Solution10_121

# Solution10_121 - Solutions to PS 10(Math 121 J Scholtz...

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Solutions to PS 10 (Math 121) J. Scholtz January 6, 2007 Question 1: 6.3/18 Let A be a n × n matrix. Then det( A * ) = det( A t ) = det( A t ) = det( A ). Question 2: 6.3/22c Since this a computational exercise I won’t show the solution, but please do not hesitate to ask me in case you are unsure about anything. Question 3: 6.3/24 Let us use our satndard definition of ’finite membered’ sequences and define the operator: T ( σ )( k ) = i = k σ ( i ) 1. This operator is linear in sums since T ( σ + τ ) = ( σ + τ ) = σ + τ = T ( σ ) + T ( τ ). 2. T ( e n )( k ) = i = k e n ( i ) But this is a series such that it is equal to 1 for any position such that k n . Hence T ( e n ) = n 1 e i . 3. Suppose that T * exists, then e n , T ( e j ) = 1 if n j . On the other hand e n , T ( e j ) = T * e n , e j . But this means that T * e n contains e j as long as j n and hence T * e n is not in our vectorspace. Contradiction. Question 4: 6.4/8 Let T be a normal operator on a finite dimensional complex vector-space and W be its subspace. Show that if W is T -invariant then it is also T * -invariant: Let’s use the hint: We know that since T

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