n
)
sin(
=n
)
A;
M =@
sin(=n) cos(=n)
we have S (x) = S (x; y ) = xM . Now we compute S 2 , S 3 , S 4 and observe the hidden rule:
S 2 (x) = S (S (x)) = S (xM ) = xM 2 ;
S 3 (x) = S (S 2 (x)) = S (xM 2 ) = xM 3 ;
S 4 (x) = S (S 3 (x)) = S (xM 3 ) = xM 4 :
That is, we observe S m (x) = xM m , where m is any natural number. So M m = I (identity
matrix) if and only if S m = I (identity transformation). Since we have found the matrix
M corresponding to S , we compute M 2 , and M 4 just as we did before:
0
1
0
1
cos2 (=n) sin2 (=n) 2 cos(=n) sin(=n) A @ cos(2=n) sin(2=n) A
2
@
M =
=
;
2 cos(=n) sin(=n) cos2 (=n) sin2 (=n)
sin(2=n) cos(2=n)
0
1
cos(4
=n
)
sin(4
=n
)
A:
M4 = @
sin(4=n) cos(4=n)
Thus, we can see
0
M 2n = @
1
0
1
cos(2n=n) sin(2n=n) A @1 0A
=
= I;
sin(2n=n) cos(2n=n)
0 1
which is the identity matrix. Therefore, we conclude S 2n = I (identity transformation).
Remark 3.
(i) Simple trigonometric formulas used above:
cos(2) = cos2 sin2 ;
sin(2) = 2 sin cos :
(ii) We recall from the example 1 on page 86 (textbook) that S in (2) is a rotation of the
plane through the angle =n while T in (1) is the rotation through the angle =2. When
we rotate a point consecutively by 4 times with the angle =2, the point should come back
to its rst place, because in the end it travels 4 (=2) = 2 = 360 on the circle. For
this geometrical reason, T 4 = I . By the same argument, rotation of a point by 2n times
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with the angle =n gives the rst place and so S 2n = I .
Besides the rotation, projection and reection are also easy to be understood when we
approach from the geometry. It is because they are geometrical concepts. So understanding
in the geometrical way is much easier than understanding via the algebra. However, as we
can see, for the rigorous proof, we need algebra.
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5. (10 points) Let T : R2 ! R2 dened by T (x; y ) = (x; y ). Show that T is an orthogonal
reection with reecting line x{axis on R2 plane.
Proof. Three properties dening a general orthogonal reection S with reecting certain
object (line/plane) are as follows:
(1) For x in one side of , S (x) should be in the other side of .
(2) The distance between x and should be equal to the distance between S (x) and .
(3) The line passing trough two points x and S (x) should be orthogonal to .
So we prove those three properties for T . Let a = (a; b) be arbitrarily chosen with b 0.
(1) Since b 0, clearly T (a) = (a; b) belongs to the lower area of x{axis.
(2) The distance between a and x{axis is b and the distance between T (a) = (a; b) and
x-axis is also j bj = b.
(3) The line passing through two points a = (a; b) and T (a) = (a; b) has the equation
x = a on the R2 plane, which is clearly orthogonal to the x{axis.
Since T satises all three properties above, T is an orthogonal reection with reecting
line x{axis on R2 plane.
6. (10 points) Let V and W be vector spaces and T : V ! W a linear transformation.
Prove that T (av + bw) = aT (v) + bT (w) if and only if T (v + w) = T (v ) + T (w) and
T (av ) = aT (v ), where v , w are arbitrary elements of V and a, b are arbitrary numbers.
Proof. We prove the following two statements.
(1) If T (av + bw) = aT (v) + bT (w), then T (v + w) = T (v ) + T (w) and T (av) = aT (v):
Putting a = 1 = b in the condition, we get T (v + w) = T (v ) + T (w). Putting b = 0 in the
condition, we get T (av) = aT (v). Thus, the statement (1) is proved.
(2) If T (v + w) = T (v) + T (w) and T (av) = aT (v), then T (av + bw) = aT (v ) + bT (w):
For v 2 V , av 2 V . (V is a vector space.) So the rst condition should work for av and
bw and the conditions implies
T (av + bw) = T (av ) + T (bw) = aT (v ) + bT (w):
6
Therefore, the statement (2) is proved.
Please memorize that all the followings have the same meaning.
\A is equivalent to B ."
\A if and only if B ."
\If A, then B . Also if B , then A."
\A is necessary and sucient to B ."
\A () B ."
Remark 4.
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Linear Algebra II – Homework 2 Solution
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Math 341 Section 51: Tuesday, September 18, 2007
Due: Tuesday, September 25, 2007
1. (10 points) Let T : R3 → R be a linear transformation defined by T (x, y, z) = x − 3y + 2z. Find
a basis for the kernel of T . What are dim(ker(T )) and dim(Im(T ))?
Solution. (1) Kernel of T : The definition of the kernel of T is as follows
{
}
ker(T ) = (x, y, z) ∈ R3 : T (x, y, z) = x − 3y + 2z = 0 .
For (x, y, z) ∈ ker(T ), since x−3y +2z = 0, so x = 3y +2z, that is, (x, y, z) becomes (3y +2z, y, z) =
y(3, 1, 0) + z(1, 0, 1). It implies that v = (3, 1, 0), w = (1, 0, 1) ∈ ker(T ) forms a basis for ker(T ).
Let us check our choice v = (3, 1, 0), w = (1, 0, 1) ∈ ker(T ) really forms a basis for ker(T ).
(i) Linear Independence: if 0 = av + bw = (3a + b, a, b) for any real numbers a, b, then clearly
a = 0 = b. So v and w ARE linearly independent.
(ii) Span ker(T ): for any (x, y, z) ∈ ker(T ), x, y, z satisfies x − 3y + 2z = 0 and so x = 3y − 2z, i.e.,
(x, y, z) = (3y − 2z, y, z) = y(3, 1, 0) + z(1, 0, 1) = yv + zw. That is, there are real numbers y and
z such that (x, y, z) = yv + zw. Thus, v and w spans ker(T ).
By (i) and (ii), v and w really forms a basis for ker(T ).
(2) Clearly dim(ker(T )) = 2, because the basis of ker(T ) has two elements v and w. Since
dim(R3 ) = 3, by the formula,
dim(R3 ) = dim(ker(T )) + dim(Im(T )),
we deduce dim(Im(T )) = 1.
2. (10 points) Let T : R3 → R3 be a linear transformation defined by T (x, y, z) = (y, 0, z). Find
bases for ker(T ) and Im(T ). What are their dimensions? Let U ⊂ R3 be the vector subspace
consisting of all vectors with y = 0 (that is, the xz–plane). Find a basis for T (U ).
3. (Each 5 points) (1) For a linear transformation T : R → R, there exists a real number a such
that T (x) = ax for all x ∈ R. Using this fact, solve the following problem: letting S : R → R be a
linear transformation satisfying S(3) = −4, calculate S(−7).
(2) For a linear transformation T : R2 → R, there exist two real numbers a and b such that
T (x, y) = ax + by for all (x, y) ∈ R2 . Using this fact, solve the following problem: letting S : R2 →
R be a linear transformation satisfying S(1, 1) = 3 and S(1, 0) = 4, calculate S(2, 1).
∗
Office Hour: 9:30 – 11:30 AM, Sunday & Tuesday and by appointment.
1
4. (Each 5 points) (1) Let T : R3 → R3 be a linear transformation defined by
T (x, y, z) = (3x, x − y, 2x + y + z).
Then T is invertible. Find a rule for T −1 like the one which defines T .
Proof.
T −1 (a, b, c) =
(a a
)
, − b, −a + b + c .
3 3
(2) For the linear transformation T in (1), prove that
(T 2 − I)(T − 3I) = 0,
where I is the identity transformation, i.e., I(v) = v, and for all v ∈ R3 ,
(T 2 − I)(T − 3I)(v) = (T 2 − I)(T (v) − 3I(v)) = T 2 (T (v)) − 3T 2 (I(v)) − T (v) + 3I(v).
5. (10 points) Let S : Rn → Rn be defined by S(a1 , a2 , . . . , an ) = (0, a1 , a2 , . . . , an−1 ). (This
transformation is called the shift operator .) Compute the dimensions of the kernel and image
of S. As usual, S k is defined by S k (v) = S(S(· · · (S(v)))) (k times) for any v ∈ Rn . Show that
S n = 0. (That is, S is a nilpotent operator as well as the shift operator.)
6. (10 points) A linear transformation T : V → W is said to be injective if T has the following
property: if T (v) = T (w) for any v, w ∈ V , then v = w. Show that T is injective if and only if
ker(T ) = {0}.
7. (10 points) Suppose that V is a finite–dimensional vector space and S : V → W and T : V → W
are linear transformations. Let v1 , v2 , . . . , vn be a basis for V . Suppose further that
S(vi ) = T (vi ),
i = 1, 2, . . . , n.
Show that S = T , i.e., S(v) = T (v) for all v ∈ V .
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