Given the elow bases for R^2 and the point at the specified coordinate in the standard basis as below, (40 points)

Wribreeminsl

Wribreeminsl

Answered question

2020-10-18

Given the elow bases for R2 and the point at the specified coordinate in the standard basis as below, (40 points)

(B1={(1,0),(0,1)}&

B2=(1,2),(2,1)}(1, 7) = 3(1,2)(2,1)
B2=(1,1),(1,1)(3,7=5(1,1)+2(1,1)
B2=(1,2),(2,1)   (0,3)=2(1,2)2(2,1)
(8,10)=4(1,2)+2(2,1)
B2 = (1, 2), (-2, 1) (0, 5) =
(1, 7) =

a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points)

 

Answer & Explanation

lamusesamuset

lamusesamuset

Skilled2020-10-19Added 93 answers

B1={(1,0)(0,1)} standard basis R2 Alternate basis B2={(1,2)(2,1)},B2={(1,1),(1,1)}B2={(1,2)(2,1)},andB2{(1,2)(2,1)}) a. (0,5)=a(1,2)+b(2,1)=(a2b,2a+b)
a2b=0 and 2a + b = 5
a=2b7b+b=55b=5b=1thena=2,
(0,5)=1(1,2)+2(2,1)(1,7)=(1,2)+b(2,1)=(a2b,2a+b)a2b=1 and 2a+b=7
2aab=25b=5b=1a=3(1,7)=3(1,2)+1(2,1)
b. B2={(1,2)(2,1)} is orthogonal since [12][21]=22=0
B2={(1,1)(1,1)}
is orthogonal since ([11][11])=11=0B2={(1,2)(2,1)} is not orthogonal since [12][21]=2+2=70.B2={(1,2)(2,1)}is orthogonal c. The vector UR2 is normal of null=1B at every vector in B2 is not normal. They are unit vectors X All bases B2 are not normal X. d. The transition matrix from B to BisTB2B=[[b1]B2[b2]B2]
b1=[10],b2=[01],B={b1,b2}
standard basis B2={(1,2)(2,1)}Then (10)=a(12)+b(21)=(10)=(a+2b2ab)a=15b=25
Therefore

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