math270 assin

math270 assin - (5 pts.) Let U := { x R 3 : x 1 + 2 x 2-x 3...

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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 denition of scalar product on complex numbers. Let A = 3 1 1 2 . Prove that the product dened 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 denition. (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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