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MAT224H1 (138)

Lengths and distrances (inner products), angles, cauchy-schwartz inequality, triangle inequality

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Martin, Burda

Lecture 14 notes by Y. Burda 1 An exercise to be used later Exercise: Prove that the determinant of a positive-denite matrix is a pos- itive number. Solution 1(due to Kirby): Let A be a positive-denite matrix. Since it is Hermitian (and every complex positive denite matrix by denition is Hermitian), it is diagonalizable. Hence detA is the product of As eigenval- ues. Since the eigenvalues of a positive-denite matrix are real and positive, their product is also positive. Solution 2(due to Kristie): Since A is positive-denite, there exists an invertible matrix B such that A = B B. Hence detA = detB detB = T detB detB = detB detB = detB detB = detB > 0 (in the last step we used the fact that B is invertible and hence its determinant is non-zero). 2 Lengths and distances in inner product spaces If V is an inner product space we can dene the length of a vector by the formula p v = (v,v) This length is also called the norm of the vector v and we use the symbol v instead of v to distinguish it from the length computed using n the standard dot-product on R . Distance between two points u,v V is dened as the length of the vector from v to u: d(u,v) = u v p 2 i Example: If (,) i th usupl dot product on C , then 1 ) = ii + 11 = 2 and d((i),( )) = i1 = i 1 + 1 i = 2 1 i 1i 2 x1 y1 Example: If (,) is the inner product on R given by (( ),( x2 y2)) = 2x1 1+ x 2 2hen the unit circle centered at origin is the set 2 2 {v R d(v,0) = 1} = {v R v = 1} = p ={( y 2x + y = 1} = {( )y2x + y = 1} 1
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