Suppose F = C and T E L(V). Then the following are equivalent: (a) T is normal. (b) V has an orthonormal basis consistin

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Suppose F C And T E L V Then The Following Are Equivalent A T Is Normal B V Has An Orthonormal Basis Consistin 1 (412.8 KiB) Viewed 43 times

Suppose F = C and T E L(V). Then the following are equivalent: (a) T is normal. (b) V has an orthonormal basis consisting of eigenvectors of T. (c) T has a diagonal matrix with respect to some orthonormal basis of V. Proof First suppose (c) holds, so T has a diagonal matrix with respect to some orthonormal basis of V. The matrix of T* (with respect to the same basis) is obtained by taking the conjugate transpose of the matrix of T; hence T* also has a diagonal matrix. Any two diagonal matrices commute; thus T commutes with T*, which means that T is normal. In other words, (a) holds. Now suppose (a) holds, so T is normal. By Schur's Theorem (6.38), there is an orthonormal basis e1, ..., en of V with respect to which T has an upper-triangular matrix. Thus we can write a 1.1 81,n 7.25 M(T, (e1, ..., en)) = C : 0 an,n = We will show that this matrix is actually a diagonal matrix. We see from the matrix above that || Te 1 || 2 = |a1,112 and || T*e 1 || 2 = |a1,112 + |a1,212 + ... + |11,1|2. Because T is normal, || T e 1 || = ||T*e1 || (see 7.20). Thus the two equations above imply that all entries in the first row of the matrix in 7.25, except possibly the first entry 01,1, equal 0. Now from 7.25 we see that || Te2||2 = |a2,212 (because a 1,2 = 0, as we showed in the paragraph above) and || T* e2 || 2 = |a2,212 + |a2,312 + ... + |2,n|2. Because T is normal, || Te2|| = ||T*e2||. Thus the two equations above imply that all entries in the second row of the matrix in 7.25, except possibly the diagonal entry a2,2, equal 0. Continuing in this fashion, we see that all the nondiagonal entries in the matrix 7.25 equal 0. Thus (c) holds. =