2.1-2.2

# 2.1-2.2 - MATH321 – HOMEWORK SOLUTIONS HOMEWORK#3 Section...

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Unformatted text preview: MATH321 – HOMEWORK SOLUTIONS HOMEWORK #3 Section 2.1: Problems 1, 2, 3, 5, 6, 9, 10, 12,18 21, 25 Section 2.2: Problems 1, 2(a)(b)(c)(d), 3, 4, 9 Krzysztof Galicki Problem 2.1.1 (See Answers to Selected Exercises ). Problem 2.1.2 T is linear because T a 1 a 2 a 3 = parenleftbigg 1 − 1 2 parenrightbigg a 1 a 2 a 3 = parenleftbigg a 1 − a 2 2 a 3 parenrightbigg is given by a matrix multiplication. We see that N ( T ) = { ( t, t, 0) , t ∈ R } and it is one- dimensional. The nullity is one so the rank is 2 and, therefore, T is onto. A basis for N ( T ) is { (1 , 1 , 0) } . A basis for R ( T ) = R 2 is { (1 , 0) , (0 , 1) } . Problem 2.1.3 T is linear because T parenleftbigg a 1 a 2 parenrightbigg = 1 1 2 − 1 parenleftbigg a 1 a 2 parenrightbigg = a 1 + a 2 2 a 1 − a 2 is given by a matrix multiplication. Here N ( T ) is trivial, hence its basis is {∅} . T is one-to-one but not onto (no linear transformation from R 2 to R 3 can be onto). R ( T ) consists of all vectors which have second coordinate equal to zero. A basis for R ( T ) is { (1 , , 1) , (0 , , 1) } , for example. Problem 2.1.4 T ( f ( x )) = xf ( x ) + f ′ ( x ) is linear because T ( f ( x ) + cg ( x )) = x ( f ( x ) + cg ( x )) + f ′ ( x ) + cg ′ ( x ) = = xf ( x ) + f ′ ( x ) + c ( g ( x ) + g ′ ( x )) = T ( f ( x ) + cT ( g ( x )) . Note that explicitly T ( a + bx + cx 2 ) = b + ( a + 2 c ) x + bx 2 + cx 3 . Hence, N ( T ) is trivial with basis {∅} . Nullity is zero so rank must be 3. T is one-to-one but not onto. R ( T ) consists of polynomials whose constant and quadratic terms have the same coefficients....
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2.1-2.2 - MATH321 – HOMEWORK SOLUTIONS HOMEWORK#3 Section...

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