MATH 110 SOLUTIONS TO THE PRACTICE FINAL EXERCISE 1. [Cauchy-Schwarz inequality]

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1 MATH 110 SOLUTIONS TO THE PRACTICE FINAL PEYAM TABRIZIAN Note: There might be some mistakes and typos. Please let me know if you find any! EXERCISE 1 Theorem: [Cauchy-Schwarz inequality] Let V be a vector space over F with inner product <, >, and define u = < u, u >. Then for all u and v in V : < u, v > u v Moreover, equality holds if and only if u is a multiple of v or v is a multiple of u Proof: 1 Fix u, v in V. First of all, the result holds for v = 0, because: < u, v >=< u, 0 >= 0 u 0 = u v And also if v = 0, then v = 0u, so v is a multiple of u Hence, from now on, for the rest of the proof, we may assume v 0. Now consider < u av, u av >, where a F is to be selected later. On the one hand, by the nonnegativity axiom of <, >, we have: < u av, u av > 0 ( ) On the other hand, expanding < u av, u av > out, we get: Date: Monday, April 29th, The proof here is different from the one given in the book, but is equally valid. Feel free to memorize either one of them 1

2 2 PEYAM TABRIZIAN < u av, u av > = < u, u > + < u, av > + < av, u > + < av, av > = < u, u > a < u, v > a < v, u > +aa < v, v > = u 2 a < u, v > a < u, v > + a 2 v 2 Combining this with ( ), we get: u 2 + a < u, v > +a < v, u > + a 2 v 2 0 ( ) Now let a = <u,v> <v,v> = <u,v> v 2 2 (which is well-defined since v 0) In particular, we get: and and a < u, v >= a< u, v > = a 2 v 2 = ( < u, v > < u, v > v 2 = < u, v > < u, v > v 2 = 2 < u, v > v 2 2 < u, v > v 2 ) 2 < u, v > v 2 v 2 < u, v > 2 = v 2 Therefore, ( ) becomes: That is: u 2 < u, v > 2 v 2 < u, v > 2 < u, v > 2 v 2 + v 2 0 u 2 Solving for < u, v >, we get: < u, v > 2 v 2 0 ( ) < u, v > 2 u 2 v 2 And taking square roots (given that all the terms are nonnegative), we have: Which is the Cauchy-Schwarz inequality! < u, v > u v 2 The idea is that we want to turn a < v, u > into <u,v> 2 v 2

3 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 3 Finally, if equality holds in the Cauchy-Schwarz inequality, that is, if < u, v > = u v then working our way backwards, then ( ) becomes an equality, and so ( ) becomes an equality becomes equalities, and in particular, ( ) becomes an equality, that is: < u av, u av > = 0 And hence, by the positivity axiom, u av = 0, that is, u = av, so u is a multiple of v. Conversely, if u = av for some a F, then: < u, v > = < av, v > = a < v, v > a < v, v > = a v 2 = a v v = av v = u v So equality holds in the Cauchy-Schwarz inequality. Similarly if v = au for some a F

4 4 PEYAM TABRIZIAN EXERCISE 2 Let (v 1,, v m ) be a basis for Nul(T ), and extend it to a basis (v 1,, v m, v m+1,, v n ) of V. Claim: (T (v m+1 ),, T (v n )) is a basis of Ran(T ). Proof: We need to show that the set spans Ran(T ) and is linearly independent Span: First of all, each T (v m+1 ), T (v n ) is in Ran(T ) (by definition of Ran(T )), and hence, because Ran(T ) is a subspace of W, Span(T (v m+1 ),, T (v n )) Ran(T ) Conversely, let w Ran(T ). Then w = T (v) for some v in V. But then, since (v 1,, v n ) is a basis for V, we have v = a 1 v 1 + +a n v n for scalars a 1,, a n. But then: T (v) = T (a 1 v a n v n ) = a 1 T (v 1 ) + + a m T (v m ) + a m+1 T (v m+1 ) + + a n T (v n ) = a a m 0 + a m+1 T (v m+1 ) + + a n T (v n ) Because v 1,, v m are in Nul(T ) = a m+1 T (v m+1 ) + a n T (v n ) Span(T (v m+1 ),, T (v n )) Hence, w = T (v) Span(T (v m+1 ),, T (v n )). Hence, since w was arbitrary, we get: And therefore: Ran(T ) Span(T (v m+1 ),, T (v n )) So T (v m+1 ),, T (v n ) spans Ran(T ). Span(T (v m+1 ),, T (v n )) = Ran(T )

5 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 5 Linear independence: Suppose a m+1 T (v m+1 ) + + a n T (v n ) = 0. Then T (a m+1 v m a n v n ) = 0 Hence a m+1 v m a n v n Nul(T ) Hence a m+1 v m a n v n = a 1 v a m v m for scalars a 1,, a m, because (v 1,, v m ) is a basis for Nul(T ). Hence a 1 v 1 a m v m + a m+1 v m a n v n = 0. However, (v 1,, v n ) is linearly independent, hence a 1 = = a m = a m+1 = = a n = 0 Hence a m+1 = = a n = 0, which is what we wanted to show. Hence (T (v m+1 ),, T (v n )) is a basis for Ran(T ), and hence dim(ran(t )) = n m But then, it follows that: dim(v ) = n = m + (n m) = dim(nul(t )) + dim(ran(t ))

6 6 PEYAM TABRIZIAN EXERCISE 3 Note: The explanations are optional, and are here to convince you why an answer is true or false. (a) VERY FALSE!!! Remember that P rop1.9 ONLY works for TWO subspaces! (Let V = R 2, U 1 = Span {(1, 0)} (the x axis), U 2 = Span {(0, 1)} (the y axis), and U 3 = Span {(1, 1)} (the line y = x). Notice that V U 1 U 2 U 3 because (0, 0) can be written in two different ways as sums of vectors in U 1, U 2, U 3, namely (0, 0) = (1, 0) + (0, 1) + (0, 0), but also (0, 0) = (1, 0) + (0, 1) + ( 1, 1). This violates the definition of direct sums on page 15) (b) FALSE (c) (d) (e) (Let V = R 3, and [ let T] L(V ) be the linear transformation whose matrix 1 0 is M(T ) = A =. Then (1, 0) and (0, 1) are eigenvectors of T, but 0 2 (1, 1) = (1, 0) + (0, 1) isn t) TRUE (See Theorem For a direct proof: Let m be the minimal polynomial of T and p be the characteristic polynomial. Then by the division algorithm for polynomials, there exist polynomials q and r with deg(r) < deg(m) such that m = qp + r. But then m(t ) = q(t )p(t ) + r(t ). But m(t ) = 0 by definition of the minimal polynomial, and p(t ) = 0 by the Cayley-Hamilton theorem, whence 0 = 0+r(T ), so r(t ) = 0, However, deg(r) < deg(m), whence r 0 (otherwise this would contradict the definition of m as the minimal polynomial). But then m = qp + r = qp + 0 = qp, so p divides m) TRUE (If F = R, then the real spectral theorem applies, and if F = C, then T T = T T = T T, so T is normal and the complex spectral theorem applies) FALSE In general, the statement is false, and the reason is that we didn t specify whether the vector spacee is over R or over C. (In the case F = C, the answer is FALSE, because for example if V = C, then T (v) = iv has only one eigenvalue, λ = i, which is not real. However, in the case F = R, the answer is TRUE, See Theorem 8.2 in section 8 of Axler s paper, or Theorem 5.26 on page 92 of the book.)

7 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 7 (f) FALSE If you take the statement as it is, it doesn t make sense. It should be generalized eigenspaces of a (given) linear operator T Also, if you correct that statement, then it is FALSE if F = R, but TRUE if F = C. (For the case F = C, this is just theorem 8.23 in the book. For the case F = R, consider V [ = R 2,] and let T L(V ) be defined by T (x, y) = (y, x). Then 0 1 M(T ) =, which has no (real) eigenvalues, and hence each eigenspace 1 0 of T is just {0}. And hence the direct sum of the generalized eigenspaces of T is just {0}, which is not equal to V = R 2 )

8 8 PEYAM TABRIZIAN EXERCISE 4 Since S is nilpotent, there exists m such that S m = 0, and since T is nilpotent, there exists n such that T n = 0. Now, because T S = ST, the binomial formula applies, that is, for every k: (technically a j = Now take k = m + n, then: (S + T ) k = k a j S j T k j j=0 j!k! (j k)!, but we won t need that here) (S + T ) m+n = = m+n j=0 a j S j T m+n j m a j S j T m+n j + j=0 n j=m+1 S j T m+n j However, if j m, then m + n j m + n m = n, so m = n j = n + l, where l 0, and so: T m+n j = T n+l = T n T l = 0T l = 0 In particular S j T m+n j = S j 0 = 0, hence all the terms in the first sum are 0. On the other hand, if j m + 1, then j = m + l, where l 0, and so: S j = S m+l = S m S l = 0S l = 0 In particular, S j T m+n j = 0T m+n j = 0, hence all the terms in the second sum are 0. Combining this, we get: Hence S + T is nilpotent. (S + T ) m+n = 0

9 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 9 EXERCISE 5 ( ) Suppose x Nul(P ) and y Ran(P ). We want to show that < x, y >= 0. Since x Nul(P ), P (x) = 0, and since y Ran(P ), y = P (z) for some z V. But then: < x, y > = < x, P (z) > = < P x, z > = < P (x), z >= < 0, z > = 0 Where in the third equality, we used the fact that P = P, since P is self-adjoint. Hence < x, y >= 0, and we re done. ( ) Suppose Nul(P ) Ran(P ). We want to show P = P. That is, for every x and y in V, we want to show that: Notice that you can write: < P x, y > = < x, P y > < P x, y > = < P x, P y + (y P y) >=< P x, P y > + < P x, y P y > However, notice that P x Ran(P ) and P (y P y) = P y P 2 y = P y P y = 0, so y P y Nul(P ). Because Nul(P ) Ran(P ) by assumption, we get that < P x, y P y > = 0. Therefore: But now, we can write: < P x, y >=< P x, P y > < P x, y > = < P x, P y > = < x + (P x x), P y > = < x, P y > + < P x x, P y > But P y Ran(P ) and P (P x x) = P 2 x P x = P x P y = 0, so P x x Nul(P ). Because Nul(P ) Ran(P ) by assumption, we get that < P x x, P y >= 0, and hence: as we wanted to show < P x, y > = < x, P y >

10 10 PEYAM TABRIZIAN EXERCISE 6 Note: Remember how to calculate the matrix of a linear transformation with respect to a basis (v 1,, v n )! For each basis vector v i, calculate T (v i ) and express your result as a linear combination of all your basis vectors (v 1,, v n ). Here: [ ] [ ] [ ] T = 3 4 [ ] 1 0 = 3 0 [ ] 1 0 = [ ] [ ] [ ] 0 1 Hence the first column of M(T ) is: Next: [ ] [ ] [ ] T = 3 4 [ ] 0 1 = 0 3 [ ] 1 0 = Hence the second column of M(T ) is: Next: [ ] [ ] [ ] 1 2 T = [ ] 2 0 = 4 0 [ ] 1 0 = [ ] [ ] [ ] [ ] [ ] 0 1 [ ] 0 1

11 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 11 Hence the third column of M(T ) is: Finally: [ ] [ ] [ ] 1 2 T = [ ] 0 2 = 0 4 [ ] 1 0 = [ ] [ ] [ ] 0 1 Hence the fourth column of M(T ) is: Putting everything together, we get: M(T ) =

12 12 PEYAM TABRIZIAN EXERCISE 7 ( ) Let (v 1,, v n ) be a basis for V. By definition, we know that D(v i ) = λ i v i for some λ i, where i = 1, n. Now fix i and define K i L(V ) by: K i (v 1 ) = 0. K i (v i 1 ) = 0 K i (v i ) = v i K i (v i+1 ) = 0. K i (v n ) = 0 That is, K i (v j ) = 0 for j i, and K i (v i ) = v i. Note that K i exists by the linear extension lemma. More explicitly (we ll need this below), if v V, then there exist a 1,, a n such that v = a 1 v a n v n (because (v 1,, v n ) is a basis for V ), and then: K i (v) = K i (a 1 v a n v n ) = a 1 K i (v 1 ) + + a i K i (v i ) + + a n K i (v n ) = a a i v i + + a n 0 = a i v i That is, K i (v) = a i v i, where v = a 1 v a n v n Now we only need to show 3 things: (1) D = n i=1 λ ik i Proof: Let v V, then v = a 1 v a n v n for scalars i = 1,, n. But then:

13 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 13 D(v) = D(a 1 v a n v n ) = a 1 D(v 1 ) + + a n D(v n ) = a 1 λ 1 v a n λ n v n = λ 1 a 1 v λ n a n v n = λ 1 K 1 (v) + + λ n K n (v) ( n ) = λ i K i v i=1 Since v was arbitrary, we get that D = n i=1 λ ik i (2) K 2 i = K i Proof: Let v V, then v = a 1 v a n v n for scalars a 1,, a n. But then: Ki 2 (v) = K i (K i v) = K i (K i (a 1 v a n v n )) = K i (a 1 K i (v 1 ) + + K i (v i ) + + a n K i (v n )) = K i (a a i v i + + a n 0) = K i (a i v i ) = a i K i (v i ) = a i v i = K i (v) Since v was arbitrary, we get K 2 i = K i for all i (3) If i j, then K i K j = 0 Proof: Let v V, then v = a 1 v a n v n for scalars a 1,, a n. But then:

14 14 PEYAM TABRIZIAN K i K j (v) = K i (K j v) = K i (K j (a 1 v a n v n )) = K i (a 1 K j (v 1 ) + + K j (v j ) + + a n K j (v n )) = K i (a a j v j + + a n 0) = K i (a j v j ) = a j K i (v j ) = 0 (where in the last line we used j i) Hence K i K j = 0, since v was arbirary. ( ) For this, we use the result of exercise 11 in chapter 8 (which was on your homework), namely if T L(V ), then: Here with T = K 1, we get: V = Ran(T n ) Nul(T n ) V = Ran(K n 1 ) Nul(K n 1 ) However, because K 2 1 = K 1, we have K n 1 = K 1 (use induction), and hence: V = Ran(K 1 ) Nul(K 1 ) Now let U 1 = Nul(K 1 ). First of all, U 1 is invariant under K 2, because if u 1 U 1, then K 1 (K 2 u 1 ) = (K 1 K 2 )u 1 = 0u 1 = 0, so K 2 u 1 Nul(K 1 ) = U 1. Hence, applying the result of exercise 11 in chapter 8 to V = U 1 and T = K 2 := (K 2 ) U1, we get: Nul(K 1 ) = U 1 = Ran(K 2) Nul(K 2) ( ) Claim: Ran(K 2) = Ran(K 2 ) Proof: Suppose v Ran(K 2 ), then v = K 2 (v ) for some v V. But then since V = Ran(K 1 ) Nul(K 1 ), we get v = v 1 + v 2, where v 1 Ran(K 1 ), so v 1 = K 1 (u 1 ) and v 2 Nul(K 1 ) = U 1. But then K 2 (v ) = K 2 (v 1 ) + K 2 (v 2 ) = K 2 (K 1 u 1 ) + K 2 (v 2 ) = 0u 1 + K 2 (v 2 ) = K 2 (v 2 ) = K 2v 2 That is: v = K 2 (v ) = K 2(v 2 ) Ran(K 2) Conversely, if v Ran(K 2), then v = K 2(v ) for some v U 1 V, so v = K 2(v ) = K 2 (v ) Ran(K 2 ).

15 MATH 110 SOLUTIONS TO THE PRACTICE FINAL 15 Claim: Nul(K 2) = Nul(K 1 ) Nul(K 2 ) Proof: Suppose v Nul(K 1 ) Nul(K 2 ), then v Nul(K 1 ) = U 1, and so K 2(v) = K 2 (v) = 0, since v Nul(K 2 ), and so v Nul(K 2). Conversely, suppose v Nul(K 2). Then v U 1 = Nul(K 1 ) (by definition of K 2)), and hence K 2 v = K 2v = 0, so v Nul(K 2 ) as well, and hence v Nul(K 1 ) Nul(K 2 ) Combining the two claims and ( ), we get: U 1 = Ran(K 2 ) Nul(K 1 ) Nul(K 2 ) So if you let U 2 = Nul(K 1 ) Nul(K 2 ), you get: And so: U 1 = Ran(K 2 ) U 2 V = Ran(K 1 ) Ran(K 2 ) U 2 Now in general, you can prove by induction on i 3 that if U i = Nul(K 1 ) Nul(K i ), then: And also by induction: U i = Ran(K i ) U i+1 V = Ran(K 1 ) Ran(K 2 ) Ran(K i ) U i And in particular, with i = n, we get: V = Ran(K 1 ) Ran(K 2 ) Ran(K n ) U n Where U n = Nul(K 1 ) Nul(K n ). That is: V = Ran(K 1 ) Ran(K n ) (Nul(K 1 ) Nul(K n )) Note: If any of the above sets are 0, just delete them from the list. Let (v i 1,, v i k i ) be a basis for Ran(K i ) (where k i = dim(ran(k i ))) and (w 1,, w p ) be a basis for Nul(K 1 ) Nul(K n ). Then because V is a direct sum of all the above spaces, we have that the whole list (v 1 1,, v 1 k 1,, v n 1,, v n k n ), w 1,, w p ) is a basis for V. To show D is diagonal, we need (as usual) to calculate D(v k j ) and D(w i): 3 do it!

16 16 PEYAM TABRIZIAN First of all: m m D(vj k ) = ( λ l K i )(vj k ) = λ i K i (vj k ) i=1 Now vj k Ran(K k), so vj k = K k(u k j ) for some uk j ), and hence: K i (vj k) = K ik k (u k j ), which is 0 if i k, and if i = k, this is K2 k (uk j ) = K k(u k j ) = vk j i=1 In other words, we get: D(v k j ) = λ k v k j Finally, for the w i, notice that for all j, K j w i = 0 (because w i is in the Nullspace of all the K i ), and so D(w i ) = 0 = 0w i. From this it follows that D is diagonal