### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Linear Algebra MATH 21

UCSC

GPA 3.69

### View Full Document

## 57

## 0

## Popular in Course

## Popular in Mathematics (M)

This 15 page Class Notes was uploaded by Sienna Bradtke III on Monday September 7, 2015. The Class Notes belongs to MATH 21 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 57 views. For similar materials see /class/182212/math-21-university-of-california-santa-cruz in Mathematics (M) at University of California - Santa Cruz.

## Similar to MATH 21 at UCSC

## Popular in Mathematics (M)

## Reviews for Linear Algebra

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/07/15

UCSC MATH 21 FALL 2007 Review Questions 3 1 Consider the matrix 1 1 1 1 6 1 2 3 4 2 15 0 A 1 2 3 3 11 3 2 1 0 2 9 4 3 2 1 1 13 1 21 Find the reduced rowechelon rm of A Step 1 subtract a Rl frorn Rj for j 2345 1 1 1 1 6 1 1 1 1 1 6 1 2 3 4 2 15 0 0 1 2 0 3 i2 1 2 3 3 11 3 gt 0 1 2 2 5 2 2 1 0 2 9 4 0 71 i2 0 i3 2 3 2 1 1 13 1 0 71 i2 i2 i5 72 Step 2 add E2 to E4 and add E3 to E5 1 1 1 1 6 1 1 1 1 1 6 1 0 1 2 0 3 i2 0 1 2 0 3 i2 0 1 2 2 5 2 gt 0 1 2 2 5 2 0 71 i2 0 i3 2 0 0 0 0 0 0 0 71 i2 i2 i5 i2 0 0 0 0 0 0 Step 3 subtract E2 from E1 and from E3 1 1 1 1 6 1 1 0 71 1 3 3 0 1 2 0 3 i2 0 1 2 0 3 i2 0 1 2 2 5 2 gt 0 0 0 2 2 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Step 4 divide is by 27 and then subtract the result from E1 1 0 71 1 3 3 1 0 71 0 2 1 0 1 2 0 3 72 0 1 2 0 3 i2 0 0 0 2 2 4 gt 0 0 0 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Fquot Find a basis for the null space of A To nd a basis for the null space of A7 we nd the solutions of the system A 6 and express the basis of the null space in terms of these solutions You can also use the short cut described in class i Drop the 0 rows from rrefA ii For each free variable7 xi insert a new jth row with a 71 in the jth position and 07s everywhere else iii The columns in the resulting rnatrix corresponding to the free variables form a basis for the null space of A 1 In this case7 the free variables are 3 5 and 6 and the algorithm above yields the matrix 1071021 0 2 0 3 72 0 0 71 0 0 0 0 0 0 1 1 2 7 0 0 0 0 71 0 0 0 0 0 0 71 and the columns in positions 37 5 and 6 of this matrix form a basis for kerA Find a basis for the column space of A The pivot columns of A form a basis for the column space of A These are the columns in A that correspond to the columns in rrefA that have pivots7 ie7 columns 17 2 and 4 of A Find a basis for the row space of A The pivot nonzero rows of rrefA form a basis for the row space of A These are the rst three rows of rrefA in this case S 3 P 2 Use your answer to 1b and the record you kept in 1a to describe the set of solutions to the system 111161951 3 2342150952 7 123311336 2102944 5 3211131955 6 6 We rewrite the system as Ax b where A is the matrix of coef cients and b is the column on the right ilf x1 is a solution of this system7 and x0 is a solution of the homogeneous system Ax 07 then x1 x0 is also a solution of the system above7 because Ax1 x0 Axl Ax0 E6E furthermore7 if x1 and x2 are both solutions of the system Ax b then x1 7x2 is a solution of the homogeneous system Ax 07 because Ax1 7x2 Ax1 iAxZ Bi 6 Conclusion Every solution of the system Ax b has the formix x0 x1 where x1 is one cced solution of the system and x0 is any solution of Ax 0 ie x0 6 kerA Since we already know what all the solutions of Ax 6 look like7 from 1b7 above7 all we need to do is nd any one particular solution of the nonhomogeneous system Ax b To do this we need to look at the reduced row echelon form of the augmented matrix Al The reduced row echelon form of Alb is rrefA that we found in 12317 augmented by the column you get by performing the same row operations that we used in part 121 on b 3 3 3 2 1 7 1 1 1 1 6 St 1 3 St 2 3 St 3 2 St 4 1 5 i1 0 161 1331 181 181 101 This means that the augmented matrix of the system is 1 0 71 0 2 1 1 0 1 2 0 3 i2 1 0 0 0 1 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Setting the free variables 3725 and 26 equal to 0 yields the particular solution 1 1 1 7 0 0 so the general solution to the system has the form 1 i1 2 1 1 2 3 72 i 0 a 1 b 0 0 0 1 0 1 2 0 0 i1 0 lol lol 01 11 3 Let W C R5 be the subspace spanned by the vectors 7 l 3 l l 1 Find a basis for WL The space lVL is the orthogonal complement of lV7 ie7 the set of all vectors in R5 that are orthogonal to all the vectors in W Now7 if X is orthogonal to W1 W2 and W3 the given spanning set of lV7 then 1 2 W1 0 1 HHMWN 2 0 W2 1 and W3 7 71 3 1W1 bW2C 3gt 1i W1 b wz6 W3 In other words lVL is equal to the set of vectors that are simultaneously perpendicular to W1 W2 and W3 If A is the matrix with these three vectors as its rows7 then lVL kerA Thus7 to nd a basis for lVL all we have to do is nd a basis for kerA First we nd the reduced row echelon form of A 1 2 0 1 73 1 0 0 759 239 2 0 1 71 3 gt 0 1 0 79 7259 2 3 72 1 1 0 0 1 19 7199 Following the algorithm described in class and again above we insert the rows 000710 000071 UCSC MATH 21 FALL 2007 Review Questions 1 Solutions 1 The identity 3m274x1 a n0 W m 1 m2 2 am2 2 bx cm 1 x lm2 2 a bm2 b cm 2a c x lm2 2 gives the system of equations below for 071 and c a b 3 b c 74 2a c l The augmented matrix for this system is A OHH HO l HgtOJ 1 1 l 0 2 A Next7 perform Gauss Jordan elimination on 1 1 0 3 1 0 0 1 1 74 3372R1HR3 0 1 1 74 2 0 1 1 0 2 175 1 10 10 71 7 0 1 1 74 gffg 7 0 1 1 4 0 72175 3 2 3 0 0 3 713 10 71 7 10 1 7 01 1 74 gtR3gtR3gt 01 1 74 0 0 3 713 0 0 17 10 71 7 10 0 R1R3HR1 i 01174 ltRRgtHR 010 g 0 0 17 2 1 2 0 0171T So the solution is a 8371 13 and c 7133 2 We want to nd the coefficients of the polynomial x mg bzz oz d satisfying f117 f0 271W 4 and f31 This leads to the system of equations for 717 110 and 1 below 7a b 7 c d l d 2 8a4b2cd 4 27a9b3cd l 1 The augmented matrix and its reduced row echelon form are 71 1 71 1 1 1 0 0 0 7 O O O 1 2 1 1 1 O 1 O O 7 8 4 2 1 4 G J elimination 0 0 1 0 g L 27 9 3 1 1 O O O 1 2J so the cubic polynomial in question is i 1 3 1 2 5 7 32 32 322 3 The reduced row echelon forms are 2 3 71 1 1 0 0 g A 1 2 4 0 G J elimination 0 1 0 g0 1 4 2 73 0 0 1 7g and 2 3 7 O 13 1 O 2 O O 2 4 8 2 15 1 1 1 O 1 1 O O B 7 2 0 4 2 g G J elimination 0 0 0 1 0 1 O 2 1 5 O O O O 1 4 The systems of equations corresponding to matrices A and B are 22 3y 2 1 A m 2y 42 0 m 4y 22 3 and 2w 32 7y 13 B 2w 42 8y 22 15 39 2w 4y 22 9 w 2y 2 5 Based on the reduced row echelon forms of the matrices that we found in problem 3 we see that system A has the unique solution x 125 y 0725 and 2 0675 and system B has no solutions since the last row of its reduced row echelon form has a pivot in the last column corresponding to the equation 0 1 0 5a The zero vector g is always a solution of a homogeneous system of equations 0 b We know that 07121 ajgyl 07321 0 and 07122 ajgyg 07322 0 for 1 2 3 since 11 I2 yl and yg are both solutions of the homogeneous system in question Now choose any 21 22 two real numbers 04 and and check 1710421 322 aj204y1 6242 aj30421 622 ajiomi ajzozyi 1730421 01715902 172592 173522 0417391951 ajzyi ajszi 50171962 ajzyz 1739322 04 0 B 0 0 UCSC MATH 21 FALL 2007 Review Questions 6 Solutions 1 Compute the determinants of the matrices below 1 2 2 Bl 71372271 detlt1 2 2 3 4 5 4 253241340735172437240 1 0 3 30871572471 J3 4 2 5J J1 4 0 3J 2 0 2 1 2 0 2 1 1 4 0 3 idet 3 4 2 5 swap rows 1 and 3 J3 2 1 7J J3 2 1 7J 1 4 0 3 7 d t 0 8 2 5 subtract multiples of row 1 7 7 e 0 78 2 74 from rows 27 3 and 4 J0 710 1 72J 1 4 0 3 7 d t 0 8 2 5 subtract multiples of row 2 7 7 e 0 0 0 1 from rows 3 and 4 0 0 715 425 1 4 0 3 0 78 2 75 712 det 0 0 715 425 swap rows 3 and 4 0 0 0 1 2 a The characteristic polynomial of D is XD 114 7 714 7 9832 i A2 7 52 1 1 eigenvalues of D are the roots of 26 1 5 and 2 2 The To nd eigenvectors for an eigenvalue of D7 we nd the basis for the space of solutions of D 7 AI 6 7 1 94 798 x i 0 1 1 1 o 1 7 5 J 32 734 y 7 0 The space of solutions of this system is spanned by V1 i 7 so V1 is an eigenvector for 1 7 34 798 x 7 0 1 1 1 o 2 7 2 J 32 794 J J y J 7 J 0 The space of solutions of this system is spanned by V2 3 J 7 so V2 is an eigenvector for A2 1 CO b The characteristic polynomial of E is XEA 7 17 A72 7 A17 A 7 2 74 22 A 417 A 7 A2 A17 A The eigenvalues of E are the roots of XE A1 17 A2 0 and A3 71 To nd eigenvectors for an eigenvalue A of E7 we nd the basis for the space of solutions of E 7 AI U 0 71 1 ac 0 0 A1 1 2 73 1 y 0 The space of solutions of this system is spanned 2 72 0 z 0 1 by V1 1 J 7 so V1 is an eigenvector for A1 1 1 71 1 ac 0 0 A2 0 2 72 1 y 0 The space of solutions of this system is spanned 2 72 1 z 0 1 by V2 1 J 7 so V2 is an eigenvector for A2 0 2 71 1 ac 0 0 A3 71 2 71 1 y 0 The space of solutions of this system is 0 2722 2 0 spanned by V3 1 7 so V3 is an eigenvector for A3 1 c The characteristic polynomial of F is Ape 7 lt3 7 w 7 A The eigenvalues of F are the roots of XF A1 3 and A2 2 Note that A2 has algebraic multiplicity 2 To nd eigenvectors for an eigenvalue A of F7 we nd the basis for the space of solutions of F 7 AI 5 0 712 2 ac 0 0 A1 3 0 71 0 y 0 The space of solutions of this system is 0 0 71 z 0 1 spanned by V1 0 7 so V1 is an eigenvector for A1 0 1 712 2 ac 0 0 A2 2 0 0 0 y 0 The space of solutions of this system is two 0 0 0 z 0 1 0 dimensional7 and is spanned by V2 2 and V3 4 807 V2 and V3 are linearly 1 independent eigenvectors for A2 This means that A2 has geometric multiplicity 2 as well The characteristic polynomial of M is XMA 72 7 A4 7 A 9 A 7 127 so M has one eigenvalue A1 17 that has algebraic multiplicity two The geometric multiplicity of A1 is the dimension of the eigenspace of Alithat is the dimension of the kernel of M 7 AI The space of solutions of 73 9 ac i 0 71 3 y T 0 UCSC MATH 21 FALL 2007 Review Questions 4 Solutions 1 Denote by A the matrix whose columns are the basis vectors in B ie7 10 A01 2 0 L0 1 1 0 01 00 2 3 Finding the coordinate vector E of a vector Y in R4 with respect to the basis 3 means solving the system AE Y Since we have to do this for three different vectors in this problem7 it makes sense to nd the inverse of A 1 0 0 1 0 0 1 l 1 2 so A 1 ONO 1 0 0 3 1 0 0 1 0 0 0 0 37 0 767 47 0 0 1 0 0 0 GausseJordan 0 1 1 0 0 0 0 1 l0 0 17 27 717 1 0 0 727 37 27 717 727 17 0 0 37 17 27 717 0 0 0 0 1 0 767 727 37 27 7 0 1 47 417 727 17 j Now7 if we denote by 6162 and 63 the coordinate vectors of Y17 Y2 and Y3 with respect to B then we have 2 a Let Y1 and Y2 be vector TaY1 big 37 17 27 611471Y1 6 7 7 727 37 47 717 727 37 17 27 521471Y2 0 1 0 767 727 37 47 717 727 37 17 27 531471Y3 7 0 1 0 767 727 37 47 717 727 aTY1 bTYz so T is a linear transformation b Note that E1 and g are linearly independent check7 so if Y l Y 2 2 5 then both Y H1 and Y g must be zero This means that Kaay eieT 6 eiv 10mm 1 717 0 27 17 717 0 27 17 717 0 27 17 1 l 2 1 0 3 4 3 2 111 171 47 1 787 107 187 7 3 7227 107 737 111 quot 67 107 s in R4 and let 711 6 R be scalars Then aY1 big 1 1 aY1 big 2 2 aY1 1 wig 1 1 aY1 2 Y2 2 2 aY1 1 1 wig 1 1 aY1 2 2 wig 2 2 IKE 1 1 Y1 2 2l b 2 1 1 Y2 2 2l Y 20 O This means that KerT is the set of solutions of the homogeneous system 1 2 0 1 0 1 1 2 x3 7 10241 A basis for this system is found by using G J elimination to reduce the matrix on the left into reduced row echelon form row 2 minus 0 2 1 2 0 1 73 0 1 1 2 0 1 1 2 7 and modifying the nonpivot columns 3 and 4 to obtain the basis vectors 72 73 Y1 ii and Y2 2 l 01 1711 of KerT On to the image of T First observe that every vector in lmT is a linear combination of E1 and g by de nition The only question is is every linear combination of H1 and g included in lmT In other words given a b E R is there a vector x E R4 such that 2 times row 1 Y 1a and Y 2b7 Finding such a vector x amounts to solving the system U where U is the 4 x 2 matrix whose columns are H1 and g Since the vectors H1 and g are linearly independent this system always has a unique solution why and so we can always nd such a vector Y In conclusion lmT span 1 2 The matrix of T with respect to the basis 3 of problem 1 BT is related to the standard matrix of T ie T s matrix with respect to the standard basis of R4 by the relation BT AilMTA where A is the matrix in problem 1 and the standard matrix of T MT is the matrix whose columns are TE1TE2TE3 and TE4 See section 43 Now a direct computation gives 1 2 0 1 2 5 1 4 MT 0 1 1 2 7 1 4 2 5 and using the matrix A and its inverse that we have already computed yayl we nd that 37 17 27 717 1 2 0 1 1 0 0 1 B i 0 1 0 0 2 5 1 4 0 1 0 0 T T 767 727 37 27 0 1 1 2 2 0 1 0 47 717 727 17 1 4 2 5 0 1 2 3 67 157 137 227 7 4 9 9 14 T 27 797 97 727 17 67 17 27 It is possible that the nal answer is off 7 I did it in my head sort of Please check 3 Let A Z Z and let 73A be the space of all polynomials in A 73Aa0Ia1AagA2anA ozjCR for j012 where I is the 2 x 2 identity matrix a F7 P 041 a d and 040 7detA 7ad 0 Note a d is called the trace of A It follows from part a that A2 is in spanA I so spanA2 A C spanA I But A3 A A2 041142 1 040A so A3 C spanA2 A C spanA I from which it follows that A3 C spanA I And so on le suppose that Ak C spanA I so that Ak 31A 1 BOI then 14k A Ak 145114 501 51142 30147 so A1111 6 spanA2 A C spanA I This means that if Ak C spanA I then so is Ak1 and and this implies that Ak C spanA I for all k And that implies that fA BOI 1 31A 1 kAk C spanA I for any polynomial fz 30 lm 1 kmk So all of73A is spanned by A and I Since 73A is at most 2 dimensional it is exactly 2 dimensional if and only if A is not a multiple of I 4 For integers k 2 0 let fkz be the polynomials de ned as follows f0z 1 f1z m and in general for k 2 1 mz71m72m7k139 Mg k where for a positive integer n n 71 71 71 71 7 2 a b P 2 1 f2m m2 7 z m3 7 z2 m and f4m 714 7 if 1 m2 7 ix Since we know that dimPn n 1 and since f0z fnz are n 1 elements in P it is enough to show that f0z are linearly independent To do this we rst observe that if m and k are integers with 0 g m lt k then fk m O and also fkk 1 Now suppose that 00131 0 are scalars satisfying 996 00f095 01f1 1an 0 ForallzCR Weneedtoshowthat0001cn0 Since gz 0 for all x then in particular 0 90 COMO 01f10 Cnfn0 007 because f00 1 and fk0 0 for k gt 0 Next it follows that 0 91 Cif11 62f21 39 39 39 Cnfn1 017 1 and fk1 0 for k gt 1 Likewise it follows that 0 92 02152 03132 CnMQ 027 because f22 1 and fk2 0 for k gt 2 Continuing in this way it follows that ck 0 for 0 g k g n which implies that f0z are linearly independent We want to nd the scalars co 01 02 Cg such that 00f095 c1f1ltzgt c2f2ltzgt csfsltzgt 7 ms z x 17 M for all m For this problem it is relatively simple to do this one coordinate at a time either by using the answer to part a or as below First since f0 1 1 Cof00 Cif10 62f20 63f30 007 because f00 1 and f10 f20 f30 0 because f1 1 UCSC MATH 21 FALL 2007 Review Questions 2 Solutions 1 951 2902 i T R2 7 R3 where T 1 2 7 3x1 Apply T to the standard unit vectors in 2 21 52 R2 to nd the columns of AT 1 2 AT 73 1 2 5 m1 m2 7 m1 ii S R3 7 R3 where S M x3 7 x2 Apply S to the standard unit vectors in R3 903 901 i 903 to nd the columns of As 71 1 0 a a l l o l H H iii R R2 7 R2 where R is rotation counter clockwise by 7r3 radians Use the formulas in Section 22 12 7112 39 2 12 iv PL R2 7 R2 where PL is the projection onto the line L with equation y 3z4 Use the AR cos7r3 7s1n7r3 sin7r3 cos7r3 formulas in Section 22 the vector u is a unit vector spanning L so APL 048 036 Note to nd i I solved the pair of equations 742 31414 and u 14 1 V BL R2 7 R2 where BL is re ection in the line L with equation y 12z5 To nd the matrix of this transformation we need to nd a unit vector in the direction of L eg 15201132 Next we apply the transformation x 7 2x 7 i see page 60 to the standard unit vectors to obtain the matrix 7119169 120169 120169 119169 39 7 064 048 ARL Vi PV R3 7 R3 where PV is projection onto the plane V with equation 2x 7 3y 1 62 0 First note that the plane V is perpendicular to the vector V0 7 this means that 6 every vector in V is perpendicular to V0 The magnitude of V0 is 7 checkl so g V0 is a unit vector that is perpendicular to V As in the 2 dimensional case the projection onto the line L spanned by Hg is given by projLY Y 0 0 1 It follows see Figure 7 on page 61 that the projection onto V is given by projVY x7 proj i x7 x 0 0 To nd the matrix of projV we apply the transformation to the three standard unit vectors in R3 to nd the three columns of the matrix 1 37 67 7127 projVe1 projVe2 projVE3 67 727 187 1 1 7127 187 7297 2 i T is not invertible since T maps R2 to R3 Remember only maps from R to itself might be invertible ii The reduced row echelon form of As is 1 0 i1 0 1 i1 0 0 0 so As and consequently S is not invertible iii The inverse of counter clockwise rotation by 7r3 is counter clockwise rotation by 77r3 so the matrix of the inverse in this case is AA 7 cos 77r3 7 sin 77173 R T sin77r3 cos77r3 7 12 2 5 1452 12 1 39 iv Projection onto a line in R2 is never invertible because it is never 1 1 Suppose that u is a unit vector parallel to L and let i be a unit vector that is orthogonal to n Then you should check that for any scalar a E R 7rL ail u Alternatively you can check that the reduced row echelon form of the matrix of the projection in 1iv is 1 0 0 0 V Re ection in a line is always invertible because if you re ect twice you return to where you started In other words the matrix in 1V is its own inverse Check Vi Projection onto a plane in R3 is not invertible You can show this in general by showing that projection is not 1 1 In the speci c case at hand you should check that the reduced row echelon form of the matrix in 1Vi is not the identity 3 Consider the matrices 1741 07111 33 A 0 371B10711andC 231 2710 11071 how 1 AB BC CA and CB are de ned The others are not ii 73 0 5 74 5 1 2 71 i AB 2 7173 4 BC 5 73 1 CA 4 0 1 7172 3 1 0 2 0 6 717 4 1 72 2 2 71 71 3 CB 4 71 71 4 173 4 3 iii 715 15 72 ABC 15 711 3 ABC 5 5 3 4 45 3A2 7A 71 7 7A5 7 3A2 7 7A In 7 A7A4 7 3A 7 71 In So A l 7A4 7 3A 7 71 5 If B is not invertible7 then we are done If B is invertible and x1 34 5 then Bil 34 6 But ABx1 ABx1 5 since AB 07 so A cannot be invertible 6 a The transformation Teye that the eye performs is given by R G B R 3 I Teye G R 7 G L B R G S B 7 7 l 2 1 Applying this transformation the the standard unit vectors in R3 gives the matrix 1 3 13 13 7 1 71 0 712 712 1 b The transformation Ty performed by the yellow lens is given by R R Ty G G B 0 The matrix of this transformation with respect to the standard unit vectors of R3 is 1 0 0 A 0 1 0 0 0 0 c The matrix of the composite transformation Tags 0 Ty is given by the product of the respective matrices 13 13 13 1 0 0 13 13 0 PA 1 71 0 0 1 0 1 71 712 712 1 0 0 0 712 712 0 d If we denote the the original l L S coordinates of light before sunglasses by the vector V0 and the R G B coordinates of light by the vector x then we have the relation V0 P X7 where P is the matrix in part a Next7 if we denote the new l L S coordinates of light after sunglasses by the vector V1 then V1 Y UCSC Equot MATH 21 FALL 2007 Review Questions 5 Solutions This amounts to nding a basis for kerAT7 where A is the matrix whose columns are the given spanning vectors for V We do this the usual way reduce AT to reduced row ecehlon form7 and modify the nonpivot columns to get the basis of the kernel 1 0 23 53 ATi 1 2 0 3 GaussJordan 0 1713 23 39 2114 The nonpivot columns are columns 3 and 4 so we insert 0 0 7 1 O and 0 0 0 7 1 as the third and fourth rows respectively7 and the third and fourth columns of the modi ed matrix give the required basis for Vi 23 53 713 23 71 and 0 0 71 To do this7 we perform the Gram Schmidt procedure to the given basis of V7 1 2 7 2 7 1 X1 0 and X2 1 7 3 4 to obtain the orthonormal basis H1 and g 7 fl 1 7 St 1 7 7 ep 111 HEM mxl so 1 7 1 2 1 7 M14 0 3 Step 2 u i2 7 Y2 1u1 i2 7 gm so 6 7 1 79 u 7 7 4 7 i 7 7 Step 3 112 7 g muz so 3 The matrix of the projection is UUT where U is the matrix whose columns are the vectors H1 and g from the previous problem UUT L 1777 72620 2058 1215 m 82 182 182 182 182 2 63 1 2 0 3 72620 4021 73087 71686 m 4182 414 414 m 182 182 182 182 0 49 42 763 49 28 2058 73087 2401 1372 4182 182 182 182 182 182 182 182 182 3 28 1215 71686 1372 901 TM 7182 182 182 182 182 Comment The products AAT and ATA are always symmetric and using this observation can save time in computing the products Namely the entries below the main diagonal are equal to the corresponding ones above the main diagonal 4 The rank of a matrix is equal to the dimension of its image among other things Since the matrix in problem 3 is the matrix of the projection onto V the subspace from problem 1 its image is equal to V and so its rank is equal to dimV 2 5 a True F7 False P False 9 True SD False 139quot True If U is an n X n orthogonal matrix and Y1Yn is an orthonormal basis of R then Uxh U7Ln is also an orthonormal basis of R Since U preserves angles between vectors and lengths it follows that ijij1 forj1n and ijka0 forjy k This means that Ux1 an is an orthonormal set of vectors and so it is a linearly independent set of vectors and since there are n of them they form an orthonormal basis of 71 If U is an n X n orthogonal matrix and V1 Vm is an orthonormal basis of the subspace V C R then UV1 UVm is also an orthonormal basis of V The set UV1 UVm is certainly still an orthonormal set of vectors but there is no guarantee that this set still spans V since U might take vectors in V out of V For example suppose that V spane1 C R2 and U 1 1 then Uel e2 which is certainly not a basis orthonormal or otherwise of V If A and B are symmetric n X n matrices then AB is symmetric For example 1 7 g 7 4 3 1 2 15 10 39 If A is an n X in matrix then AAT is a symmetric matrix AATT ATTAT AAT If B is an n X in matrix whose columns form an orthonormal set then BBT is an orthogonal matrix If m n then B is an orthogonal matrix and BBT I which is certainly an orthogonal matrix But in general ie m lt n fewer columns than rows BB is not orthogonal For example the matrix UUT from problem 3 is not an orthogonal matrix because its rank is 2 so it is not invertible but orthogonal matrices are invertible If A is an n X in matrix and kerA 6 then ATA is invertible ATA is a square in X m matrix and kerATA kerA 5 so ATA is invertible

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.