Lecture #4 Notes

# Then n s r dimej j 1 r mj n j 1 which

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Unformatted text preview: ) &lt; mk . Assume A is diagonalizable. Then there exists a set S of n linearly independent eigenvectors of A. For each eigenvalue λj , we know that at most dim(Eλj ) elements in S are eigenvalues corresponding to λj . Then n = |S | ≤ r ￿ dim(Eλj ) &lt; j =1 r ￿ mj = n j =1 which is a contradiction. Thus, A is not diagonalizable. 1.5 Simultaneous Diagonalization 1.5.28 Deﬁnition. Two matrices A, B ∈ Mn are said to be simultaneously diagonalizable if ∃S ∈ Mn such that S is invertible and both S −1 AS and S −1 BS are diagonal matrices. 1.5.29 Remark. If A, B ∈ Mn are simultaneously diagonalizable, then AB = BA. Proof. Because diagonal matrices commute, we have AB = S (S −1 AS )(S −1 BS )S −1 = S (S −1 BS )(S −1 AS )S −1 = BA 1.5.30 Question. Is the converse true? That is, if A, B ∈ Mn with A and B diagonalizable and AB = BA, are A and B simultaneously diagonalizable? To answer this, we need a lemma. 2 1.5.31 Lemma. Two matrices A ∈ Mn and B ∈ Mm are diagonalizable iﬀ ￿ ￿ A0 C= ∈ Mn+m is diagonalizable. 0B Proof. First we assume A ∈ Mn and B ∈ Mm are diagonalizable. Then ∃S1 ∈ Mn , S2 ∈ Mm such that S1 and S2 are invertible and S −1 AS and S −1 BS are diagonal matrices. ￿ ￿ ￿ −1 ￿...
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