Homework Solutions Week of November = 1(1) + 0(x 1) + 0(x 1) 2 x = 1(1) + 1(x 1) + 0(x 1) 2 x 2 = 1(1) + 2(x 1) + 1(x 1) 2

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1 Math 34/84: Matrix heor Dr S Cooper Fall 8 Section 63: Homework Solutions Week of November 8 (7) he sets B { x (x ) } C { x x } are both bases for P We need to find [p(x) B Observe that () + (x ) + (x ) x () + (x ) + (x ) x () + (x ) + (x ) hus B heorem 6 P B C [ [ B [x B [x B [p(x) B P B C [p(x) C We conclude that the alor polnomial of p(x) about a is p(x) 8(x ) 5(x ) (8) he sets B { x + (x + ) } C { x x } are both bases for P We need to find [p(x) B Observe that () + (x + ) + (x + ) x () + (x + ) + (x + ) x 4() 4(x + ) + (x + ) hus P B C [ [ B [x B [x B 4 4 B heorem 6 [p(x) B P B C [p(x) C We conclude that the alor polnomial of p(x) about a is p(x) 3 + (x + ) 5(x + )

2 Section 64: () is not a linear transformation For example [ [ [ ( [ [ [ ) (3) is a linear transformation Let A C M nn α be a scalar hen (A + C) (A + C)B AB + CB (A) + (C) (αa) (αa)b α(ab) α (A) (5) he transformation : M nn R defined b a a a n a a a n (A) tr(a) a + a + + a nn a n a n a nn is linear Suppose that A [a ij B [b ij are both n n matrices α is a scalar hen (A + B) ([a ij + b ij ) (a + b ) + (a + b ) + + (a nn + b nn ) (a + a + + a nn ) + (b + b + + b nn ) tr(a) + tr(b) (A) + (B) (αa) tr(αa) αa + αa + + αa nn α(a + a + + a nn ) αtr(a) α (A)

3 (7) is not linear For example let hen We see that B A A + B [ [ [ (A) + (B) rank(a) + rank(b) + 4 (A + B) rank(a + B) (8) is not linear For example (6) Using the fact that is linear we have (7) First note that ( + x + x ) ( + x + x ) 4 + 4x + 4x (( + x + x )) ( + x + x ) 3 + 3x + 3x (6 + x 4x ) 6 () + (x) 4 (x ) 6(3 x) + (4x x ) 4( + x ) 8x 9x (a + bx + cx ) a () + b (x) + c (x ) So using the fact that is linear we have a(3 x) + b(4x x ) + c( + x ) (3a + c) + ( a + 4b)x + ( b + c)x 4 x + 3x ( + x) (x + x ) + 4( + x ) (4 x + 3x ) (( + x) (x + x ) + 4( + x )) ( + x) (x + x ) + 4 ( + x ) ( + x ) (x x ) + 4( + x + x ) 4 + 3x + 5x o find (a + bx + c ) in general we need to write a + bx + cx as a linear combination of + x x + x + x : a + bx + cx c ( + x) + c (x + x ) + c 3 ( + x ) (c + c 3 ) + (c + c )x + (c + c 3 )x 3

4 Comparing coefficients this gives the linear sstem c + c 3 a c + c b c + c 3 c We form the augmented matrix row-reduce: a a b b a c c b + a Solving the sstem we obtain hus using the linearit of we have c a + b c c a + b + c c 3 a b + c (a + bx + cx (( ) ( ) ( ) a + b c a + b + c a b + c ( + x) + (x + x ) + ( ) ( ) ( a + b c a + b + c a b + c ( + x) + (x + x ) + ( ) ( ) ( a + b c a + b + c a b + c ( + x ) + (x x ) + ( ) 3a b c a + cx + x (9) Let B {E E E E } be the stard basis for M hen [ w x we z + xe + E + ze ) ( + x ) ) ( + x ) ) ( + x + x ) Since : M R we know that the images of the stard basis vectors are simpl real numbers Let (E ) a (E ) b (E ) c (E ) d where a b c d are real numbers hen b the linearit of we have [ w x (we z + xe + E + ze ) w (E ) + x (E ) + (E ) + z (E ) aw + bx + c + dx () Observe that

5 Since x ( + x) 4( x + x ) + x 4x the transformation with the given properties cannot be linear (5) We have [ (S ) ( [ ) S [ 5 S [ 4 6 [ x Finall ( S) [ x (S ) [ x is not defined since S ( [ ) x S [ x + S [ x x + is a matrix but the domain of is R (9) We have [ x (S ) ( [ ) x S [ x S 3x + 4 [ 4(x ) + ( 3x + 4) 3(x ) + ( 3x + 4) [ x 5

6 [ x ( S) ( [ ) x S [ 4x + 3x + [ (4x + ) (3x + ) 3(4x + ) + 4(3x + ) [ x B definition since S I R S I R S are inverses Section 65: () (a) (i) Since (b) [ 3 the given matrix is not in ker( ) (ii) Since [ 4 the given matrix is in ker( ) (iii) Since [ 3 3 the given matrix is not in ker( ) [ 3 [ 3 3 [ [ [ (i) An matrix in range( ) must have zeros for the () () entries hus the given matrix is not in range( ) (ii) For the same reason as in (i) the given matrix is not in range( ) (iii) Since the () () of the given matrix are this matrix is in range( ) In fact [ 3 3 [ 3 3 (c) We have ker( ) {[ [ [ } a b a b : c d c d {[ [ [ } a b a : c d d {[ } a b : a d c d {[ } b c 6

7 range( ) { [ } a b c d {[ } a d (3) (a) (i) Since (b) ( + x) the given polnomial is not in ker( ) (ii) Since (x x ) the given polnomial is not in ker( ) (iii) Since the given polnomial is in ker( ) (i) Since the given vector is in range( ) (ii) Since the given vector is in range( ) (iii) Since (c) We have the given vector is in range( ) ker( ) [ [ ( + x x ) ( + x x ) ( + x x ) ( + x) [ [ [ [ [ [ { [ a + bx + cx : (a + bx + cx ) { [ a + bx + cx a b : b + c { a + bx + cx : a b b c } { a + ax ax } [ } range( ) { (a + bx + cx ) : a + bx + cx } P {[ } a b b + c R 7 }

8 (5) From Exercise we have that ker( ) {[ } b c { [ c ([ span Suppose we have scalars c c such that c [ + c [ [ + b [ [ It is eas to see that c c hus the spanning set is also linearl independent We conclude that {[ [ } B is a basis for ker( ) so nullit( ) Similarl range( ) {[ } a d { [ a ([ span Suppose we have scalars c c such that c [ + c [ } [ + d [ [ It is eas to see that c c hus the spanning set is also linearl independent We conclude that {[ [ } B is a basis for range( ) so rank( ) B the Rank heorem (7) From Exercise 3 we have that ) } ) dim(m ) 4 + nullit( ) + rank( ) ker( ) {a + ax ax } {a( + x x )} span( + x x ) hus since there is onl onl spanning vector in this case we see that B { + x x } 8

9 is a basis for ker( ) so nullit( ) Since range( ) R we can take the stard basis B {[ [ as a basis for range( ) We conclude that rank( ) B the Rank heorem } dim(p ) 3 + nullit( ) + rank( ) () We have {[ [ [ [ } a b a b ker( ) : c d c d {[ [ [ } a b a b a + b : c d c d c + d {[ } a b : a b c d c d {[ } a a c c { [ [ } a + c ([ [ ) span Furthermore if c c are scalars such that c [ + c [ [ then clearl c c which shows that the spanning set is also linearl independent hus {[ [ } B is a basis for ker( ) hus nullit( ) B the Rank heorem dim(m ) 4 nullit( ) + rank( ) + rank( ) rank( ) 4 (7) (a) We appl heorem 6 to conclude that the transformation is not - o see that ker( ) {} observe that + x + x + x + x ( + x + x ) 9

10 (b) Since dim(p ) 3 dim(r 3 ) is not - we appl heorem 6 to conclude that is also not onto () he vector space has basis B V D 3 a b c (ou should check that these matrices span D 3 are linearl independent) hus dim(d 3 ) 3 dim(r 3 ) B heorem 65 the vector spaces D 3 R 3 are isomorphic Define the linear transformation : D 3 R 3 b a b c Note that so a b c a b c ker( ) which shows that is - a Also if x b c is an vector in R 3 then a b c which shows that range( ) R 3 hus is onto We conclude that is an isomorphism a b c a b c x (3) Note that V {A M 33 : A A} W {B M : B B}

11 We observe that V a b c b e f c f s span It is also eas to see that these spanning matrices are linearl independent hence form a basis for V Similarl W span b c b f c f It is also eas to see that these spanning matrices are linearl independent hence form a basis for W We conclude that dim(v ) 6 dim(w ) 3 isomorphic So b heorem 65 V W are not

MATH 323 Linear Algebra Lecture 12: Basis of a vector space (continued). Rank and nullity of a matrix.

MATH 323 Linear Algebra Lecture 12: Basis of a vector space (continued). Rank and nullity of a matrix. Basis Definition. Let V be a vector space. A linearly independent spanning set for V is called a basis.