6.4.14 Solution
April 17, 2019
6.4.14 For any eigenspace Eλ of T , we know that W is also U -invariant since
for any w ∈ W ,
T (U w) = U (T w) = λU w
Thus, by exercise 6.4.7 we know W ⊥ is also both T -invariant and U invariant subspace. Then, by induction on n = dimV ,
(a) If n = dimV = 1, and let 0 6= w ∈ W , then W = span{w}. Since
W is also U -invariant, we know U w ∈ W and thus U w = σw for
some σ 6= 0. Therefore, T and U are simultaneously diagonalizable.
(b) Now, suppose for any vector space V with 1 ≤ dimV ≤ k − 1,
and two self-adjoint operator T and U from V to V such
that TU = UT, there exists a orthonormal basis such that [T ]β
and [U ]β are both diagonal matrices. Now, for any vector space
V with dimV = k, and let W = Eλ be any eigenspace of T .
If W = V , then the fact that T is self-adjoint allows us to find a
orthonormal basis β = {v1 , · · · , vn } for V consisting eigenvector
of T . Then, for every vj , 1 ≤ j ≤ n, by (1), U (vj ) ∈ span{vj },
which tells us that vj are also eigenvectors of U for 1 ≤ j ≤ n.
Hence, [U ]β is a diagonal matrix.
If W is a proper subspace of V , (i.e dimW ≤ k − 1), since both
TW and UW are operators from W to W , self-adjoint and
TW UW = UW TW , thus we can use induction hypothesis to say
that there is an orthogonal basis β1 for W consisting eigenvectors of T and U simultaneously. That is, [TW ]β1 and [UW ]β1 are
both diagonal matrices. Also, W ⊥ is a proper subspace of V (i.e
dimW⊥ ≤ k − 1), and both TW⊥ and UW⊥ are operators
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from W⊥ to W⊥ , self-adjoint and TW⊥ UW⊥ = UW⊥ TW⊥ .
Thus, we can use induction hypothesis again: there is an orthogonal basis β2 for W consisting eigenvectors of T and U simultaneously. That is, [TW ⊥S
]β2 and [UW ⊥ ]β2 are both diagonal matrices.
Finally, let β = β1 β2 , and we yield [T ]β and [U ]β are both
diagonal matrices.
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