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midterm-solns

# midterm-solns - MATH 110 SOLUTIONS TO MIDTERM LECTURE 1...

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MATH 110 - SOLUTIONS TO MIDTERM LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ 1. Give two equivalent definitions (or characterizations) of each of the following: (a) A list ( v 1 , . . . , v n ) is linearly dependent (b) An operator T ∈ L ( V ) can be represented by an upper-triangular matrix (c) A scalar λ F is an eigenvalue of an operator T ∈ L ( V ) (d) A vector space V is the direct sum of subspaces U and W Solution. Here are some possible answers. (a) One of the v i can be written as a linear combination of the rest; or, there exist scalars a 1 , . . . , a n F , not all zero, such that a 1 v 1 + · · · + a n v n = 0 (b) There exists a basis ( v 1 , . . . , v n ) of V with respect to which M ( T ) is upper-triangular; or, there exists a basis ( v 1 , . . . , v n ) of V such that for each k , span( v 1 , . . . , v k ) is T -invariant (c) There exists a nonzero v V such that Tv = λv ; or, the operator T - λI is not injective (d) V = U + W and U W = { 0 } ; or, for each v V , there is a unique way to write v as v = u + w with u U and w W 2. Determine whether the following statements are true or false. You do not need to provide justifications for your answers. There will be 1 point subtracted for each incorrect response, so do not guess just for the sake of guessing. (a) Any operator on a real vector space has a 1-dimensional invariant subspace. (b) If V = U W and p ( z ) = z 2 , then p ( P U,W ) = P U,W . (c) If v 1 , . . . , v n V , then span( v 1 , . . . , v n ) is n -dimensional. (d) Any operator on a 10-dimensional complex vector space has a 4-dimensional invariant subspace. Solution. Justifications were not required, but we will provide some below. (a) False. This is the same as saying that any operator on a real vector space has an eigenvector, which is not true; for example, take the rotation by 90 operator on R 2 . (b) True. The projection operator P U,W satisfies P 2 U,W = P U,W . (c) False. This is true if and only if ( v 1 , . . . , v n ) is linearly independent; for example, take each v i to be zero — then their span is 0-dimensional. (d) True. This follows from the last homework; explicitly, let ( v 1 , . . . , v 10 ) be a basis of V with respect to which the operator is upper-triangular — then span( v 1 , . . . , v 4 ) is 4-dimensional and invariant. 3. Let n be a positive integer and let U be the subset of P n ( R ) defined by U = { p ( x ) ∈ P n ( R ) | p 00 (1) - p 0 (2) + p (3) = 0 } . (a) Prove that U is a subspace of P n ( R ). (b) Find a linear map T : P n ( R ) R such that null T = U , and use this to compute dim U . (You do not need to prove that the map you write down is in fact linear) Date : July 16, 2009.

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2 GSI: SANTIAGO CA ˜ NEZ Solution. (a) First, the zero polynomial clearly satisfies the defining properties of U , so 0 U . Now, let p ( x ) , q ( x ) U . Then p 00 (1) - p 0 (2) + p (3) = 0 and q 00 (1) - q 0 (2) + q (3) = 0 , so ( p + q ) 00 (1) - ( p + q ) 0 (2)+( p + q )(3) = ( p 00 (1) - p 0 (2)+ p (3))+( q 00 (1) - q 0 (2)+ q (3)) = 0+0 = 0 . Thus ( p + q )( x ) U so U is closed under addition.
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