B 1 5 3 2 10 1 4 1 5 1 5 Sol By performing the Gaussian elimination we see that

Linear Algebra with Applications (3rd Edition)

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B = 1 5 0 - 3 2 10 1 - 4 - 1 - 5 1 5 . Sol. By performing the Gaussian elimination we see that the pivot columns in B are: column 1 and column 3. Hence, 1 2 - 1 and 0 1 1 form a basis for the column space of B . 4. [ 20 marks] Let V be the plane in R 3 defined by (the set of solutions of) the equation x - y + z = 0. Find the matrix B of the linear transformation T : ( V, B ) ( V, B ), with respect to the basis B = { 0 1 1 , 1 0 - 1 } of V , which describes the orthogonal projection onto the line spanned by the vector 1 1 0 . Sol. A unit direction vector of the line is u = 1 2 1 1 0 . Let v 1 = 0 1 1 and v 2 = 1 0 - 1 . The orthogonal projection onto the line with direction vector u is defined by: T ( x ) = ( x · u ) u , x V. In particular, we have T ( v 1 ) = ( v 1 · u ) u = 1 2 ( 0 1 1 · 1 1 0 ) 1 1 0 = 1 2 1 1 0 = 1 2 v 1 + 1 2 v 2 , [ T ( v 1 )] B = 1 2 1 2 . Similarly, T ( v 2 ) = ( v 2 · u ) u = 1 2 ( 1 0 - 1 · 1 1 0 ) 1 1 0 = 1 2 1 1 0 = 1 2 v 1 + 1 2 v 2 , [ T ( v 2 )] B = 1 2 1 2 . It follows that B = 1 2 1 2 1 2 1 2 .

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