6.36.46.5 - Extend the given vector to an arbitrary basis...

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Homework 13 - Solutions 6.3 Problem 10 (a) If h T ( x ) ,T ( y ) i = h x, y i ,thentak ing y = x ,weget || T ( x ) || = || x || . (b) If || T ( x ) || = || x || for all x V , then by the polar identity, h T ( x ) ,T ( y ) i = 1 4
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6.4 Problem 9 x N ( T ) T T ( x )=0 0= h T T ( x ) ,x i = h TT ( x ) ,x i since T is normal = h T ( x ) ,T ( x ) i = || T ( x ) || 2 This is if and only if x N ( T ). By the above argument and the fact that orthogonal complement of a subspace is is unique, we get N ( T ) = N ( T ) . By 6.3 Problem 12, N ( T ) = R ( T )a n d N ( T ) = R ( T ∗∗ )= R ( T ) Therefore, R ( T )= R ( T ). 6.4 Problem 10 Expand || T ( x )+ ix || 2 as h T ( x )+ ix.T ( x )+ ix i and verify the identity. Using the identity, we get that x ker( T ) if and only if || x || = 0. Therefore, T iI has a trivial kernel and is invertible. For any invertible linear transformation S on a ±nite-dimensional vector space, ( S 1 ) =( S ) 1 .U s e this with the fact that T is self-adjoint to prove the ±nal assertion.
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Unformatted text preview: Extend the given vector to an arbitrary basis of R 3 and perform Gram-Schmidt, starting with the given vector as the ±rst basis vector. 6.4 Problem 17 Since A is unitary, its columns { v 1 , . . . , v n } form an ONB of C n . Consider the product A ∗ A = I . We can compute the (1 , j ) entry of the product by expressing it as the inner product of v 1 and v j . Using the fact that A is upper triangular, we now deduce that all entries in the ±rst row of A are zero ecept for a 11 . A similar argument using the inner product of v j and v k for k > j is used to prove every entry in the j th row of A are zero except for the entry on the diagonal. (Hint: Use induction on j and k )....
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6.36.46.5 - Extend the given vector to an arbitrary basis...

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