260S0708M2 - V × V → R by A | B = x 1 y 1 3 x 2 y 2 5 x...

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Instructor Time Duration Signature : : : : : : Last Name Name ID Number M E T U Department of Mathematics Math 260 Basic Linear Algebra Exam 2 01.08.2008 13: 00 100 minutes 4 QUESTIONS ON 4 PAGES TOTAL 30 POINTS 1 2 3 4 (1) (3 + 3 + 1 = 7 pts.) Let W = { ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ) R 5 : x 1 - 2 x 2 = 0 , x 3 + 4 x 4 = 0 } . (a) Show that W is a subspace of R 5 . (b) Find a basis for W. (c) Find the dimension of W.
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2 (2) (4 + 4 + 1 = 9 pts.) Let V be the space of polynomials with degree at most 3, and W be the subspace of V spanned by the polynomials p 1 = 1 + x - x 2 - x 3 , p 2 = x - x 3 , p 3 = - 2 + 3 x + 2 x 2 - 3 x 3 , p 4 = x 2 . (a) Find a basis for W. (b) Extend the basis you found in part (a) to a basis for V. (c) Extend the basis you obtained in part (b) to a basis for the space of polynomials with degree at most 5.
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3 (3) (3 + 4 = 7 pts.) Let V = R 2 × 2 be the space of 2 × 2 real matrices. Define the function ( ·|· ) :
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Unformatted text preview: V × V → R by ( A | B ) = x 1 y 1 + 3 x 2 y 2 + 5 x 3 y 3 + 7 x 4 y 4 for A = " x 1 x 2 x 3 x 4 # and B = " y 1 y 2 y 3 y 4 # . (a) Show that ( ·|· ) is an inner product in V. (b) Let W be the subspace spanned by " 1 0 0 0 # and " 0 1 1 0 # . Find a basis for the orthogonal complement, W ⊥ , of W. 4 (4) (2 + 5 = 7 pts.) Let V = R 3 with the inner product ( u | v ) = x 1 y 1-3 x 1 y 2-3 x 2 y 1 + 10 x 2 y 2 + x 3 y 3 , for u = ( x 1 ,x 2 ,x 3 ) and v = ( y 1 ,y 2 ,y 3 ) . (a) Show that B = { (1 , , 1) , (2 , 1 ,-1) , (1 , 1 , 1) } is a linearly independent set. (b) Apply Gram-Schmidt orthogonalization process to B to obtain an orthonormal basis for V....
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This note was uploaded on 05/26/2010 for the course MATHEMATIC 260 taught by Professor Uguz during the Spring '10 term at Middle East Technical University.

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260S0708M2 - V × V → R by A | B = x 1 y 1 3 x 2 y 2 5 x...

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