Homework 5 Solutions

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Alexis Jordan
6 years ago
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1 Q Homework Soltions We know that the colmn space is the same as span{a & a ( a * } bt we want the basis Ths we need to make a & a ( a * linearly independent So in each of the following problems we row redce the matrix to find pivot colmns and pivot elements to determine which colmn vectors are independent Non-pivot colmns are linearly dependent on the pivot colmns so we can discard those colmn vectors to form or basis for the colmn space For example in part (a) colmn (a non-pivot) is a linearly dependent on two pivot colmns: (colmn ) (colmn ) a b c 9 row redces to We see that colmn and colmn are linear combinations of colmn and colmn Ths the colmn space basis is row redces to We see that colmn and colmn are linear combinations of the remaining colmns Ths the colmn space basis is row redces to Only colmn is a linear combination of the remaining colmns Ths the colmn space basis is Q a x & x ( x & 8x & 9x ( 8

2 b x & x ( x < x & x ( x < Q a x & x ( x < 8 b x & x ( x < Q a b 8 Not linearly independent The determinant of the only x sbmatrix which is the matrix itself is so the rank is less than (Note how the last row is a mltiple of the st row * -) This shows the set of vectors are not linearly dependent Note: A sbmatrix formed by removing the last row and last colmn is whose determinant is Since we fond a x sbmatrix with non-zero determinant the rank of the matrix is so two vectors are linearly independent 8 Not linearly independent The determinant of the only x sbmatrix which is the matrix itself is so the rank is less than (Note how the second row is ( st row * -) ( rd row / ) which is noticeable when converting to redced row echelon form) This shows the set of vectors are not linearly dependent Note: A sbmatrix formed by removing the last row and last colmn is whose determinant is Since we fond a x sbmatrix with non-zero determinant the rank of the matrix is so two vectors are linearly independent Q A matrix eqation comes in the form Ax B x & 8 a x 8 ( x <

3 b z & z ( z < z > Q a Agmented matrix: Row redced echelon form (to find soltion as a vector): mltiply st row by - and add to nd row à mltiply st row by and add to rd row à mltiply nd row by - and add to rd row à divide last row by - à divide nd row by à mltiply rd row by - and add to nd row à mltiply rd row by and add to st row à mltiply nd row by - and add to st row à

4 Right colmn is the soltion vector: b Agmented matrix: Row redced echelon form: mltiply st row by and add to nd row à mltiply st row by - and add to rd row à mltiply nd row by and add to rd row à divide rd row by à add rd row to nd row à 8 add rd row to st row à 8 mltiply nd row by - and add to st row à 8 divide nd row by à

5 Right colmn is the soltion vector: Q a ~ 9 8 The basic variables are x & 9x < 8x > x ( x < x > so the parametric form is 9s 8t s t s t b x & x ( x A x < x A so the parametric form is x B x A ~ s s t The basic variables are Q8 Becase t t t t space R < H span{ a vector v in R < wold be } is also a sbspace of R < Becase is in vector Q9 Becase b c b c respectively Becase a sbspace of R < b and c vectors and v wold be is in vector space R < W span{ and } is also Q a Becase both H and K are closed nder vector addition that means that a vector in H K may be written as v So H K ( & v & ) ( ( v ( ) & ( (v & v ( ) where the rearranged terms are in H and K respectively Becase both H and K are closed nder scalar mltiplication that means that a vector in H K may be written as c v for some scalar c This is eqivalent to c cv which are in H and K respectively Therefore H K mst be a sbspace of V

6 b One property of sbspaces is that the zero vector of V is in H and K Ths since a vector in H can be represented as where is from H and is from K then H is a sbspace of H K By the same note K is a sbspace of H K as shown by v v Q a False: while this is part of the definition of a sbspace there are two other remaining properties that mst hold namely that H is closed nder additional and scalar mltiplication b Tre: the three properties in the definition of sbspaces garantee that a sbspace is a vector space c Tre: converse of the answer for part (b) d False: R ( is not a sbspace of R < as R < has an additional entry over R ( Q Let vector a be a vector in W Using scalars c & c N vector a is represented as follows: a c & v & c ( v ( c N v N Vectors are orthogonal when the dot prodct is zero Find the dot prodct of vector a and x: a x c & v & c ( v ( c N v N x c & x v & c ( x v ( c N x v N (x is orthogonal to each v S ) Since the dot prodct of an arbitrary vector a and x is x is orthogonal to every vector in W Q Ux ( ( ( x ( ( So Ux x

7 Q Let a and v Orthogonal projection of a onto the line throgh v and the origin is the orthogonal projection of a onto v: a U V & X& Y(X& <) v V V v X& Y(X<) X& X& Y(< <) & Y(Z) v X> X( v &[ B / / Q Let y 9 and Orthogonal projection of y onto the line throgh and the origin is the orthogonal projection of y onto : y ^ _ X< & Y(Z () _ _ X< Y(&`) & & Y(( () & Y(>) &B B The distance from y to the line passing throgh and the origin is given by: y y ( ( Q a Tre: special case where the zero vector is in the orthogonal set otherwise it is false b False: S is an orthogonal set only To be orthonormal the magnitdes of all vectors mst be ; the vectors mst be nit vectors c Tre: Ax x (Theorem in textbook) d Tre: ^ av cv ab (^ V) ^ V v v which is the projection of y onto v av av a b (V V) V V e Tre: the vectors in an orthogonal matrix are linearly independent and the matrix is a sqare matrix Therefore it is invertible Q Verify that the vectors are orthonormal: / & which is not to be orthonormal < / & which is not to be orthonormal ( Verify that the set of vectors are orthogonal: / X& & so this is orthogonal A A /

8 So normalize the vectors: & < / & < & < v & & < / &/< &/< / v ( X& ( [ & ( X& ( [ & ( / / The orthonormal set is v v Q8 Verify that the vectors are orthonormal: However the vectors do not form an orthogonal set: Ths we cannot make an orthonormal set which is defined to be an orthogonal set of nit vectors Q9 Verify that the vectors are orthonormal: Verify that the vectors are orthogonal: Becase the set of vectors are nit vectors and orthogonal the vectors given form an orthonormal set

9 Q Verify that the vectors are orthonormal: > & > Z Z Z & > B which is not to be a nit vector Z Z Z Verify that the set of vectors are orthogonal: X( ( Z Z We only need to normalize the second vector into a nit vector: v & v ( & < ( < [ & < ( < [ /9 /9 The orthonormal set is v v Q Verify that the vectors are orthonormal: ( ([ ([ ([ &` ([ & ([ & ([ & ( & ( Verify that the vectors are orthogonal:

10 A ([ < ([ < ([ & ( ([ & ( ([ < ( ([ < ( ([ Becase the set of vectors are nit vectors and orthogonal the vectors given form an orthonormal set Q Verify that the vectors are orthonormal: & &` &` &` 8 8 & ( & ( > Z & Z > Z Verify that the vectors are orthogonal: 8 8 & & A A 8 X( < ( ( < ( X( < &` > < &` X( < &`

11 Becase the set of vectors are nit vectors and orthogonal the vectors given form an orthonormal set Q a & ( so they are orthogonal / ^ _ e _ & X<X(Y&( f e _ e ZY&Y> & &> & & / ( ^ _ b _ ( X&X(X&( b _ b &Y&Y> ( X&B A ( X&B A / / / / / / b & ( so they are orthogonal ^ _ e _ & X(>X>Y& X(f e _ e &AY&Y& & &` & < / ( / ^ _ b _ ( [Y>Y& b _ b [Y&Y& ( B ( ( B ( / / / / / / Q a U and U g Compte U g U: U g U Compte UU g : UU g

12 b proj l y y & & y ( ( becase vectors & and ( are orthonormal proj l y 8 8 This redces to Using UU g from part (a): UU g y /9 /9 /9 Q a The vectors w & w N in W are orthogonal to each other and the vectors v & v n are orthogonal to W Ths any w and v in the set are orthogonal to each other b Using the Orthogonal Decomposition Theorem we can rewrite a vector y R * as y y z where y is in W and z is in W p This can be rewritten as y c & w & c N w N d & v & d n v n Becase vector y can be written as a linear combination of vectors in R * the set given in part (a) spans R * c The set given in (a) is orthogonal and linearly independent so the set given in (a) is a basis of R * Ths the dimension of R * n p q dim W dim W p Q a Let x ( 9 9 and v & x & v ( x ( y b V e 9 9 v & x ( X&BXZX&`X< v ZY&Y>Y& & x ( X>B v &B & Answer: v & v (

13 b Let x ( and v & x & v ( x ( y b V e v & x ( fy(`y[y& &Y&AY[Y& v & x ( <A &` v& 8 Answer: v & v ( Q a v & x & x ( x < v ( x ( y b V e v & x ( (X&X>X>Y( v &Y&Y&Y&Y& & x ( XB v B & v < x < y z V e v & y z V b V b V b v ( x < BY>Y<YfY& &Y&Y&Y&Y& v & &BY[XZX(&Y< ZY[YZYZYZ v ( x < v & &( <A v ( & < Answer: v & v ( v <

14 b v & x & x ( x < 8 v ( x ( y b V e v & x ( <Y<Y[YBYB &Y&Y[Y&Y& v & x ( &A > v & v < x < y z V e v & y z V b V b V b v ( x < BX&Y[Y(Y` &Y&Y[Y&Y& v & XBY&YAY(Y` &Y&Y>Y&Y& v ( x < &> > v & &( ` v( 8 f ( < ( 8 / / / / / / / / Answer: v & v ( v < Q8 a v & x & x ( x < x > v ( x ( y b V e v & x ( X>[[ >[[ v & x ( v &

15 v < x < y z V e v & y z V b V b V b v ( x < X([[ >[[ v & X>` <A v ( x < & ( v & > < v ( & ( > < 8 v > x > x > v & v v & v & x > v ( v & v ( v ( x > v < v ( v < v < x > < v & v ( 8 v < 8 Answer: v & v ( v < v >

Geometry of Span (continued) The Plane Spanned by u and v

Geometry of Span (continued) The Plane Spanned by u and v Geometric Description of Span Geometr of Span (contined) 2 Geometr of Span (contined) 2 Span {} Span {, } 2 Span {} 2 Geometr of Span (contined) 2 b + 2 The Plane Spanned b and If a plane is spanned b