Spectral theorem over C
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Exercise:
Let V be a finite dimensional inner product space over mathbbC and T:Vrightarrow V a linear map. Then T is orthogonally diagonalizable iff T is normal.
Solution:
Proof. We only need to prove the direction T is normal Longrightarrow T is orthogonally diagonalizable. By a previous theorem T is trigonalizable because KmathbbC i.e. exists a basis mathcalC for V s.t. T_mathcalC^mathcalC is an upper triangular matrix. By applying Gram-Schmidt to the basis mathcalC we obtain a new basis mathcalBv_...v_n which is orthonormal and s.t. T_mathcalB^mathcalB is still upper-triangular. We claim that T_mathcalB^mathcalB is actually diagonal. To see this write T_mathcalB^mathcalBa_ij_leq ileq n leq jleq n pmatrix a_ & a_ & s & a_n & a_ & s & a_n & & ddots & vdots & & s & a_nn pmatrix Since mathcalB is orthonormal we have ||Tv_||^ |a_|^. Since T^*_mathcalB^mathcalB leftT_mathcalB^mathcalBright^T we also have: ||T^*v_||^ _k^n|overlinea_k|^ |a_|^+...+|a_n|^ As T is normal ||Tv_||||T^*v_|| hence a_...a_n. This shows that T_mathcalB^mathcalB pmatrix a_ & & s & & a_ & s & a_n & & ddots & vdots & & s & a_nn pmatrix ||Tv_||^ |a_|^ and ||T^*v_||^ |a_|^+|a_|^+...+|a_n|^. As ||Tv_||||T^*v_|| we obtain a_...a_n. We can continue this proof by induction and show that forall leq k n we have a_k k+...a_kn hence T_mathcalB^mathcalB is diagonal.
Let V be a finite dimensional inner product space over mathbbC and T:Vrightarrow V a linear map. Then T is orthogonally diagonalizable iff T is normal.
Solution:
Proof. We only need to prove the direction T is normal Longrightarrow T is orthogonally diagonalizable. By a previous theorem T is trigonalizable because KmathbbC i.e. exists a basis mathcalC for V s.t. T_mathcalC^mathcalC is an upper triangular matrix. By applying Gram-Schmidt to the basis mathcalC we obtain a new basis mathcalBv_...v_n which is orthonormal and s.t. T_mathcalB^mathcalB is still upper-triangular. We claim that T_mathcalB^mathcalB is actually diagonal. To see this write T_mathcalB^mathcalBa_ij_leq ileq n leq jleq n pmatrix a_ & a_ & s & a_n & a_ & s & a_n & & ddots & vdots & & s & a_nn pmatrix Since mathcalB is orthonormal we have ||Tv_||^ |a_|^. Since T^*_mathcalB^mathcalB leftT_mathcalB^mathcalBright^T we also have: ||T^*v_||^ _k^n|overlinea_k|^ |a_|^+...+|a_n|^ As T is normal ||Tv_||||T^*v_|| hence a_...a_n. This shows that T_mathcalB^mathcalB pmatrix a_ & & s & & a_ & s & a_n & & ddots & vdots & & s & a_nn pmatrix ||Tv_||^ |a_|^ and ||T^*v_||^ |a_|^+|a_|^+...+|a_n|^. As ||Tv_||||T^*v_|| we obtain a_...a_n. We can continue this proof by induction and show that forall leq k n we have a_k k+...a_kn hence T_mathcalB^mathcalB is diagonal.