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math270 assin

# math270 assin - (5 pts Let U:= x ∈ R 3 x 1 2 x 2-x 3 = 0...

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Assigment for Applied Linear Algebra - Math 270 January 28, 2011 The four exercises amount to 15 points. Obtaining 13 points out of 15 guarantees full marks. The deadline is set to February 14. (Please see details in the Outline of the course, on WebCT.) Problem 1 (2 pts.) Show that u 1 = (1 , 1 , 1) T , u 2 = (1 , - 1 , 0) T , u 3 = (1 , 1 , - 2) T , form a basis of R 3 . Write v = (1 , - 2 , 5) in the coordinates of the basis { u 1 ,u 2 ,u 3 } . Problem 2 (3 pts.) Recall the deﬁnition of linear subspace of a vector space. Which of these subsets are linear subspaces ? Explain your answer. (i) { (1 , 1 , 1) } ⊂ R 3 , (ii) { (0 , 0 , 0) } ⊂ R 3 , (iii) { ( x 1 ,x 2 ) : x 1 = x 2 2 } ⊂ R 2 , (iv) { ( x 1 ,x 2 ,x 3 ) : x 1 + 2 x 3 = 0 } ⊂ R 3 , (v) { ( x 1 ,x 2 ) : x 3 1 = x 3 2 } ⊂ R 2 , (vi) { ( x 1 ,x 2 ,x 3 ,x 4 ) : x 1 0 } ⊂ R 4 . Problem 3
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Unformatted text preview: (5 pts.) Let U := { x ∈ R 3 : x 1 + 2 x 2-x 3 = 0 } , and V := { x ∈ R 3 : x 1-2 x 2 + 2 x 3 = 0 } . Find a basis for ( i ) U, ( ii ) V, ( iii ) U ∩ V, ( iv ) span U ∪ V. Problem 4 (5 pts.) Recall a deﬁnition of scalar product on complex numbers. Let A = ± 3 1 1 2 ² . Prove that the product deﬁned as u · v := u T A ¯ v = 2 X i,j =1 u i a ij ¯ v j is a scalar product on C 2 , according to the deﬁnition. (At some point, it could be useful to remember that 2( ac ) ≥ -a 2-c 2 for all a,c ∈ R .)...
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