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Dartmouth Linear Algebra Final

# Dartmouth Linear Algebra Final - Linear Algebra Final Exam...

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Linear Algebra Final Exam 1:00–3:00, Sunday, June 2 Bradley 102 1 Let T : R 3 -→ R 3 be a linear transformation with the property that T T T = 0 (we’ll refer to T T T as T 3 for the rest of this problem). (a) What exactly does this mean? That is, what is the practical upshot when it comes to plugging in vectors to T 3 ? Solution It means that T 3 ( v ) = 0 for every vector v R 3 . (b) Suppose that x R 3 is such that T 2 ( x ) = T ( T ( x )) 6 = 0 . If z = cT 2 ( x ), then what is T ( z )? Let y = c 1 T ( x ) + c 2 T 2 ( x ). What is T 2 ( y )? Solution Well, T ( z ) = T ( cT 2 ( x )) = cT 3 ( x ) = 0 and T 2 ( y ) = T 2 ( c 1 T ( x ) + c 2 T 2 ( x )) = c 1 T 3 ( x ) + c 2 T 3 ( T ( x )) = 0 . (c) Let’s say b 1 = x b 2 = T ( x ) , and b 3 = T 2 ( x ) . Show that b 1 is not a linear combination of b 2 and b 3 . Solution From what we’ve seen in (b), any linear combination y of b 2 and b 3 has the property that T 2 ( y ) = 0 . Since T 2 ( b 1 ) = T 2 ( x ) 6 = 0 , b 1 can’t be a linear combination of b 2 and b 3 . 1

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(d) Explain why the set B = { b 1 , b 2 , b 3 } is a basis for R 3 . ( Hint: Some of the work you’ve already done might help. ) Solution We know that T ( b 2 ) = T 2 ( x ) 6 = 0 and T ( b 3 ) = 0 , so b 2 is not a multiple of b 1 . And we showed in (c) that b 1 is not a linear combination of b 2 and b 3 , so the whole set must be linearly in- dependent. Since we are working in
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