Math 108B - Home Work # 1
Due: Friday, April 11, 2008
1. Let T :R2 →R3 be the linear transformation given by the matrix
⎛
⎝ 1 −1
2 2
0 3
⎞
⎠
with respect to the standard bases. Find bases for R2 and R3 in which the matrix of
T is ⎛
⎝ 1 0 0 1 0 0
⎞
⎠
2. The matrix
4 −1
2 4
represents a linear transformationT :R2 →R2 with respect to the basis{v1, v2}where v1 = (1,1) and v2 = (−1,1). Find the matrix of T with respect to the basis {w1, w2} where w1 = (1,2) and w2 = (0,1).
3. Let T :V →W be a linear transformation, and let{v1, . . . , vn}be a basis forV. Show that T is invertible if and only if {T v1, . . . , T vn} is a basis for W.
4. The trace of an n×n matrix A is defined as the sum of all the entries on the main diagonal of A. That is,
tr(A) = n
i=1
Aii,
where Aij denotes the entry ofA in the ith row and jth column.
(a) Show that for any two n×n matrices A and B,tr(AB) = tr(BA).
(b) Use (a) to show that if X and Y are similar matrices then tr(X) = tr(Y).
5. LetV be an inner-product space, and letW be a subspace ofV. Define theorthogonal complement of W by
W⊥ ={v ∈V | v, w= 0 ∀w∈W}.
Show thatW⊥ is a subspace ofV.
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Delhi School of Economics
Introductory Math Econ.
Problem Sets - Part A
-
Instructor: SUGATA BAG
Summer 2010
Delhi School of Economics
Introductory Math Econ.
Problem Sets - Part B
Instructor: SUGATA BAG
-
Summer 2010
Page | 2
7. Let x1 ,………, xn be positive real numbers. Prove that (Hint: Use the Cauchy- Schwarz inequality) –
Delhi School of Economics
Introductory Math Econ Problem Sets - Part C
Instructor: SUGATA BAG
-
Page | 3 1. On P2(R), consider the inner product given by
Apply the Gram-Schmidt procedure to the basis (1, x, x2) to produce an orthonormal basis of P2(R).
2. Find an orthonormal basis of P2(R) (with inner product as in Exercise 1) such that the differentiation operator (the operator that takes p to pƍ ) on P2(R) has an upper-triangular matrix with respect to this basis.
3. Suppose U is a subspace of V. Prove that dimU٣= dimV dim U.
4. In R4, let U = span {(1, 1, 0, 0), (1, 1, 1, 2)}.
Find u א U such that || u (1, 2, 3, 4)|| is as small as possible.
5. (Do not turn in) For Exercise 1, does anything change if you apply the Gram- Schmidt Process to the basis {1, x, x2} for P2(C) with the inner product
1
, 0 ( ) ( ) p q
³
p x q x d x ?6. If U is a subset of an inner product space V (but not necessarily a subspace), we can still define –
Math 108B - Home Work # 1 Solutions 1. For T to have the matrix ⎛
⎝ 1 0 0 1 0 0
⎞
⎠
with respect to a basis {u1, u2} of R2 and a basis {v1, v2, v3} for R3, means simply that T u1 =v1 and T u2 =v2. Hence {u1, u2} can remain the standard basis, and then v1 = (1,2,0) and v2 = (−1,2,3) will be the columns of the given matrix for T. Since v1 and v2 are linearly independent, we can complete them to a basis. To do this we just need to find a third vector of R3 that is not a linear combination ofv1 andv2. For instance, v3 =e3 = (0,0,1) works.
2. We must multiply the given matrix on the right by the change of basis matrixC whose columns are the coordinates of the new basis w1, w2 in the old basis {v1, v2}, and we must multiply it on the left by the change of basis matrix C−1 whose columns are the coordinates of v1, v2 in the new basis {w1, w2}. To find C, note that w1 = (1,2) =
3
2(1,1) +12(−1,1) = 32v1+12v2 andw2 = (0,1) = ((1,1) + (−1,1))/2 = 12v1+12v2. Hence C =
3/2 1/2 1/2 1/2
.
To get C−1, note v1 = (1,1) = (1,2)−(0,1) = w1−w2 and v2 = (−1,1) =−(1,2) + 3(0,1) =−w1+ 3w2. Hence
C−1 =
1 −1
−1 3
,
and the matrix for T in the new basis is C−1AC =
1 −1
−1 3
4 −1
2 4
3/2 1/2 1/2 1/2
=
13/2 3/2
−11/2 3/2
3. LetT :V →W be a linear transformation, and let{v1, . . . , vn}be a basis forV. Show that T is invertible if and only if {T v1, . . . , T vn} is a basis for W.
Solution. ⇐: Suppose {T v1, . . . , T vn} is a basis for W, and write wi =T vi for each i. Then we can define a linear transformation L:W →V by
L(c1w1+· · ·+cnwn) = c1v1+· · ·cnvn, ∀ c1, . . . , cn∈F.
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Delhi School of Economics
Introductory Math Econ Solution to Problem Set - Part A
SB
Lis well-defined since the vectors w1, . . . , wn are linearly independent, and since these vectors span W, L is defined for all vectors in W. Clearly LT vi = Lwi = vi and T Lwi =T vi =wi for every i. Using linearity of T and L it follows that LT v =v and T Lw =w for all v ∈V and all w∈W. ThusL is the inverse ofT and T is invertible.
⇒: Let L = T−1. We first show that T v1, . . . , T vn span W. Let w ∈ W and write Lw =c1v1+· · ·+cnvn inV. Applying T, we get
w=T Lw=c1T v1+· · ·+cnT vn.
To show thatT v1, . . . , T vnare linearly independent, suppose thatc1T v1+· · ·+cnT vn= 0 for scalars ci. Applying L, we get
0 =L(0) =c1LT v1+· · ·+cnLT vn =c1v1+· · ·cnvn. Since v1, . . . , vn are linearly independent, we must have ci = 0 for all i.
4. The trace of an n×n matrix A is defined as the sum of all the entries on the main diagonal of A. That is,
tr(A) = n
i=1
Aii,
where Aij denotes the entry ofA in the ith row and jth column.
(a) Show that for any two n×n matrices A and B,tr(AB) = tr(BA).
(b) Use (a) to show that if X and Y are similar matrices then tr(X) = tr(Y).
Solution. (a)
tr(AB) = n
i=1
(AB)ii= n
i=1
n j=1
AijBji = n
j=1
n i=1
BjiAij = n j=1
(BA)jj =tr(BA).
(b) If X and Y are similar matrices, then X =C−1Y C for some invertible matrix C. Thus
tr(X) = tr(C−1(Y C)) =tr((Y C)C−1) = tr(Y).
5. LetV be an inner-product space, and letW be a subspace ofV. Define theorthogonal complement of W by
W⊥ ={v ∈V | v, w= 0 ∀w∈W}.
Show thatW⊥ is a subspace ofV. 2
Solution. Clearly, 0 ∈ W⊥ since 0, w = 0 for any w ∈ W. If v ∈ W⊥ and a ∈ F, then av∈W⊥ since av, w=av, w= 0 for any w∈W. Finally, if u, v ∈W⊥, then u+v ∈W⊥ since u+v, w=u, w+v, w= 0 + 0 = 0 for any w∈ W. ThusW⊥ is a subspace of V.
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Math 108B - Home Work # 2 Solutions 1. Letb1, . . . , bn be positive real numbers. Check that the form
z, w=b1z1w¯1+· · ·bnznw¯n
defines an inner product on Fn, where z = (z1, . . . , zn) and w = (w1, . . . , wn). (In particular, the dot product onCn is an inner product.)
Solution. We must check that z, w is (1) linear in z; (2) positive definite; and (3) conjugate symmetric.
(1) Leta, c∈F and z, z, w∈Fn. Then az+cz, w =
n i=1
bi(azi+czi)wi
= n
i=1
abiziwi+ n
i=1
cbiziwi
= az, w+cz, w.
(2) Letz∈Fn. Then
z, z= n
i=1
bizizi = n
i=1
bi|zi|2≥0,
since allbi >0. Furthermore, equality holds if and only if |zi|= 0 for alli. That is, if and only if,z= 0.
(3) Letz, w∈Fn. Then w, z=
n i=1
biwizi = n
i=1
biwizi =z, w.
2. Let V be an F-vector space with basis {v1, . . . , vn}, and let B = (bij) be the n×n matrix with entriesbij =vi, vj ∈F. Show that
(a) bii >0 for 1≤i≤n; and
(b) B= ¯Bt, i.e., bij = ¯bji for all 1≤i, j≤n.
Solution. (a) By definition, bii = vi, vi > 0 since the inner product is positive definite.
(b) By definition, bij = vi, vj = vj, vi = bji by conjugate symmetry of the inner product.
1
Delhi School of Economics
Introductory Math Econ.
Solution to Problem Set - Part B
SB
3. Give an example of a 2×2 matrixBsatisfying (a) and (b) above that does not define an inner product on F2 with ei, ej = bij for 1 ≤ i, j ≤ 2. ({e1, e2} is the standard basis for F2.) Hint: Construct the matrix B so that there is a vector v whose norm would be negative with respect to the corresponding inner product.
Solution. Suppose B =
a b b d
for a, d > 0 defines an inner product −,−. If (x, y)∈R2, we would then have
(x, y),(x, y)=x2e1, e1+ 2xye1, e2+y2e2, e2=ax2+ 2bxy+dy2.
We can obtain a contradiction by exhibiting somea, b, d, x, y ∈Rsuch that the above expression is negative or zero, since that will imply that this inner product is not actually positive definite. To get an example, let y = 1, and solve ax2+ 2bx+d= 0 for x using the quadratic formula. We get x = (−2b+√
4b2−4ad)/2a, and this is a real number as long as 4b2−4ad≥0. So for instance, we may take a=d= 1, b= 2 and then (x, y) = (−1,1) would have a negative inner-product with itself.
4. 4. If ||u|| = 3,||u +v|| = 4, and ||u − v|| = 6, we can solve for ||v|| using the parallelogram identity.
||v||2 = (||u+v||2+||u−v||2−2||u||2)/2 = 17. Thus ||v||=√
17.
5. The norm ||(x, y)||=|x|+|y|does not come from an inner product onR2, since it does not satisfy the parallelogram identity. For example, let u= (1,0) and v = (0,1) then
||u+v||2+||u−v||2= 22+ 22= 8, but
2(||u||2+||v||2) = 2(12+ 12) = 4.
6. Let u, v∈V, where V is an inner product space overR. We have
||u+v||2 =u+v, u+v=||u||2+ 2u, v+||v||2, and
||u−v||2=u−v, u−v=||u||2−2u, v+||v||2.
Thus, we can solve for u, vby subtracting the second equation from the first to get u, v= ||u+v||2− ||u−v||2
4 .
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5. (Bonus) Letx1, . . . , xn be positive real numbers. Prove that (x1+· · ·+xn)
1 x1
+· · ·+ 1 xn
≥n2. (Hint: Use the Cauchy-Schwarz inequality.)
Solution. Sincexi>0, we may writexi=a2i for real numbers ai >0. Then (x1+· · ·+xn)(1
x1
+· · ·+ 1 xn
) = (a21+· · ·+a2n)(1
a21 +· · ·+ 1 a2n)
= ||(a1, . . . , an)||2· ||(a−11 , . . . , a−1n )||2
≥ |(a1, . . . , an)·(a−11 , . . . , a−1n )|2
= n2,
where the inequality is by (the square of) the Cauchy-Schwarz inequality for the stan- dard dot product inRn.
3 7.
Math 108B - Home Work # 3 Solutions 1. LADR Problems. 10. We have||1||= 1
0 12 dx= 1, so we can take e1 = 1. Now let u2 =x− x, e1e1=x− x,11 =x−
1
0
xdx=x−1/2. Since||u2|| = 1
0(x−1/2)2 dx=
1/12, we let e2=√
12x−√
12/2. Now let u3 = x2− x2, e1e1− x2, e2e2
= x2− 1
0
x2 dx− 1
0
x2(√
12x−√
12/2) dx(√
12x−√ 12/2)
= x2−1/3−12(1/12)(x−1/2)
= x2−x+ 1/6 Since ||u3|| = 1
0(x2−x+ 1/6)2 dx = 1
0(x4−2x3+ 4x2/3−x/3 + 1/36)dx = 1/180, let e3 = √
180u3 =√
180(x2−x+ 1/6). Now {e1, e2, e3} is an orthonormal basis.
14. According to corollary 6.27, we need to simply apply the Gram-Schmidt process to any basis for P2(R) in which the differentiation operator already has an upper triangular matrix. Since this is the case for the basis {1, x, x2}, it will also be true for the orthonormal basis constructed in exercise 10. It is also easy to check directly that differentiation has an upper triangular matrix with respect to the basis{e1, e2, e3} found in 10.
15. By Theorem 6.29, V =U ⊕U⊥. If {u1, . . . , um} is a basis forU and {v1, . . . , vn} is a basis for U⊥ then it is easy to see that {u1, . . . , um, v1, . . . , vn} is a basis for V. Hence dimV =m+n= dimU + dimU⊥. Alternatively, use Theorem 2.18.
21. We know that ||u −(1,2,3,4)|| is minimized for u = PU(1,2,3,4). In order to calculate PU(1,2,3,4), we first need to find an orthonormal basis for U. Let e1 = (1,1,0,0)/||(1,1,0,0)|| = (1/√
2,1/√
2,0,0). Let u2 = (1,1,1,2)−(1,1,1,2)· (1/√
2,1/√
2,0,0)e1= (1,1,1,2)−√
2e1= (0,0,1,2). Thene2=u2/||u2||= (0,0,1/√ 5,2/√
5).
NowPU(1,2,3,4) = (1,2,3,4)·(1/√ 2,1/√
2,0,0)e1+(1,2,3,4)·(0,0,1/√ 5,2/√
5)e2= 3e1/√
2 + 11e2/√
5 = (3/2,3/2,0,0) + (0,0,11/5,22/5) = (3/2,3/2,11/5,22/5).
2. No, nothing changes. We get the same orthonormal basis forP2(C) in this case.
3. If U is a subset of an inner product space V (but not necessarily a subspace), we can still define
U⊥={v∈V | v, u= 0∀u∈U}.
(a) Prove thatU⊥ = span(U)⊥. (Recall, that span(U) is the subspace of V consisting of all finite F-linear combinations of vectors inU.)
1 Solution:
2. Solution: Accordingly
3. ,
. .
4.
5.
6.
Delhi School of Economics
Introductory Math Econ.
Solution to Problem Set - Part C
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Solution. Let v ∈ span(U)⊥. Then u, v = 0 for all u ∈ U since U ⊆ span(U).
Hence v ∈ U⊥. Conversely, supposev ∈ U⊥. If u ∈ span(U), then u = n
i=1ciui for scalars ci ∈F and vectors ui ∈ U. Hence u, v=n
i=1ciui, v=n
i=1ciui, v= 0.
Thus v∈span(U)⊥.
(b) Use (a) to prove that (U⊥)⊥ = span(U).
Solution. Since span(U) is a subspace of V, we can apply Corollary 6.33 to get span(U) = (span(U)⊥)⊥ = (U⊥)⊥, where the second equality follows from (a).
In particular, this exercise implies that if {u1, . . . , um} is a basis for the subspaceU, then
U⊥ ={v∈V | v, ui= 0∀i}.
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