Written Solution to Questions #1,2,5,6,7

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Written Solution to Questions #1,2,5,6,7 Written Solution to Questions #1,2,5,6,7 Written Solution to Questions #1,2,5,6,7 Written Solution to Questions #1,2,5,6,7 Written Solution to Questions #1,2,5,6,7

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11 1. Let U be the subspace of R4 consisting of all vectors x= 12 13 such that 14 X1 + x2 + x3 + x4 = 0 Find a basis for the orthogonal complement, Ut. 2. Suppose A is an n x n symmetric matrix and that x and y are two vectors in R™ such that Ax = 2x and Ay = 3y. Show that xTy = 0. (Note: This illustrates something that is true in general – symmetric matrices have orthogonal eigenvectors.) Given the data set {(-2,0), (–1, 2), (0,3), (1,5), (2,6)}, find the least squares regression line for the data, then graph the points and the line on the same set of coordinate axes. Use the Gram-Schmidt orthonormalization process to transform the given set of vectors into an orthonormal basis for R3. 4. { 2 0 2 5. Consider P4 and P3 (the vector spaces consisting of polynomials of degree 4 or less and degree 3 or less, respectively), and let Tı: P4 → P3 and T2: P3 P4 be defined as follows: T1(P(x)) = p' (2) and T2(p(x)) = [” pct) dt (a) Find the matrices A1 and A2 that represent these transformations with respect to the standard bases for P4 and P3. (b) Compute the products A1 A2 and A2A, and compare these, using what you know about differentiation and integration to describe what tranformations these products represent. 6. Find the matrix representation for the tranformation T: R3 R2 defined by T(x, y, z) = (x - y, y - 2) relative to the basis B = {(1,1,1),(1,1,0), (0,1,1)} for R3 and B' = {(1,2), (1,1)} for R2. 6.) 2 38 7. Let T: M2x2 + M2x2 be the transformation defined by Tabc der shi Iben fi T(A) = AT Find the matrix representation for T relative to the standard basis for M2x2, {[B]: 8:1 1
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Explanation & Answer

attached is my answer

Question 1:
We are given that



 x1 


x 


2
U =  x =   : x1 + x2 + x3 + x4 = 0 
 x3 


 


x
 4


In U , we can re-write x1 = − x2 − x3 − x4 so that x can be expressed by

 x1 
x 
x =  2
 x3 
 
 x4 
 − x2 − x3 − x4 


x2

x=


x3


x4


 −1
 −1
 −1
1
0
0




 x = x2
+x
+x  
0 31 40
 
 
 
0
0
1

  −1  −1  −1 
      
0
0 
 1
Hence, the basis for U is BU =    ,   ,    , and the dimension of U is 3 .
 0   1   0  
  0   0   1  
By definition of orthogonal component, we have

 −1 1 0 0 y = 0

y  U ⊥   −1 0 1 0 y = 0
 −1 0 0 1 y = 0


 y1 
y 
2
Now, if we let y =   where y1 , y2 , y3 , y4  R then from the above matrix equations we have
 y3 
 
 y4 

− y1 + y2 = 0

 − y1 + y3 = 0
− y + y = 0
 1 4
 y1 = y2 = y3 = y4

 y1 
y 
2
which means that for y =    U ⊥ , we have y1 = y2 = y3 = y4 .
 y3 
 
 y4 
Hence,



 y1 


y 


2

U⊥ =  y =
: y1 = y2 = y3 = y4 
 y3 


 


y
 4




1


1...


Anonymous
Awesome! Perfect study aid.

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