x + y + z = 1 2x + 3y + 4z = 0 5x + 6y + 7z = 3


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1 Math 24 FINAL EXAM (2/9/9  SOLUTIONS ( Find the general solution to the system of equations ( r 2 2r r 2 r 5r r x + y + z 2x + y + 4z 5x + 6y + 7z So x z + y 2z 2 and z is free. ( r r 2 r r r 2 r 2 2 (2 Let V R and define a product on V by (x y z x 2 y 2 z 2 x x 2 + y y 2. Show that this is not an inner product by finding a property of inner products that fails to hold. The property that fails to hold is (u u u. For example if u then u but (u u. ( Let A 2 7 (a Find a basis for the nullspace of A. Reduce the augmented matrix A to RREF: 2 7 ( r 2 r r 2 r r r r r 2 4 So the solutions are x x 2 x x 4 x x + x 5 4x 5 x x 5 ( r 2r r r 2 +r r 2 x ( r r 2 r r +r 2 r 4 + x so { T 4 T } is a basis for the nullspace of A. (b Find a basis for the columnspace of A. From above the RREF of A has leading ones in columns 2 and 4. Therefore the first second and fourth columns of A form a basis: { T 2 T { T }. (c Find a basis for the rowspace of A. From above the three rows of the RREF of A have leading ones so they form a basis for the rowspace of A: { 2 4 }. (d What is the rank of A? Explain. rank A since by part (b that is the dimension of the columnspace. (or since by part (c that is the dimension of the rowspace.
2 (4 Let L : V W be a linear transformation. (a Prove that L(. L( L( L(. OR: L( L( +( L( + ( L( L( L(. (b Prove that the range of L is a subspace of W. Let w w 2 range L. Then w L(v for some v V and w 2 L(v 2 for some v 2 V. So w + w 2 L(v + L(v 2 L(v + v 2 range L since v + v 2 V. Also let w range L c R then w L(v for some v V so c w c L(v L(c v range L since c v V. Thus the range is closed under addition and under scalar multiplication so it is a subspace. (5 Let V be an inner product space and let u v be vectors in V. Prove that u + v 2 + u v 2 2 u v 2 u + v 2 + u v 2 (u + v u + v + (u v u v (u u + (u v + (v u + (v v + (u u (u v (v u + (v v 2(u u + 2(v v 2 u 2 + v 2 (6 Let V be an inner product space let w w 2 be linearly independent vectors in V and let W span{w w 2 }. Let u be a nonzero vector in V which is an element of W the orthogonal complement of W in V. Prove that w w 2 u are linearly independent. Supose c w +c 2 w 2 +c u. Then (u c w +c 2 w 2 +c u (u. So c (u w + c 2 (u w 2 + c (u u. But since u W (u w (u w 2 so c (u u and since u this implies c. Therefore c w +c 2 w 2. Since w w 2 are linearly independent this implies c c 2. Thus c c 2 c proving that w w 2 u are linearly independent. (7 Let V P the space of all polynomials of degree. Define an inner product on V by (p(t q(t p(tq(tdt. Let W be the subspace of V with basis {t t 2 }. Find an orthogonal basis for W (you do not need to find an orthonormal basis. ( Hint: tn dt n+ We use the GS process: p (t t and p 2 (t t 2 (t2 t So {t t 2 4t} is an orthogonal basis for W. R (tt t t2 t dt R t2 dt t t2 /4 / t t2 4 t.
3 (8 Let V M 22 and let W R 2. Define a function L : V W as follows. Let b 5 and for any matrix A in V define L(A A b. (a Prove that L is a linear transformation. Let A A 2 V. Then L(A + A 2 (A + A 2 b A b +A 2 b L(A + L(A 2. Also if A V c R then L(cA (ca b c(a b cl(a. So L is linear. (b Find the representation of L with respect to the standard basis for V : S { and the standard basis for W : T ( L 5 ( L 5 { } ( L ( L So the matrix is 5 5 (9 Let A (a Find the eigenvalues of A. The characteristic polynomial is det(ti A det t 2 t t } (t (t + 4t + + ( 2(2 (t + 4( 2 (t ( 2(2 t + t 2 4t 4 + 2t t 4 t + t 2 + 2t t(t + 2(t +. So the eigenvalues are 2.
4 ( Let (b For each eigenvalue of A find a corresponding eigenvector. For the eigenvalue λ we find the nullspace of I A: So x z y z and z is free so T is a corresponding eigenvector. For the eigenvalue λ we find the nullspace of I A: So x 2 z y z and z is free so 2 2T is a corresponding eigenvector. For the eigenvalue λ 2 we find the nullspace of 2I A: So x z y 4 z and z is free so 4 T is a corresponding eigenvector. (c Find the eigenvalues of A 2. If A v λ v then A 2 v A(λ v λ 2 v. Therefore the eigenvalues of A 2 are 2 ( 2 and ( A (a Show that the determinant of A is equal to. One option is to make one rowreduction step (r 4 r r 4 and then expand along the first column: det(a det det ((2 ((4( (b Use the result of part (a to show that A 2A 2. det(a while det(2a So A 2A 2 since their determinants are different. ( Let V P 2 the vector space of all polynomials of degree at most 2. Let S {+t t t 2 } and T { + t + t 2 } be ordered bases for V. Let p(t be a polynomial such that p(t S 2 (a Find p(t p(t ( + t + 2 ( t + t + t 2.
5 (b Find the transition matrix P T S between S and T. The matrix has columns + t T t T t 2 T. We can find these simultaneously using the rowreduction: Therefore (c Find p(t T 2 P T S p(t T P T S p(t S (2 Let L : V W be a linear transformation with ker L {}. Let v... v n be linearly independent vectors in V. Prove that the vectors L(v... L(v n are also linearly independent. Suppose c L(v c n L(v n. Then L(c v c n v n. So c v c n v n ker L {} and so c v +... c n v n. Since v... v n are linearly independent this implies c... c n. Thus L(v... L(v n are linearly independent. MAKEUP EXAM QUESTIONS: 4(b Prove that the kernel of L is a subspace of W. Let v v 2 ker L. Then L(v + v 2 L(v + L(v 2 + so v + v 2 ker L. Also if v ker L c R then L(c v cl(v so c v ker L. Thus the kernel is closed under addition and under scalar multiplication so it is a subspace. (5 Let V be an inner product space and let u v be vectors in V. Prove that (u v 4 u + v 2 4 u v 2 (u + v 2 (u v 2 (u + v u + v (u v u v (u u + (u v + (v u + (v v ( (u u + (u v + (v u + (v v 2(u v + 2(v u 4(u v. Dividing by 4 the identity follows. (2 Let L : V W be a linear transformation with range L W. Let v... v n be vectors in V and assume that v... v n are a spanning set for V. Prove that L(v... L(v n are a spanning set for W. Let w W. Then w range L so w L(v for some v V. Since v... v n are a spanning set for V v c v +... c n v n. So w L(v L(c v +... c n v n c L(v +...+c n L(v n so w span{l(v... L(v n }. Thus L(v... L(v n are a spanning set for W.
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