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Exercise:
If T is orthogonally diagonalizable then Tcirc T^* T^*circ T.

Solution:
Proof. Let mathcalB be an orthonormal basis for V s.t. all the vectors in mathcalB are eigenvectors of T Longrightarrow T_mathcalB^mathcalB diagonal matrix. We also know that T^*_mathcalB^mathcalBleftoverlineT_mathcalB^mathcalBright^T diagonal matrix. But if AB are diagonalizable matrices then they must commute ABBA. Longrightarrow T_mathcalB^mathcalB T^*_mathcalB^mathcalBT^*_mathcalB^mathcalB T_mathcalB^mathcalB Tcirc T^*_mathcalB^mathcalB T^*circ T_mathcalB^mathcalB Tcirc T^* T^*circ T.
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Exercise:
If T is orthogonally diagonalizable then Tcirc T^* T^*circ T.

Solution:
Proof. Let mathcalB be an orthonormal basis for V s.t. all the vectors in mathcalB are eigenvectors of T Longrightarrow T_mathcalB^mathcalB diagonal matrix. We also know that T^*_mathcalB^mathcalBleftoverlineT_mathcalB^mathcalBright^T diagonal matrix. But if AB are diagonalizable matrices then they must commute ABBA. Longrightarrow T_mathcalB^mathcalB T^*_mathcalB^mathcalBT^*_mathcalB^mathcalB T_mathcalB^mathcalB Tcirc T^*_mathcalB^mathcalB T^*circ T_mathcalB^mathcalB Tcirc T^* T^*circ T.
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eth, fs23, lineare algebra, orthogonal, proof
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Difficulty
(3, default)
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0 (default)
Language
ENG (English)
Type
Proof
Creator rk
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