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Practice Exam on Chapter 14 The Elements of Eigenvalue and Eigenvector Theory
Math 341 Linear Algebra II Section 51
UID No: Name: Date: November, 2007 Score: Exam problems will be chosen from this set. The solutions to some problems are not provided. I believe you can solve them. If you cannot, please stop by my oce.
- (5 points) Write down the necessary and sucient condition(s) for a linear transformation T : V! V of the nite dimensional vector space V to be diagonalizable. Answer. (1) The characteristic polynomial of T is a product of linear factors.
(2) For each eigenvalue of T , the geometric and algebraic multiplicities are the same. X
- (5 points) Let T : R^2! R^2 be de ned by T (x; y) = (x + y; y). Is T diagonalizable? Justify your answer. Rough Answer. (1) T has the characteristic polynomial () = ( 1)^2 = ^2 2 + 1. By the theorem, the root of the polynomial is the eigenvalue of T. So T has the eigenvalue = 1 whose algebraic multiplicity is 2. (2) From ([ T ]E I) (x; y)t^ = (0; 0)t^ with = 1, we have two equations, 0 x + 1 y = 0; 0 x + 0 y = 0; i.e., x is anything and y = 0 and so (x; y) = (x; 0) = x(1; 0) which gives the eigenspace V=1 = Spanf (1; 0) g: Since dim(V=1) = 1, so we deduce that = 1 has the geometric multiplicity 1.
Since the algebraic and geometric multiplicities for = 1 are di erent, by the Theorem, T is not diagonalizable. X
- (5 points) Let T : R^2! R^2 be de ned by T (x; y) = (x + 3y; x + y). Is T diagonalizable? Justify your answer. Rough Answer. (1) T has the characteristic polynomial () = ^2 2 2. By the theorem, the root of the polynomial is the eigenvalue of T. So T has the eigenvalue = 1 p 3 ; = 1 + p 3 ; and each has the algebraic multiplicity 1. (2) From ([ T ]E I) (x; y)t^ = (0; 0)t^ with = 1 p3, we have two equations, p 3 x + 3 y = 0; 1 x + p 3 y = 0; i.e., y = x=p3 and so (x; y) =
x; px 3
= x
1 ; p^13
which gives the eigenspace
V=1 p 3 = Span
1 ; p^13
Since dim(V=1 p 3 ) = 1, so we deduce that = 1 p 3 has the geometric multiplicity 1. From ([ T ]E I) (x; y)t^ = (0; 0)t^ with = 1 + p3, we have two equations, p 3 x + 3 y = 0; 1 x p 3 y = 0; i.e., y = x=p3 and so (x; y) =
x; px 3
= x
1 ; p^13
which gives the eigenspace
V=1+p 3 = Span
1 ; p^13
Since dim(V=1+p 3 ) = 1, so we deduce that = 1 + p3 has the geometric multiplicity 1.
Since the algebraic and geometric multiplicities for each eigenvalue are the same, by the Theorem, T is diagonalizable. X
- (5 points) Let T : R^3! R^3 be de ned by T (x; y; z) = (2x; 2 x + y; 3 y + z). Is T diagonalizable? Justify your answer.
(Continued)
Answer. T has the matrix [ T ]E relative to the standard basis E:
[ T ]E =
and it has the characteristic polynomial, () = det ([ T ]E I) = ( 1)^2 ( 2) which is a product of linear factors and from which we have the eigenvalues of T : = 1 (algebraic multiplicity 2) and = 2 (algebraic multiplicity 1). For = 1, [ T ]E I becomes [ T ]E I and so 2 4
5 = ([ T ]E I)
x y z
x y z
x 2 x 3 y
That is, we have x = 0 = y and z is anything. It implies (x; y; z) = (0; 0 ; z) = z(0; 0 ; 1). So the eigenvector associated to = 1 is generated by (0; 0 ; 1) and we have the eigenspace, V=1 = Span f (0; 0 ; 1) g : Since dim (V=1) = 1, so = 1 has the geometric multiplicity 1, while = 1 has the algebraic multiplicity 2. Since two
multiplicities are di erent, by the Diagonalization Theorem, T is not diagonalizable. X
- (5 points) Let T : R^3! R^3 have the following matrix relative to the standard basis E:
[ T ]E =
cos 0 sin 0 1 0 sin 0 cos
Is T diagonalizable? Justify your answer. Answer. T has the characteristic polynomial, () = det ([ T ]E I)
=
cos 0 sin 0 1 0 sin 0 cos
(^5) = ( 1) ^2 2 cos + 1^ :
Since the determinant of the quadratic function ^2 2 cos + 1 is cos^2 1 < 0 for any 2 (0; ), so the quadratic function does not have any real root. It implies () cannot be a product of linear factors. Thus by the Diagonalization Theorem,
T is not diagonalizable. X
- (Total 25 points) Consider the linear transformation T : R^3! R^3 de ned by
T (x; y; z) = (3x z; y x + 2z; 4 z): (a) (5 points) Find the matrix of T relative to the standard basis. Answer. [ T ]E =
where E is the standard basis for R^3. X
(b) (5 points) Find the characteristic polynomial of T. Answer. () = det ([ T ]E I)
= det
5 = ( 1)( 3)( 4): X
(Continued)
- (Total 25 points) Consider the linear transformation T : R^3! R^3 de ned by
T (x; y; z) = (3x + 2y + 2z; x + 4y + z; 2 x 4 y z): (a) (5 points) Find the matrix of T relative to the standard basis. Rough Answer. [ T ]E =
5 : X
(b) (5 points) Find the characteristic polynomial of T. Rough Answer.
() = ^3 + 6^2 11 + 6 = ( 1)( 2)( 3): X
(c) (10 points) Find the eigenvalues and eigenvectors and eigenspaces of T , if possible. Rough Answer. (1) By the theorem, the root of the characteristic polynomial is the eigenvalue and so T has the eigenvalues = 1, = 2 and = 3. Each eigenvalue has the algebraic multiplicity 1. (2) From ([ T ]E I) (x; y; z)t^ = (0; 0 ; 0)t^ with = 1, we have three equations, 2 x + 2y + 2z = 0; x + 3y + z = 0; 2 x 4 y 2 z = 0: Solving the system, we get z = x and y = 0 and so (x; y; z) = (x; 0 ; x) = x(1; 0 ; 1) which gives the eigenspace V=1 = Span f (1; 0 ; 1) g : Since dim(V=1) = 1, so we deduce that = 1 has the geometric multiplicity 1. From ([ T ]E I) (x; y; z)t^ = (0; 0 ; 0)t^ with = 2, we have three equations, x + 2y + 2z = 0; x + 2y + z = 0; 2 x 4 y 3 z = 0: Solving the system, we get z = 0 and x = 2 y and so (x; y; z) = ( 2 y; y; 0) = y( 2 ; 1 ; 0) which gives the eigenspace V=2 = Span f ( 2 ; 1 ; 0) g : Since dim(V=2) = 1, so we deduce that = 2 has the geometric multiplicity 1. From ([ T ]E I) (x; y; z)t^ = (0; 0 ; 0)t^ with = 3, we have three equations, 0 x + 2y + 2z = 0; x + y + z = 0; 2 x 4 y 4 z = 0: Solving the system, we get x = 0 and y = z and so (x; y; z) = (0; z; z) = z(0; 1 ; 1) which gives the eigenspace V=3 = Span f (0; 1 ; 1) g :
Since dim(V=3) = 1, so we deduce that = 3 has the geometric multiplicity 1. X
(d) (5 points) Diagonalize T , if possible. Rough Answer. Using the basis E^0 = f (1; 0 ; 1); ( 2 ; 1 ; 0); (0; 1 ; 1) g, we deduce the matrices,
[ T ]E 0 =
5 ; [ I ]E 0 E =
5 ; [ I ]EE 0 =
Therefore, we have the equation,
2 [^ T^ ]E^ = [^ I^ ]E^0 E^ [^ T^ ]E^0 [^ I^ ]EE^0 4
5 : X
- (Total 25 points) Consider the linear transformation T : R^2! R^2 de ned by
T (x; y) = (y; x): (a) (5 points) Find the matrix of T relative to the standard basis.
(Continued)
Rough Answer. [ T ]E =
: X
(b) (10 points) Find the characteristic polynomial of T. Rough Answer.
() = ^2 1 = ( 1)( + 1): X
(c) (5 points) Find the eigenvalues and eigenvectors and eigenspaces of T , if possible. Rough Answer. (1) Since the polynomial has root = 1 and the root of the polynomial is the eigenvalue, so the eigenvalue is = 1 with algebraic multiplicity 2. (2) From ([ T ]E I) (x; y)t^ = (0; 0)t^ with = 1, we have two equations, x + y = 0; x y = 0: Solving the system, we get x = y and so (x; y) = (x; x) = x(1; 1) which gives the eigenspace V=1 = Span f (1; 1) g :
Since dim(V=1) = 1, so we deduce that = 1 has the geometric multiplicity 1. X
(d) (5 points) Diagonalize T , if possible. Rough Answer. Since dim (V=1) = 1, so = 1 has the geometric multiplicity 1, while = 1 has the algebraic
multiplicity 2. Since two multiplicities are di erent, by the Diagonalization Theorem, T is not diagonalizable. X
- (5 points) Consider the 3 3 matrix L whose elements are real numbers,
L =
d 2 0 f e 3
This kind of matrix is called the lower triangular matrix. (a) Find the characteristic polynomial of L. Rough Answer.
() = ( 1)( 2)( 3): X
(b) Compute the eigenvalues of L.
Rough Answer. Eigenvalues are = 1, = 2, and = 3. X
(c) Diagonalize L, if possible. Rough Answer. Eigenspaces to each eigenvalue 1 = 1, 2 = 2 and 3 = 3 are as follows: V 1 =1 = Span
1 ; d; f^2 +^ de
V 2 =2 = Span f (0; 1 ; e) g ; V 3 =3 = Span f (0; 0 ; 1) g : It implies [ I ]E (^0) E =
^ f +dde^1 2 e^1
5 ; [ I ]EE 0 =
f +^ dde^1 2 e^1
We observe [ I ]E (^0) E and [ I ]EE 0 are also lower triangular matrices. Therefore, we deduce
2 [^ T^ ]E^ = [^ I^ ]E^0 E^ [^ T^ ]E^0 [^ I^ ]EE^0 4
d 2 0 f e 3
^ f +dde^1 2 e^1
f +^ dde^1 2 e^1
If a matrix is lower or upper triangular, it is very easy to diagonalize it (if possible). So one may ask a question: how can we decompose a matrix into a product of triangular matrices, e.g., for a given M , what are lower and upper triangular
matrices, L and U , such that M = LU? X
Solution to Practice Exam on Chapter 15 Inner Product Spaces
Math 341 Linear Algebra II Section 51
As usual, exam problems will be chosen from this set.
1 (5 points) Concept Problems.
( 1 ) Write down the de nition of an inner product (or scalar product).
Answer. If a function h ; i : V V! R; where V is a vector space and R is the set of real numbers, satis es the following properties, then it is called the inner product or scalar product. (1) Commutativity: for all v and w in V, h v; w i = h w; v i. (2) Bi{Linearity: for all vectors v; w and z in V and any numbers a and b in R, h av; w i = a h v; w i = h v; aw i ; h v + w; z i = h v; z i + h w; z i ; h v; w + z i = h v; w i + h v; z i :
Simply, h av + bw; z i = a h v; z i + b h w; z i = h v; az i + h w; bz i : h v; aw + bz i = a h v; w i + b h v; z i = h av; w i + h bv; z i :
(3) Self{Nonnegativity: for all v in V, h v; v i 0 and h v; v i = 0 (number zero) if and only if v = 0 (vector zero). X
( 2 ) Write down the de nition of the norm.
Answer. For an inner product space (V; h ; i), the functional k k : V! R de ned by kvk = h v; v i^1 =^2 = p h v; v i;
is called the norm of v or length of v. X
( 3 ) Write down the Cauchy{Schwartz Inequality.
Answer. For an inner product space (V; h ; i) and for any pair of vectors v and w in V, the following inequality is called the Cauchy{Schwartz inequality,
jh v; w ij kvk kwk : X
( 4 ) Write down the de nition of an orthonormal set of vectors.
Answer. For an inner product space (V; h ; i) and for a set S V, if (i) 0 2 S= and (ii) for any v 6 = w 2 S, h v; w i = 0, then S is called the orthogonal set. Moreover, if S is orthogonal and if any vector v 2 S has the norm 1, i.e., kvk = 1,
then S is called to be orthonormal. X
( 5 ) Write down the 1st^ and 2th^ and nth^ orthogonal vectors, 1 ; 2 ; n, deduced from a basis f f 1 ; f 2 ; : : : ; fng by Gram{
Schmidt process. Answer. Gram{Schmidt process implies
1 = f 1 ; 2 = f 2 h^ fk^2 ; ^1 i 1 k^2
n = fn h^ fn; n ^1 i k n 1 k^2 n 1 h^ fn; ^1 i k 1 k^2
1 : X
Exam on Chapter 15 (Continued)
( 6 ) Write down the de nition of an isometric embedding.
Answer. For two inner product spaces ( V; h ; iV ) and ( W; h ; iW ), if a linear transformation T : V! W satis es
h T (v); T (w) iW = h v; w iV ;
for all v; w 2 V, then T is called to be an isometric embedding. X
( 7 ) Write down the necessary and sucient condition for two nite{dimensional inner product spaces V and W to be
isometric. Answer. The theorem says
dim(V) = dim(W) () V and W are isometric. X
( 8 ) Write down the Riesz Representation Theorem.
Answer. For a nite{dimensional inner product space (V; h ; i) and a linear functional T : V! R, there exists a unique vector vT 2 V (depending on T ) such that for all w 2 V,
T (w) = h w; vT i : X
Exam on Chapter 15 (Continued)
= a h p(x); r(x) i + b h q(x); r(x) i : h p(x); aq(x) + br(x) i = p(x 1 )(aq(x 1 ) + br(x 1 )) + p(x 2 )(aq(x 2 ) + br(x 2 )) + + p(xN +1)(aq(xN+1) + br(xN+1)) = a (p(x 1 )q(x 1 ) + p(x 2 )q(x 2 ) + + p(xN+1)q(xN +1))
- b (p(x 1 )r(x 1 ) + p(x 2 )r(x 2 ) + + p(xN +1)r(xN +1)) = a h p(x); q(x) i + b h p(x); r(x) i :
(3) Self{Nonnegativity: for p(x) 2 PN (R), h p(x); p(x) i = p(x 1 )^2 + p(x 2 )^2 + + p(xN +1)^2 0 :
So it implies h p(x); p(x) i = 0 =) p(x 1 ) = 0 = p(x 2 ) = = p(xN+1): (F) We want to prove the statement: h p(x); p(x) i = 0 () p(x) = 0: One direction is clear to see: if p(x) = 0, then h p(x); p(x) i = 0. To get the converse direction, we develop the following argument based on the statement (F). If f(x) = c (constant) has a root, i.e., the graph of f crosses the x{axis at one point, then c must be 0, i.e., f(x) = 0. That is, if a polynomial of degree 0 has ONE root, then the polynomial should be 0. If g(x) = ax + b (a 6 = 0) has two roots, i.e., the graph of g crosses the x{axis at two points, then a and b must be 0, i.e., g(x) = 0. That is, if a polynomial of degree 1 has TWO roots, then the polynomial should be 0. By arguing in the same way, we obtain the statement on general case, if a polynomial of degree N has N + 1 roots, then the polynomial should be 0. So by the argument, when a polynomial p(x) of degree N satis es p(x 1 ) = 0 = p(x 2 ) = = p(xN +1), i.e., when the polynomial p(x) of degree at most N has N + 1 roots, the polynomial p(x) should be 0. That is, with (F), we deduce
h p(x); p(x) i = 0 =) p(x 1 ) = 0 = p(x 2 ) = = p(xN+1) =) p(x) = 0: X
4 Let R^2 be the standard inner product space and v = (1; 2); w = ( 1 ; 1) 2 R^2. If u = (a; b) 2 R^2 satis es h v; u i = 1 and
h w; u i = 3, nd u = (a; b). Answer. 1 = h v; u i = h (1; 2); (a; b) i = a + 2b; a + 2b = 1 ; 3 = h w; u i = h ( 1 ; 1); (a; b) i = a + b; a + b = 3: Hence, we deduce a = 73 ; b =^23 ; u = (a; b) =
=^13 ( 7 ; 2) : X
5 Let h ; i : R^2 R^2! R be a functional de ned by
h (x; y); (z; w) i =
h x y
i " a b c d
z w
; M =
a b c d
where the 2 2 matrix M is xed. Find the necessary and sucient condition on M for h ; i to be an inner product.
Too complicated! Not in the Exam!
Answer. h ; i is an inner product if and only if a > 0, d > 0, b = c and det(M ) > 0. X
Exam on Chapter 15 (Continued)
Proof of Answer. (1) Commutativity: for (x; y); (z; w) 2 R^2 ,
h (x; y); (z; w) i =
h x y
i " a b c d
z w
= axz + bxw + cyz + dyw
= azx + bzy + cwx + dwy =
h z w
i " a b c d
x y
= h (z; w); (x; y) i :
where the equality axz + bxw + cyz + dyw = azx + bzy + cwx + dwy occurs if and only if b = c. (2) Bi{Linearity: for (x; y); (z; w); (u; v) 2 R^2 and any constant ; ,
h (x; y) + (z; w); (u; v) i = h ( x + z; y + w); (u; v) i =
h x + z y + w
i " a b c d
u v
= a( x + z)u + b( x + z)v + c( y + w)u + d( y + w)v = (axu + bxv + cyu + dyv) + (azu + bzv + cwu + dwv) =
h x y
i " a b c d
u v
h z w
i " a b c d
u v
= h (x; y); (u; v) i + h (z; w); (u; v) i h (x; y); (z; w) + (u; v) i = h (x; y); ( z + u; w + v) i =
h x y
i " a b c d
z + u w + v
= ax( z + u) + bx( w + v) + cy( z + u) + dy( w + v) = (axz + bxw + cyz + dyw) + (axu + bxv + cyu + dyv) =
h x y
i " a b c d
z w
h x y
i " a b c d
u v
= h (x; y); (z; w) i + h (x; y); (u; v) i :
(3) Self{Nonnegativity: for (x; y) 2 R^2 ,
h (x; y); (x; y) i =
h x y
i " a b c d
x y
= ax^2 + (b + c)xy + dy^2 :
The condition, self{nonnegativity, should be satis ed for any (x; y) 2 R^2. So for (x; 0) 2 R^2 , we have
h (x; 0); (x; 0) i = ax^2 0 () a 0 :
Moreover, a > 0 is required if and only if the statement holds
h (x; 0); (x; 0) i = ax^2 = 0 () x = 0; i:e:; (x; 0) = (0; 0):
By the same argument, d > 0 is required if and only if (0; y) should satisfy the self{nonnegativity condition. Taking b = c (required for the commutativity) and a > 0 and d > 0 (required for the self{nonnegativity of (x; 0) and (0; y)), we get
h (x; y); (x; y) i = ax^2 + (b + c)xy + dy^2 = a
x + (^) a by
2 a^2 y
2
The far right{hand side makes us to choose a particular vector ( b; a). A simple computation shows
h ( b; a); ( b; a) i = a( b)^2 +2b( ab)+d(a^2 ) = a
b + ba a
2 a^2 a
2
= a ad b^2 ^ 0 () det(M ) = ad b^2 0 :
Exam on Chapter 15 (Continued)
Section 15.2 Inner Product Spaces
7 Let V be an inner product space. Prove that the parallelogram law, for any v; w 2 V,
kv + wk^2 + kv wk^2 = 2 kvk^2 + 2 kwk^2 : Proof.
kv + wk^2 + kv wk^2 = h v + w; v + w i + h v w; v w i = h v; v i + h v; w i + h w; v i + h w; w i + h v; v i h v; w i h w; v i + h w; w i
= 2 h v; v i + 2 h w; w i = 2 kvk^2 + 2 kwk^2 : X
8 Let F(R) be the standard inner product space of continuous real functions, where the standard inner product is de ned by
h f(x); g(x) i =
Z 1
1 f(x)g(x) dx:
( 1 ) Given a h(x) 2 F(R) with kh(x)k = 1, compute the maximum value of the integral I,
I =
Z 1
1 x^3 h(x) dx:
Answer. By the Cauchy{Schwartz Inequality, we have
jIj =
Z 1
1 x^3 h(x) dx = x^3 ; h(x) x^3 kh(x)k = p h x^3 ; x^3 i kh(x)k = p h x^3 ; x^3 i (because kh(x)k = 1)
=
sZ (^1)
1 x^3 x^3 dx =
r 2 7 :
That is, I has the maximum value p 2 =7. X
( 2 ) Find the minimum value of the integral J,
J =
Z 1
1 (ax^4 + bx^3 + cx^2 )^2 dx;
where a; b; c are xed positive constants. Answer. Let p(x) = ax^4 + bx^3 + cx^2. Then we observe
0 J =
Z 1
1 (ax^4 + bx^3 + cx^2 )^2 dx =
Z 1
1 p^2 (x) dx = h p(x); p(x) i = kp(x)k^2 :
By the Cauchy{Schwartz Inequality, we have jh p(x); 1 ij kp(x)k k 1 k ; jh p(x); 1 ij^2 kp(x)k^2 k 1 k^2 : It implies Z (^1) 1
p(x) dx
2 = jh p(x); 1 ij^2 kp(x)k^2 k 1 k^2 = J k 1 k^2 = J h 1 ; 1 i = 2J; i:e:; (^12)
Z 1
1
p(x) dx
2 J:
Exam on Chapter 15 (Continued)
A simple computation shows Z (^1) 1
p(x) dx =
Z 1
1
ax (^4) + bx (^3) + cx 2 (^) dx = 2^ ^ a 5 +^
c 3
Z 1
1
p(x) dx
2 =^12
h 2
(^) a 5 +^
c 3
i 2 = 2
(^) a 5 +^
c 3
Therefore, J has the minimum value, 2
(^) a 5 +^
c 3
J: X
○^ Aside: If we compute the integral J by hands/computer, we get
J = 2
(^) a 2 9 +^
b^2 + 2ac 7 +^
c^2 5
and the di erence between the right{hand side and our result is positive,
2
(^) a 2 9 +^
b^2 + 2ac 7 +^
c^2 5
(^) a 5 +^
c 3
4 a 15 +
2 c 7
2 7 +
16 c^2 2205
It implies that our result is a minimum value........................................................... End of Aside
9 Let (V; h ; i) be an inner product space with the orthonormal basis E = f e 1 ; e 2 ; : : : ; en g. For a xed vector v 2 V with
kvk < 1 , let ci = h v; ei i, i = 1; 2 ; : : : ; n. Show that c^21 + c^22 + + c^2 n kvk^2 :
This inequality is called the Bessel Inequality. Proof. Let n = 2 and we prove c^21 + c^22 kvk^2 : By using the same argument, one can prove the inequality for a general n. We compute the square norm of the vector v (c 1 e 1 + c 2 e 2 ),
0 kv (c 1 e 1 + c 2 e 2 )k^2 = h v (c 1 e 1 + c 2 e 2 ) ; v (c 1 e 1 + c 2 e 2 ) i = h v; v i 2 h v; c 1 e 1 + c 2 e 2 i + h c 1 e 1 + c 2 e 2 ; c 1 e 1 + c 2 e 2 i = kvk^2 2 (c 1 h v; e 1 i + c 2 h v; e 2 i) + c^21 h e 1 ; e 1 i + c^22 h e 2 ; e 2 i (because of the linearity of inner product and the orthogonality of ei's) = kvk^2 2 c^21 + c^22 ^ + c^21 ke 1 k^2 + c^22 ke 2 k^2 (by the given setup ci = h v; ei i) = kvk^2 2 c^21 + c^22 ^ + c^21 + c^22 (by the normality of ei's)
= kvk^2 c^21 + c^22 ^ ; i:e:; 0 kvk^2 c^21 + c^22 ^ ; c^21 + c^22 kvk^2 : X
○^ Aside: In fact, for the nite{dimensional vector space V, we have the equality, c^21 + c^22 + + c^2 n = kvk^2. Since E is the basis
for V, there are constants a 1 ; a 2 ; : : : ; an such that v = a 1 e 1 + a 2 e 2 + + anen:
When we apply the inner product with each ei to the equation, we get ci = h v; ei i = h a 1 e 1 + a 2 e 2 + + anen; ei i = a 1 h e 1 ; ei i + a 2 h e 2 ; ei i + + an h en; ei i (by linearity of inner product)
Exam on Chapter 15 (Continued)
gives the orthonormal set from fe 1 = (1; 0 ; 1 ; 1); e 2 = (2; 3 ; 1 ; 2)g when the standard inner product is used: f^ 1 = p^1 3 (1;^0 ;^ ^1 ;^ 1)^ ;^ f 2 = p^1 87 (1;^9 ;^2 ;^ 1)
which is di erent than our result. It's because the built{in function in Mathematica uses the algorithm \Gram{Schmidt" process. You will see in the following examples.............................................................. End of Aside
11 Let R^3 be the standard inner product space. Given the set f e 1 = (3; 0 ; 4); e 2 = ( 1 ; 0 ; 7); e 3 = (2; 9 ; 11) g R^3 , deduce an
orthonormal set by using the Gram{Schmidt process. Answer. Step 1 Orthogonal Set (Gram{Schmidt): 1 = e 1 = (3; 0 ; 4) 2 = e 2 h^ ek^2 ; ^1 i 1 k^2
= ( 1 ; 0 ; 7) h^ ( ^1 ;k^0 (3;^ ;7) 0 ;;^ 4)(3k; 2 0 ;^ 4)^ i(3; 0 ; 4) = ( 1 ; 0 ; 7) (3; 0 ; 4) = ( 4 ; 0 ; 3)
3 = e 3 h^ ek^3 ; ^2 i 2 k^2
2 h^ ek^3 ; ^1 i 1 k^2
= (2; 9 ; 11) h^ (2;^9 ;^ 11);^ ( ^4 ;^0 ;^ 3)^ i k( 4 ; 0 ; 3)k^2 ( 4 ; 0 ; 3) h^ (2;^9 ;^ 11);^ (3;^0 ;^ 4)^ i k(3; 0 ; 4)k^2
So we have the orthogonal set f 1 = (3; 0 ; 4); 2 = ( 4 ; 0 ; 3); 3 = (0; 9 ; 0) g :
Step 2 Normalization: A simple computation shows k 1 k^2 = 25 = 5^2 ; k 2 k^2 = 25 = 5^2 ; k 3 k^2 = 81 = 9^2 : By de ning, 1 =^
5 (3;^0 ;^ 4);^2 =^
5 ( ^4 ;^0 ;^ 3);^3 =^
9 (0;^9 ;^ 0) = (0;^1 ;^ 0);
nally, we obtain the orthonormal set 1 =^15 (3;^0 ;^ 4);^2 =^15 ( ^4 ;^0 ;^ 3);^3 = (0;^1 ;^ 0)
: X
○^ Aside: We can get any orthonormal set by using the built{in function in Mathematica: the code
ß
Orthogonalize[{{3,0,4},{-1,0,7},{2,9,11}},Dot] ¶
gives the orthonormal set from f e 1 = (3; 0 ; 4); e 2 = ( 1 ; 0 ; 7); e 3 = (2; 9 ; 11) g when the standard inner product is used: 1 =
5 (3;^0 ;^ 4);^2 =
5 ( ^4 ;^0 ;^ 3);^3 = (0;^1 ;^ 0)
which is exactly same as our result.......................................................................... End of Aside
12 Let R^3 be the standard inner product space. Applying the Gram{Schmidt process to the set f e 1 = (1; 0 ; 1); e 2 =
(1; 0 ; 1); e 3 = (0; 3 ; 4) g, deduce an orthonormal basis for R^3.
Exam on Chapter 15 (Continued)
Answer. Step 1 Orthogonal Set (Gram{Schmidt):
1 = e 1 = (1; 0 ; 1) 2 = e 2 h^ ek^2 ; ^1 i 1 k^2
= (1; 0 ; 1) h^ (1;^0 k;^ (1 1); 0 ;;^ 1)(1k; 2 0 ;^ 1)^ i(1; 0 ; 1) = (1; 0 ; 1) 0(1; 0 ; 1) = (1; 0 ; 1)
3 = e 3 h^ ek^3 ; ^2 i 2 k^2
2 h^ ek^3 ; ^1 i 1 k^2
= (0; 3 ; 4) h^ (0;^3 k;(1^ 4); 0 ;^ ;(1 ;1)^0 k;^ 2 1)^ i(1; 0 ; 1) h^ (0;^ k^3 (1;^ 4); 0 ;^ ; (11);k^02 ; 1)^ i(1; 0 ; 1) = (0; 3 ; 4) + 2(1; 0 ; 1) 2(1; 0 ; 1) = (0; 3 ; 0):
So we have the orthogonal set f 1 = (1; 0 ; 1); 2 = (1; 0 ; 1); 3 = (0; 3 ; 0) g :
Step 2 Normalization: A simple computation shows
k 1 k^2 = 2; k 2 k^2 = 2; k 3 k^2 = 9: By de ning,
1 =^ p ^1 2 =^ p^1 2 (1;^0 ;^ 1);^2 =^ p ^2 2 =^ p^1 2 (1;^0 ;^ 1);^3 =^
3 (0;^3 ;^ 0) = (0;^1 ;^ 0);
nally, we obtain the orthonormal set 1 =^ p^12 (1;^0 ;^ 1);^2 =^ p^12 (1;^0 ;^ 1);^3 = (0;^1 ;^ 0)
: X
○^ Aside: We can get any orthonormal set by using the built{in function in Mathematica: the code
ß
Orthogonalize[{{1,0,1},{1,0,-1},{0,3,4}},Dot] ¶
gives the orthonormal set from f e 1 = (1; 0 ; 1); e 2 = (1; 0 ; 1); e 3 = (0; 3 ; 4) g when the standard inner product is used: 1 =^ p^12 (1;^0 ;^ 1);^2 =^ p^12 (1;^0 ;^ 1);^3 = (0;^1 ;^ 0)
which is exactly same as our result.......................................................................... End of Aside
13 Let P 2 (R) be the inner product space, where the inner product is de ned by
h p(x); q(x) i = p( 1)q( 1) + p(0)q(0) + p(1)q(1); for all p(x); q(x) 2 P 2 (R). (This inner product was already mentioned in the previous section.) Applying the Gram{Schmidt process to the set f e 1 = 1; e 2 = x; e 3 = x^2 g, deduce an orthonormal basis for P 2 (R). Answer. Step 1 Orthogonal Set (Gram{Schmidt): 1 = e 1 = 1 2 = e 2 h^ ek^2 ; ^1 i 1 k^2
1 = x h^ x;k 1 k^12 i1 = x 0 = x
Exam on Chapter 15 (Continued)
gives the rst ve orthonormal set from 1 ; x; x^2 ; x^3 ; x^4 when the Sobolev inner product is used. They are as follows:
1 =^ p^12 ;^2 =^12
r 3 2 x;^3 =
r 5 2
x^2 (^13)
r 7 302
x^3 910 x
; 5 = 2 p^1052102
x^4 3328 x^2 + 14027
If we use the standard inner product, h f(x); g(x) i = R^ ^11 f(x)g(x) dx, then the code ® ß
Orthogonalize[{1, x, x^2, x^3, x^4}, Integrate[#1*#2, {x, -1, 1}] &] ¶
gives the rst ve orthonormal set from 1 ; x; x^2 ; x^3 ; x^4 :
1 =^ p^12 ;^2 =
r 3 2 x;^3 =
r 5 2
x^2 (^13)
4 =^52
r 7 2
x^3 35 x
; 5 = 8105 p 2
x^4 67 x^2 + 353
In my computer, it took 0:234 second to get the result above. Nice computer! :P............................ End of Aside
Exam on Chapter 15 (Continued)
Section 15.3 Isometries
15 Let R^3 be a standard inner product space and T : R^3! R^3 a linear transformation de ned by T (x; y; z) = (0; x; y). T is
called the shift operator. Is T an isometric embedding? Justify your answer. Answer. For any (x; y; z); (u; v; w) 2 R^3 ,
h T (x; y; z); T (u; v; w) i = h (0; x; y); (0; u; v) i = xu + yv 6 = xu + yv + zw = h (x; y; z); (u; v; w) i :
Thus T is not an isometric embedding. X
16 Let F(R) be an inner product space of real functions integrable on [0; 1 ), where the inner product is de ned by
h f(x); g(x) i =
Z 1
0 f(x)g(x) dx:
Let T : F! F be a linear transformation de ned by
T (f)(x) =
f(x 1); x > 1 ; 0 ; x 1 : Show that T is an isometric embedding. In fact, T is an isometry. If we assign to F(R) a di erent inner product,
h f(x); g(x) i =
Z 2
0 f(x)g(x) dx;
then is the transformation T above still an isometric embedding? The answer is NO. Proof. For any f(x); g(x) 2 F(R),
h T (f)(x); T (g)(x) i =
Z 1
0 T (f)(x)T (g)(x) dx
=
Z 1
0 T (f)(x)T (g)(x) dx +
Z 1
1 T (f)(x)T (g)(x) dx =
Z 1
0 0 dx +
Z 1
1 f(x 1)g(x 1) dx =
Z 1
1 f(x 1)g(x 1) dx =
Z 1
0 f(t)g(t) dt (by substitution t = x 1) = h f(x); g(x) i :
Thus T is an isometric embedding. X
Section 15.4 Riesz Representation Theorem
17 Let RN^ be the standard inner product space, where RN^ = f (x 1 ; x 2 ; : : : ; xN 1 ; xN ) j x 1 ; x 2 ; : : : ; xN 1 ; xN 2 R g.
( 1 ) Let T : RN^! R be a linear functional de ned by
T (x 1 ; x 2 ; : : : ; xN 1 ; xN ) = x 1 + 2x 2 + + (N 1)xN 1 + NxN : Find the vector v 2 RN^ such that T (w) = h w; v i for all w 2 RN^. Answer. For any w = (w 1 ; w 2 ; : : : ; wN ) 2 RN^ ,
T (w 1 ; w 2 ; : : : ; wN ) = w 1 + 2w 2 + + NwN = h (w 1 ; w 2 ; : : : ; wN ); (1; 2 ; : : : ; N ) i :
Therefore, for v = (1; 2 ; : : : ; N ), we have T (w) = h w; v i for all w 2 RN^. X
