Math 250B Midterm II Information Spring 2019 SOLUTIONS TO PRACTICE PROBLEMS

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1 Math 50B Midterm II Information Spring 019 SOLUTIONS TO PRACTICE PROBLEMS Problem 1. Determine whether each set S below forms a subspace of the given vector space V. Show carefully that your answer is correct: (a): V P 3 (R) and S {x 3 + bx + a : a, b R}. SOLUTION: NO. This set S fails to be closed under addition. For instance, x 3 + x + 3 and x 3 x + 5 both belong to S, but (x 3 + x + 3) + (x 3 x + 5) 4x 3 x + 8 does not belong to S (since the coefficient of x 3 went up to 4). Notice that to show that S was not a subspace, we simply needed to illustrate a specific example of how the closure property was invalid. (b): V M 3 (R) and S is the set of all 3 3 lower triangular matrices. SOLUTION: YES. We will verify (using general matrices) that S is closed under addition and scalar multiplication: Closure under Addition: Suppose A and B are elements of S. Write a b A a a 3 0 and B b b 3 0, a 4 a 5 a 6 b 4 b 5 b 6 which is the general form for the lower triangular matrices. Then observe that a 1 + b A + B a + b a 3 + b 3 0 a 4 + b 4 a 5 + b 5 a 6 + b 6 is still lower-triangular. addition. Thus, A + B S, which means that S is closed under Closure under Scalar Multiplication: Let A be as above, and let k be a scalar. Then ka k A ka ka 3 0, ka 4 ka 5 ka 6

2 which is still lower-triangular. Thus, k A S, which means that S is closed under scalar multiplication. Since S is closed under addition and scalar multiplication, it forms a subspace of V. (c): V {f : [a, b R} and S is the set of all functions in V such that f(a) f(b). SOLUTION: YES. We will verify that S is closed under addition and scalar multiplication: Closure under Addition: Suppose f and g are elements of S. This means that f(a) f(b) and g(a) g(b). We must verify that f + g meets this same condition: (f + g)(a) f(a) + g(a) f(b) + g(b) (f(b) + g(b)) (f + g)(b), which shows that f + g S. Closure under Scalar Multiplication: Let f be as above and let k be a scalar. Then we must verify that the function k f still belongs to S: which shows that k f S. (k f)(a) kf(a) k(f(b)) (kf(b)) ((k f)(b)), Since S is closed under addition and scalar multiplication, it forms a subspace of V. Problem. Which of the following sets of vectors is linearly dependent? For those that are, write a linear dependency among the vectors. (a): 6 x and 1 + x + 4x SOLUTION: LINEARLY INDEPENDENT. To prove this, suppose that c 1 (6 x ) + c (1 + x + 4x ) 0. Then (6c 1 + c ) + c x + ( c 1 + 4c )x 0. Hence, 6c 1 + c c c 1 + 4c 0, from which it follows immediately that c 1 c 0. Hence, the vectors are linearly independent. (b): 1 + 3x + 3x, x + 4x, 5 + 6x + 3x, 7 + x x. SOLUTION: LINEARLY DEPENDENT. The given polynomials all belong to P (R), the 3-dimensional vector space consisting of polynomials of degree at most.

3 Since we have 4 > 3 vectors in this set, they cannot possibly be independent. This part of the answer is quick! Next, we must derive a linear dependency of the form Therefore, Hence, a(1 + 3x + 3x ) + b(x + 4x ) + c(5 + 6x + 3x ) + d(7 + x x ) 0. (a + 5c + 7d) + (3a + b + 6c + d)x + (3a + 4b + 3c d)x 0. a + 5c + 7d 0, 3a + b + 6c + d 0, 3a + 4b + 3c d 0. This gives rise to a linear system with augmented matrix The row echelon form is Setting d t, we use back-substitution to find that Putting t 1, we find that This gives us the linear dependency. a 17 4 t, b 5 4 t, c 9 t, d t. 4 a 17 4, b 5 4, c 9 4, d (1 + 3x + 3x ) 5 4 (x + 4x ) 9 4 (5 + 6x + 3x ) + (7 + x x ) 0. (c): (3, 0, 4), (5, 1, ), and (1, 1, 3).

4 SOLUTION: LINEARLY INDEPENDENT. Suppose that a(3, 0, 4) + b(5, 1, ) + c(1, 1, 3) (0, 0, 0). We wish to solve for a, b, c. To do this, we need to solve the linear system But, we can quickly show that det , so the matrix is 4 3 invertible. This means that the corresponding linear system we are considering has only the trivial solution. That is a b c 0, as desired. Problem 3. Let V : P 5 (R). Show that {1, 1 + x, x + x, x + x 3, x 3 + x 4, x 4 + x 5 } is a basis for V. SOLUTION: We need only show that this set of given vectors is linearly independent (since there are the right number of vectors, they will automatically span the space V once we show they are linearly independent). To this end, suppose that c 1 (1) + c (1 + x) + c 3 (x + x ) + c 4 (x + x 3 ) + c 5 (x 3 + x 4 ) + c 6 (x 4 + x 5 ) 0. We must show that c 1 c c 3 c 4 c 5 c 6 0. Re-combining the terms, we have (c 1 + c ) + (c + c 3 )x + (c 3 + c 4 )x + (c 4 + c 5 )x 3 + (c 5 + c 6 )x 4 + c 6 x 5 0. Comparing powers of x on each side of the equation, we conclude that c 1 + c 0, c + c 3 0, c 3 + c 4 0, c 4 + c 5 0, c 5 + c 6 0, c 6 0. A quick back-substitution argument now shows that c 1 c c 3 c 4 c 5 c 6 0, Problem 4. Suppose that {v 1, v, v 3 } is linearly independent. Let w 1 v 1. Let w v 1 + 3v. Let w 3 4v 1 + 5v + 6v 3. Show that {w 1, w, w 3 } is linearly independent. SOLUTION: Suppose that c 1 w 1 + c w + c 3 w 3 0.

5 We wish to show that c 1 c c 3 0. This means that or c 1 v 1 + c (v 1 + 3v ) + c 3 (4v 1 + 5v + 6v 3 ) 0, (c 1 + c + 4c 3 )v 1 + (3c + 5c 3 )v + 6c 3 v 3 0. Since {v 1, v, v 3 } is linearly independent, we conclude that c 1 + c + 4c 3 0, 3c + 5c 3 0, 6c 3 0. An easy back-substitution argument now shows that c 1 c c 3 0, as desired. Problem 5. Verify that the function T is a linear transformations. (a): T : R 3 M (R) defined by [ a b b c T (a, b, c) : a + b b + c. SOLUTION: We must check that T respects addition and scalar multiplication. Let (a, b, c) R 3 and let (x, y, z) R 3. T respects addition: We have T ((a, b, c) + (x, y, z)) T (a + x, b + y, c + z) [ a + x b y b + y c z a + x + b + y b + y + c + z [ [ a b b c x y y z + a + b b + c x + y y + z T (a, b, c) + T (x, y, z). T respects scalar multiplication: Next, let k be a scalar. We have T (k(a, b, c)) T (ka, kb, kc) [ ka kb kb kc ka + kb kb + kc [ a b b c k a + b b + c kt (a, b, c).

6 (b): T : P (R) R 3 defined by T (a 0 + a 1 x + a x ) : a 0 a 1 a 0 + a 1 3a a a 0. SOLUTION: We must check that T respects addition and scalar multiplication. Let a 0 + a 1 x + a x and b 0 + b 1 x + b x belong to P (R). T respects addition: We have T ((a 0 + a 1 x + a x ) + (b 0 + b 1 x + b x )) T ((a 0 + b 0 ) + (a 1 + b 1 )x + (a + b )x ) (a 0 + b 0 ) (a 1 + b 1 ) (a 0 + b 0 ) + (a 1 + b 1 ) 3(a + b ) (a + b ) (a 0 + b 0 ) a 0 a 1 b 0 b 1 a 0 + a 1 3a a a 0 + b 0 + b 1 3b b b 0 T (a 0 + a 1 x + a x ) + T (b 0 + b 1 x + b x ), T respects scalar multiplication: Next, let k be a scalar. We have T (k(a 0 + a 1 x + a x )) T (ka 0 + ka 1 x + ka x ) (ka 0 ) (ka 1 ) ka 0 + ka 1 3(ka ) ka ka 0 a 0 a 1 k a 0 + a 1 3a a a 0 kt (a 0 + a 1 x + a x ), (c): T : P (R) R defined by T (p(x)) : [ p(0) p(1).

7 SOLUTION: We must check that T respects addition and scalar multiplication. Let p(x) and q(x) belong to P (R). T respects addition: We have T (p(x) + q(x)) T ((p + q)(x)) [ (p + q)(0) (p + q)(1) [ p(0) + q(0) p(1) + q(1) [ [ p(0) q(0) + p(1) q(1) T (p(x)) + T (q(x)), T respects scalar multiplication: Next, let k be a scalar. We have T (k p(x)) T ((kp)(x)) [ (kp)(0) (kp)(1) [ k p(0) k p(1) [ p(0) k p(1) k T (p(x)), (d): T : R 3 W defined by [ a + b + c b c T (a, b, c) : b c a c, where W is the vector space of all symmetric matrices. SOLUTION: We must check that T respects addition and scalar multiplication. Let (a, b, c) and (x, y, z) belong to R 3.

8 T respects addition: We have T ((a, b, c) + (x, y, z)) T (a + x, b + y, c + z) [ (a + x) + (b + y) + (c + z) (b + y) (c + z) (b + y) (c + z) (a + x) (c + z) (a + b + c) + (x + y + z) (b c) + (y z) (b c) + (y z) (a c) + (x z) [ [ a + b + c b c x + y + z y z + b c a c y z x z Next, let k be a scalar. Then we have T (a, b, c) + T (x, y, z), T (k(a, b, c)) T (ka, kb, kc) [ ka + kb + kc kb (kc) kb (kc) ka kc [ k(a + b + c) k(b c) k(b c) k(a c) [ a + b + c b c k b c a c kt (a, b, c), (e): T : R 4 R 3 defined via T (x) Ax, where A SOLUTION: We can express T as a matrix transformation as T (x) Ax, where A

9 Such transformations are always linear: and for all scalars k. T (x + y) A(x + y) Ax + Ay T (x) + T (y) T (kx) A(kx) k(ax) kt (x) Problem 6. Can there be a linear transformation T : R 3 P (R) such that T (, 1, 0) 1 + x, T (3, 0, ) x + x, and T (0, 6, 8) + x? SOLUTION: NO. Note that so that 6(, 1, 0) 4(3, 0, ) (0, 6, 8) (0, 0, 0), 0 T (0, 0, 0) T (6(, 1, 0) 4(3, 0, ) (0, 6, 8)) 6T (, 1, 0) 4T (3, 0, ) T (0, 6, 8) 6(1 + x) 4( x + x ) ( + x ) 10x 6x. But obviously 10x 6x 0, so this is an impossible contradiction! Therefore, the answer is NO. Problem 7. (a): Let S : Span{(1, 1, 1, 0), ( 1, 0, 1, 1), (, 1,, 1)}. Find an orthonormal basis for S relative to the standard inner product on R 4. SOLUTION: Let v 1 (1, 1, 1, 0), v ( 1, 0, 1, 1), and v 3 (, 1,, 1). Set w 1 v 1 (1, 1, 1, 0) and note that w 1 3. Now v, w 1 and w v v, w 1 w 1 w 1 1 ( 1,, 1, 3). 3 Thus, 15 w 3. Next, we have v 3, w 1 1 and v 3, w 1. Thus, 3 w 3 v 3 v 3, w 1 w 1 w 1 v 3, w w w 4 (3, 1,, 1) 5

10 and Thus, an orthonormal basis is { 3 (1, 1, 1, 0), 3 w ( 1,, 1, 3), } (3, 1,, 1). (b): Let S : Span{(1, 0, 1, 0), (1, 1, 1, 0), ( 1, 1, 0, 1)}. Find an orthonormal basis for S relative to the standard inner product on R 4. SOLUTION: Let v 1 (1, 0, 1, 0), v (1, 1, 1, 0) and v 3 ( 1, 1, 0, 1). w 1 v 1 (1, 0, 1, 0), w v, w 1 (1, 1, 1, 0), (1, 0, 1, 0) ( 1)( 1) w v v, w 1 w 1 v 1 (1, 1, 1, 0) (1, 0, 1, 0) (0, 1, 0, 0). w v 3, w 1 ( 1, 1, 0, 1), (1, 0, 1, 0) ( 1) ( 1) v 3, w ( 1, 1, 0, 1), (0, 1, 0, 0) ( 1) w 3 v 3 v 3, w 1 w 1 w 1 v 3, w w w ( 1, 1, 0, 1) + 1 (1, 0, 1, 0) (0, 1, 0, 0) 1 ( 1, 0, 1, ); w Hence, an orthonormal basis is: { } 6 (1, 0, 1, 0), (0, 1, 0, 0), ( 1, 0, 1, ). 6 (c): Let S : Span{1, x }. Find an orthogonal basis for S relative to the inner product given by f, g 1 1 f(t)g(t)dt. SOLUTION: Let v 1 1 and v x. Then set w 1 v 1 1. Now, We have w w v v, w 1 w 1 w 1. dx and v, w x dx 3.

11 Therefore, w x 1 3. The vectors {w 1, w } forms an orthogonal set. Problem 8. Prove that in any inner product space V, if u and v are vectors in V, we have u + v + u v u + v. SOLUTION: We have u + v + u v u + v, u + v + u v, u v [ u, u + u, v + v, u + v, v + [ u, u u, v v, u + v, v u + v. Problem 9. Suppose that T : V W is a linear transformation. We define (This is called the kernel of T.) Ker(T ) {v V : T (v) 0 W }. (a): Show that Ker(T ) is a subspace of V. SOLUTION: This is part (1) of Theorem in the text (p. 363). (b): Let T : R n R m be the matrix transformation given by T v Av, for a fixed m n matrix A. Show that Ker(T ) nullspace(a). SOLUTION: Note that v Ker(T ) if and only if T (v) 0 W if and only if Av 0 if and only if v belongs to nullspace(a). Thus, this is easily proved already. (c): Use part (b) to find a basis and dimension for Ker(T ) for Problem 5(e). SOLUTION: Now, Ker(T ) is the null space of A. A row-echelon form of A is ,

12 with corresponding equation x 3y + 4z + w 0. Set w t, z s, y r and x 3r 4s t. So Ker(T ) {(3r 4s t, r, s, t) : r, s, t R} {r(3, 1, 0, 0) + s( 4, 0, 1, 0) + t(, 0, 0, 1) : r, s, t R}, with basis {(3, 1, 0, 0), ( 4, 0, 1, 0), (, 0, 0, 1)}. Ker(T ) is 3-dimensional. (d): Find a basis and dimension for Ker(T ), where T is given in Problem 5(b). SOLUTION: Note that a 0 + a 1 x + a x belongs to Ker(T ) if and only if a 0 a 1 a 0 + a 1 3a a a 0 0. A short calculation shows that a t is free, then a 0 t, and a 1 t. So Ker(T ) {t + (t)x + tx : t R}, which has basis (by taking t 1) of {1 + x + x }. In this case, Ker(T ) is 1-dimensional. (e): Find a basis and dimension for Ker(T ), where T is given in Problem 5(c). SOLUTION: In this case, p(x) a 0 + a 1 x + a x belongs to Ker(T ) if and only if p(0) p(1) 0. Now p(0) a 0 and p(1) a 0 + a 1 + a. Here, we find that a 0 0, a t is free, and a 1 t. Thus, p(x) x + x is a basis vector for Ker(T ), which is 1-dimensional.