20D - Homework Assignment 5 - UCSD …

1 Brian Bowers (TA for Hui Sun)MATH 20 DHomework Assignment 5 November 8, 201320D - Homework Assignment 5 First, I present the list of all matrix row operations. We use combinations of these steps to row reducematrices. Note that I will userito represent theith row of the matrix.(1)Swap:We are allowed to swap two entire rows of a matrix. [I will notate swapping rowsriandrjby ri rj .](2)Multiply:We are allowed to multiply any one row of a matrix by any nonzero scalar expression (like1, ,et t2+ 7, etc.) [I will notate multiplying rowribykas ri=kri .](3)Combine:Given two rowsriandrj, we are allowed to replaceriby (ri+krj) for any scalark. [I willnotate this as ri=ri+krj .]For our purposes, row reduction is the process of trying to get a matrix to have 1s along the diagonal and have 0s above and below the diagonal.

2 In other words, we re trying to get our matrix to look like this: 1 ? ? ? 0 1 ? ? 0 0 1 ? .. and then like this: 1 0 0 ? 0 1 0 ? 0 0 1 ? ..Note that we can t always get our matrix to look exactly like the ones above, but our goal is to always getas close as #13,14In each problem, determine whether the members of the given set of vectors are linearlyindependent for < t < . If they are linearly dependent, find the linear relation among them. [Thevectors are written as row vectors to save space.](13) x(1)(t) = (e t,2e t),x(2)(t) = (e t,e t),x(3)(t) = (3e t,0).(14) x(1)(t) = (2 sint,sint),x(2)(t) = (sint,2 sint).A set of (column) vectors{x(1),x(2).}

3 ,x(n)}(each withkentries) is considered to be linearly dependentif and only if there exist scalars{c1,c2,..,cn}(where at least one of these constants is not 0) such thatc1x(1)+c2x(2)+..+cnx(n)= could rewrite the above equation in matrix form as(x(1)x(2) x(n)) = ,where the 0 vector at the end of the equation the above setup as motivation, we construct the following steps to determine ifnvectors (eachwithkentries) are linearly independent or linearly dependent:1(1)Augment:Write the augmented matrix below:(x(1)x(2) x(n)0),where0represents the vector withkentries, all of which are 0.(2)Row Reduce:Row reduce the above matrix.(3)Conclude:There are two conclusions to be reached here:(i) If the row reduction gives us a matrix that looks like 1 0 0 000 1 0 000 0 1 0 0 10 ,then we conclude that the vectors are linearly independent.

4 (ii) Otherwise, we conclude that the vectors are linearly dependent. If we re deciding between depen-dent/independent, then we re , sometimes we re asked to find a linear relation among the vectors if they are lin-early dependent. In this case, each row of the row-reduced matrix gives us an equation (involvingc1,c2,..,cn) that we can use to find such a linear relation among the vectors. For example, thematrix(1 0200 1 10)gives us the system of equations{1c1+ 0c2+ 2c3= 00c1+ 1c2 1c3= 0We can solve the equations as{c1= 2c3c2= all that remains is to choose a value forc3. Recall that we want at least one of thec1,c2,orc3to be nonzero. So, as a general rule, when I get to choose a value for a constant, I just choose1.}}

5 (Note that choosingc3= 1 means thatc3is nonzero!) If we choosec3= 1, we getc1= 2 andc2= 1. Looking way back to our original equation, we can fill in our constants:c1x(1)+c2x(2)+..+cnx(n)= 0= 1x(1) 2x(2)+ 1x(n)= 0 Okay - let s do the problems (13)We follow the steps from above to make an augmented matrix and row reduce it:(e te t3e t02e te t00)r1=etr1 (11302e te t00)r2=r2 2e tr1 (11300 e t 6e t0)r2= etr2 (1 1 300 1 60)r1=r1 r2 (1 0 300 160)Since the left part of the matrix is not an identity matrix, this is case (ii) from our steps above. Inother words, we determined that the vectors are linearly dependent. Moreover, we can read off twoequations from the matrix we got:{1c1+ 0c2 3c3= 00c1+ 1c2+ 6c3= 0= {c1= 3c3c2= 6c3[Next, choosec3= 1]= c1= 3c2= 6c3= 1 Thus, we can write the following linear relation:3x(1) 6x(2)+x(3)= 0.}}

6 (14)Once again, we follow the steps from above to make an augmented matrix and row reduce it:(2 sintsint0sint2 sint0)r1=12 sintr1 (11/20sint2 sint0)r2=r2 (sint)r1 (11/200 (3 sint)/20)r2=23 sintr2 (1 1/20010)r1=r1 12r2 (1 000 10)This is case (i) from our steps above because the left part of the matrix is an identity matrix. In otherwords, the vectors are linearly #16,17,18,21In each problem, find all eigenvalues and eigenvectors of the given matrix.(16)(5 131)(17)(3 24 1)(18)( 211 2)(21)( 3 3/4 51)Quick reminder:det(a bc d)=ad find the eigenvalues of a matrixA, we simply solve the equationdet(A I) =0for . [Note thatA Iis the same asA, except we subtract from each entry along the diagonal ofA.]

7 ] Wename these solutions 1, 2, Next, to find the eigenvector corresponding to i(we call this eigenvectorx(1)), we row reduce the augmented matrix(A iI0)and use the resulting matrix to spit out a system of equations, just like we did to find linear s do some problems - I think examples are clearer for this than general (16)We solve the equationdet(A I) = 0:det(A I) = 0= det(5 131 )= 0= (5 )(1 ) ( 1)(3) = 0= 2 6 + 8 = 0= ( 4)( 2) = 0= = 4,2 Thus, we have the eigenvalues 1= 4 and 2= 2. Now we row reduce the appropriate augmentedmatrices.(i) 1= 4:(5 4 1031 40)=(1 103 30)r2=r2 3r1 (1 10000)Thus, the first row gives us the equationx1 x2= 0, which we can solve to getx1=x2. I willselectx2= 1, which yieldsx1= 1.

8 Thus, we get the eigenvectorx(1)=(x1x2)=(11).(ii) 2= 2:(5 2 1031 20)=(3 103 10)r2=r2 1r1 (3 10000)Thus, the first row gives us the equation 3x1 x2= 0, which we can solve to getx2= 3x1. I willselectx1= 1, which yieldsx2= 3. Thus, we get the eigenvectorx(2)=(x1x2)=(13).5(17)We solve the equationdet(A I) = 0:det(A I) = 0= det(3 24 1 )= 0= (3 )( 1 ) ( 2)(4) = 0= 2 2 + 5 = 0= = ( 2) ( 2)2 4(1)(5)2(1)= = 1 2iThus, we have the eigenvalues 1= 1 + 2iand 2= 1 2i. Now we row reduce the appropriateaugmented matrices.(i) 1= 1 + 2i:(3 (1 + 2i) 204 1 (1 + 2i)0)=(2 2i 204 2 2i0)r1=12 2ir1 (1 22 2i04 2 2i0)[We need to rationalize that denominator]=(1 22 2i 2+2i2+2i04 2 2i0)=(1 1 i204 2 2i0)r2=r2 4r1 (1 1 i20000)Thus, the first row gives us the equationx1+ 1 i2x2= 0, which we can solve to getx1=1+ will selectx2= 1, which yieldsx1=1+i2.

9 Thus, we get the eigenvectorx(1)=(x1x2)=(1+i21).(ii) 2= 1 2i:(3 (1 2i) 204 1 (1 2i)0)=(2 + 2i 204 2 + 2i0)r1=12+2ir1 (1 22+2i04 2 + 2i0)[We need to rationalize that denominator]=(1 22+2i 2 2i2 2i04 2 + 2i0)=(1 1+i204 2 + 2i0)r2=r2 4r1 (1 1+i20000)Thus, the first row gives us the equationx1+ 1+i2x2= 0, which we can solve to getx1=1 will selectx2= 1, which yieldsx1=1 i2. Thus, we get the eigenvectorx(2)=(x1x2)=(1 i21).6(18)We solve the equationdet(A I) = 0:det(A I) = 0= det( 2 11 2 )= 0= ( 2 )( 2 ) (1)(1) = 0= 2+ 4 + 3 = 0= ( + 3)( + 1) = 0= = 3, 1 Thus, we have the eigenvalues 1= 3 and 2= 1. Now we row reduce the appropriate augmentedmatrices.(i) 1= 3:( 2 + 3101 2 + 30)=(1 101 10)r2=r2 r1 (1 100 00)Thus, the first row gives us the equationx1+x2= 0, which we can solve to getx1= x2.

10 I willselectx2= 1, which yieldsx1= 1. Thus, we get the eigenvectorx(1)=(x1x2)=( 11).(ii) 2= 1:( 2 + 1101 2 + 10)=( 1101 10)r1=( 1)r1 (1 101 10)r2=r2 r1 (1 10000)Thus, the first row gives us the equationx1 x2= 0, which we can solve to getx1=x2. I willselectx2= 1, which yieldsx1= 1. Thus, we get the eigenvectorx(2)=(x1x2)=(11).7(21)We solve the equationdet(A I) = 0:det(A I) = 0= det( 3 3/4 51 )= 0= ( 3 )(1 ) (3/4)( 5) = 0= 2+ 2 + 3/4 = 0= ( + 3/2)( + 1/2) = 0= = 3/2, 1/2 Thus, we have the eigenvalues 1= 3/2 and 2= 1/2. Now we row reduce the appropriateaugmented matrices.(i) 1= 3/2:( 3 + 3/23/40 51 + 3/20)=( 3/2 3/40 55/20)r1= 23r1 (1 1/20 55/20)r2=r2+5r1 (1 1/20000)Thus, the first row gives us the equationx1 12x2= 0, which we can solve to getx1=12x2.