Proof assume that l m m if x m and z m then since l x

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Proof. Assume that L ( M ) M . If x M and z M , then since L ( x ) M we have 0 = ( z | L ( x )) = ( L * ( z ) | x ) = ± ( L ( z ) | x ) . Since this holds x M , it follows that L ( z ) M . 3.3 Polarization and Isometries Real inner product on V : ( x + y | x + y ) = ( x | x ) + 2( x | y ) + ( y | y ) ( x | y ) = 1 2 (( x + y | x + y ) - ( x | x ) - ( y | y )) = 1 2 ( || x + y || 2 - || x || 2 - || y || 2 ) . Complex inner products (are only conjugate symmetric) on V : ( x + y | x + y ) = ( x | x ) + 2 Re ( x | y ) + ( y | y ) Re ( x | y ) = 1 2 ( || x + y || 2 - || x || 2 - || y || 2 ) . Re ( x | iy ) = Re ( - i ( x | y )) = Im ( x | y ) . In particular, we have Im ( x | y ) = 1 2 ( || x + iy || 2 - || x || 2 - || iy || 2 ) . We can use these ideas to check when linear operators L : V V are 0. First note that L = 0 ( L ( x ) | y ) = 0 , x, y V . To check the part, let y = L ( x ) to see that || L ( x ) || 2 = 0 , x V . Theorem. Let L : V V be self-adjoint. L = 0 ( L ( x ) | x ) = 0 , x V . Proof. If L = 0 ( L ( x ) | x ) = 0 , x V . Assume that ( L ( x ) | x ) = 0 , x V . 0 = ( L ( x + y ) | x + y ) = ( L ( x ) | x ) + ( L ( x ) | y ) + ( L ( y ) | x ) + ( L ( y ) | y ) = ( L ( x ) | y ) + ( y | L * ( x )) = ( L ( x ) | y ) + ( y | ( L ( x ))) = 2 Re ( L ( x ) | y ) . Now insert y = L ( x ) to see that 0 = Re ( L ( x ) | L ( x )) = || L ( x ) || 2 .
Linear Algebra Igor Yanovsky, 2005 14 Theorem. Let L : V V be a linear map on a complex inner-product space. Then L = 0 ( L ( x ) | x ) = 0 , x V . Proof. If L = 0 ( L ( x ) | x ) = 0 , x V . Assume that ( L ( x ) | x ) = 0 , x V . 0 = ( L ( x + y ) | x + y ) = ( L ( x ) | x ) + ( L ( x ) | y ) + ( L ( y ) | x ) + ( L ( y ) | y ) = ( L ( x ) | y ) + ( L ( y ) | x ) 0 = ( L ( x + iy ) | x + iy ) = ( L ( x ) | x ) + ( L ( x ) | iy ) + ( L ( iy ) | x ) + ( L ( iy ) | iy ) = - i ( L ( x ) | y ) + i ( L ( y ) | x ) 1 1 - i i ‚ • ( L ( x ) | y ) ( L ( y ) | x ) = 0 0 . Since the columns of the matrix on the left are linearly independent the only solution is the trivial one. In particular ( L ( x ) | y ) = 0. 3.4 Unitary and Orthogonal Operators A linear transformation A is orthogonal is AA T = I , and unitary if AA * = I , i.e. A * = A - 1 . Theorem. L : V W is a linear map between inner product spaces. TFAE: 1) L * L = I V , ( L is unitary) 2) ( L ( x ) | L ( y )) = ( x | y ) x, y V , ( L preserves inner products) 3) || L ( x ) || = || x || x V . ( L preserves lengths) Proof. (1) (2). L * L = I V ( L ( x ) | L ( y )) = ( x | L * L ( y )) = ( x | Iy ) = ( x | y ) , x V . Also note: L takes orthonormal sets of vectors to orthonormal sets of vectors. (2) (3). ( L ( x ) | L ( y )) = ( x | y ) , x, y V || L ( x ) || = p ( L ( x ) , L ( x )) = p ( x, x ) = || x || . (3) (1). || L ( x ) || = || x || , x V ( L * L ( x ) | x ) = ( L ( x ) | L ( x )) = ( x | x ) = ( Ix | x ) (( L * L - I )( x ) | x ) = 0 , x V . Since L * L - I is self-adjoint (check), L * L = I . Two inner product spaces V and W over F are isometric , if we can find an isom- etry L : V W , i.e. an isomorphism such that ( L ( x ) | L ( y )) = ( x | y ). Theorem. Supposet L is unitary, then L is an isometry on V . Proof. An isometry on V is a mapping which preserves distances. Since L is unitary, || L ( x ) - L ( y ) || = || L ( x - y ) || = || x - y || . Thus L is an isometry.
Linear Algebra Igor Yanovsky, 2005 15 3.5 Spectral Theorem Theorem. Let L : V V be a self-adjoint operator on a finite dimensional inner product space. Then we can find a real eigenvalue λ for L . Spectral Theorem. Let L : V V be a self-adjoint operator on a finite dimensional inner product space. Then there exists an orthonormal basis e 1 , . . . , e n of eigenvectors, i.e. L ( e 1 ) = λ 1 e 1 , . . . , L ( e n ) = λ n e n . Moreover, all eigenvalues λ 1 , . . . , λ n are real.

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