Integrating Total Physical Response Strategy in a Level I Spanish ...
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- cours - matière : spanish
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MATH 203, Section 001 Fall, 2011 Exam 4
Name
Solutions
1. LetP1denote the vector space of polynomials of degree at most 1.Let a 3 T:P1→Rbe given byT(ax+b) =ab. DetermineifTis a linear a transformation. Giveclearreasons for your answer. 10 2 T(5(2x=+ 3))T(10x+ 15)= 150, but5T(2x6 == 5+ 3) 10 2 10 306=T(5(2x+ 3)), soTis not a linear transformation. 10 (Of course, there are many other examples that work.)
Tofmrarsn.ntaoiiseartalin
× T.onorsftimaraennartonsiilat
2 2. SupposethatBis the basis forRsuch that the change of coordinate 4 1 matrixPBfromBto the standard basis is given byPB=.If 3 0 2 [v~]Bfind= ,~v. −3 4 12 5 v~=PB[~v]B= =. 3 0−3 6 Answer: 5 6
2 3. SupposethatBis the basis forRsuch that the change of coordinate 4 1 matrixPBfromBto the standard basis is given byPB=. 3 0 Find the change of basis matrix from the standard basis toB.
−1 The change of basis matrix from the standard basis toBisP= B 1 0−110−1 0 1 3 ( )= () =−4. det(PB)−3 −3 4−3 41 3
Answer: 1 0 3 −4 1 3
1 1 2 4. Letthe basesBandCforRbe given byB={,}and 1−1 2 1 P C={,}. Findthe change of basis matrixC←B. 1 0 1 1 P P The matrixC←Bis given byC←B=, so we must 1−1 C C 1 21 12 solve the equations=x+yand=x+ 1 10−1 1 1 y. Wecan do both of these simultaneously if we row-reduce 0 the matrix whose first two columns are the elements ofCand whose second two columns are the elements ofB. [Ofcourse, it is fine to do these systems separately.] 2 1 11 10 1−0 11 1−1 ∼ ∼. 1 0 1−1 21 01 11−1 3 1−1 P Therefore,C←B=. −1 3
Answer: 1−1 −1 3
1 1 1 3 5. Suppose thatBis the basis forRgiven by{1,1,0} 1 0 0 3P andCis a basis forRsuch that the change of basis matrixC←Bis 1 0 23 1 1−Find 2.1 . 3−2 11 C 3 11 13 1 Since+ 1 + 02 = 1,12 =. Therefore, 1 10 01 1 B 3 10 21 3 2 =1 1−1 1= 1. 1 3−1 22 1 C
Answer: 3 1 2 1 01 6. Oneeigenvalue of the matrixA= 1−1 0Find an eigen-is 2. 1 01 vector corresponding to the eigenvalue 2. x We want a non-zero vector~v=ysuch thatA~v= 2v~, that z is, we want a non-trivial solution of the homogeneous system ~ Av~−2~v= 0find such a solution, we row-reduce the coefficient. To −1 0 1−1 01 10−1 1 matrix.1−3 0∼0−3 1∼0 1−.Therefore, 3 1 0−0 01 000 0 z zis free, withx=zandy=one eigenvector is. Hence, 3 3 1. 3
Answer: 3 1 3
0−2 0 7. Findthe characteristic polynomialp(λ) of the matrixA= 23−1 . 0 1 1 λ2 0 p(λ) =det(λI−A) =det(−2λ−) =3 1λ[(λ−3)(λ−1) + 0−1λ−1 3 2 1] + 2[2(λ−1)] =λ−4λ+ 8λ−4.
3 2 p(λ) =:λ−4λ+ 8λ−4
3−2 8. LetA= .Find all eigenvalues ofA. −1 4
The characteristic polynomial ofAisp(λ) =det(λI−A) = λ−2 2 2 det=λ+ 7λ= (+ 10λ−2)(λ−5). Therefore, 1λ−4 the eigenvalues are2and5.
Answer: 2and5
3 5 01 2−11 2 9. LetA= .Then the characteristic polynomial ofAis 1 2−1 0 6 3 02 4 32 54 3 2 p(λ) =λ−3λ−26λ−15λ+7. FindA−3A−26A−15A+A. [Hint: The easiest way to do this is to use the Cayley-Hamilton Theorem.]
4 32 By the Cayley-Hamilton Theorem,p(A) =A−3A−26A−15A+ 7I= 0. 5 43 2 Long division gives thatλ−3λ−26λ−15λ+λ=λp(λ)−6λ, 5 43 2 soA−3A−26A−15A+A=AP(A)−6A=A∙0−6A= −18−30 0−6 −12 6−12−6 . −6−12 6 0 −36−18 0−12
Answer: −18−30 0−6 −12 6−12−6 −6−12 6 0 −36−18 0−12 3 4 10. LetAThen the eigenvalues of= .Aare−1 and 5, with 2 1 1 2 corresponding eigenvectorsand .Find a matrixPsuch −1 1 −1 thatP APis a diagonal matrixDand findD.
We takePto bePBwhereBis a basis consisting of eigenvectors 1 2 ofAlinearly independent eigenvectors,. Sinceand are −1 1 1 2 we can takeP=matrix. TheDhas the corresponding −1 1 −1 0 eigenvalues on the main diagonal, soD=. 0 5
1 2 P=: −1 1
−1 0 D=: 0 5
