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# tat03 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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Unformatted text preview: T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS Summer School Math2061 2010 Tutorial 3 (week 3) — Solutions 1. a ) Let X = braceleftBigparenleftBig 1 parenrightBig , parenleftBig 2 1 parenrightBig , parenleftBig 3 2 1 parenrightBigbracerightBig . Show that X is a linearly independent subset of R 4 . b ) Let Y = { p 1 ( x ) , p 2 ( x ) , p 3 ( x ) } , where p 1 ( x ) = 1 , p 2 ( x ) = 2 x − 1 and p 3 ( x ) = (2 x − 1) 2 . Show that Y is a linearly independent subset of P 2 . Solution a ) Suppose a parenleftBig 1 parenrightBig + b parenleftBig 2 1 parenrightBig + c parenleftBig 3 2 1 parenrightBig = parenleftBig parenrightBig , for some a, b, c ∈ R . Then a + 2 b + 3 c = 0 b + 2 c = 0 c = 0 There is only one solution of this linear system, namely a = b = c = 0 (this is obvious if you write out the augmented matrix for this system of equations). Therefore X is a linearly independent set in R 4 . b ) Suppose a 1 + b (2 x − 1) + c (2 x − 1) 2 = 0 , for all x . Then ( a − b + c ) + (2 b + 4 c ) x + 4 cx 2 = 0 , for all x. Since { 1 , x, x 2 } is a linearly independent set, each of these coefficients must be zero. Comparing the coefficients of x , x and x 2 on the left and right hand sides of this equation we see that we need to solve the following system of linear equations: a − b + c = 0 2 b + 4 c = 0 4 c = 0 Using Gaussian elimination, the only solution is a = b = c = 0 . Hence, X is a linearly independent subset of P 2 . 2. a ) Let V = Span parenleftBigparenleftBig 1 parenrightBig , parenleftBig 2 1 parenrightBig , parenleftBig 3 2 1 parenrightBigparenrightBig . Determine dim V and whether or not V = R 4 . b ) Let W = Span ( p 1 ( x ) , p 2 ( x ) , p 3 ( x ) ) , where p 1 ( x ) = 1 , p 2 ( x ) = 2 x − 1 and p 3 ( x ) = (2 x − 1) 2 . Determine dim W and whether or not W = P 2 ....
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tat03 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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