There is a missing case of equal eigenvalues and showing that we can still find perpendicular eigenvectors. It is trivial if we consider they lie in orthogonal eigenspace to other eigenvectors with different eigenvalues.
Hi , this is really well presented but i feel like the difficulty is hidden in shur's theorem at 5:30, it would be interessting to develop this point. Thank you for your work
Yes, the hard work in proving the Complex Spectral Theorem was done in proving Schuur's Theorem (see the proof on page 218 of Linear Algebra Done Right). This use of Schuur's Theorem leads to a clean, easy proof of the Complex Spectral Theorem.
4:24 so if the matrices of any two operators with respect to some basis commute, then the operators itself commute as well? (tried to prove this, seems reasonable, but not at all obvious)
Yes, that's correct. To prove it, suppose S and T are the operators in question. Then because the matrices of S and T (with respect to some basis) commute, ST applied to each basis vector equals TS applied to the same basis vector. Thus ST = TS.
right, I tried kind of the same reasoning (or really the same -_-) that uses the fact from the chapter 3: the matrix of ST = M(ST) = M(S)M(T) = M(T)M(S) = M(TS) = the matrix of TS, thus ST = TS, because an operator is determined by its matrix w.r.t. some basis (= act on a basis). Thank you professor, appreciate your work :)
Or you could simply look at the adjoint of V*TV. V*TV is diagonal as of c) (V*TV)* has to be diagonal too (V*TV)* = (TV)*V = V*T*V Two diagonal matrices always commute V*TVV*T*V = V*T*VV*TV because VV* = 1 V*TT*V = V*T*TV ==> TT* = T*T
I was following the proof of theorem 7.27 until I got to the words, "Thus the equation above implies that m > 0..." I can follow before and after this point, but I don't understand what implies that m > 0.
I do not possess the subscript of capital M, so this shall be proceeded by replacing M with n. if m=0, 0 = c(T²-b₁T+c₁)...(T²-bₙT+cₙ)v T is self adjoint, so T²-bⱼT+cⱼ is invertible ∀j∈[1,n], j∈ℕ and c≠0 whence v=0, contradicting the fact that v≠0.
I love the victory music xD
Your book is awesome sir
There's a typo at the 1:50 mark: "For which operators on V is there *is* an orthonormal basis basis of V"
There is a missing case of equal eigenvalues and showing that we can still find perpendicular eigenvectors. It is trivial if we consider they lie in orthogonal eigenspace to other eigenvectors with different eigenvalues.
Hi , this is really well presented but i feel like the difficulty is hidden in shur's theorem at 5:30, it would be interessting to develop this point. Thank you for your work
Yes, the hard work in proving the Complex Spectral Theorem was done in proving Schuur's Theorem (see the proof on page 218 of Linear Algebra Done Right). This use of Schuur's Theorem leads to a clean, easy proof of the Complex Spectral Theorem.
Thank you for your answer and reference, this appears clearer now. Have a nice day ! :)
4:24 so if the matrices of any two operators with respect to some basis commute, then the operators itself commute as well? (tried to prove this, seems reasonable, but not at all obvious)
Yes, that's correct. To prove it, suppose S and T are the operators in question. Then because the matrices of S and T (with respect to some basis) commute, ST applied to each basis vector equals TS applied to the same basis vector. Thus ST = TS.
right, I tried kind of the same reasoning (or really the same -_-) that uses the fact from the chapter 3: the matrix of ST = M(ST) = M(S)M(T) = M(T)M(S) = M(TS) = the matrix of TS, thus ST = TS, because an operator is determined by its matrix w.r.t. some basis (= act on a basis).
Thank you professor, appreciate your work :)
Or you could simply look at the adjoint of V*TV.
V*TV is diagonal as of c)
(V*TV)* has to be diagonal too
(V*TV)* = (TV)*V = V*T*V
Two diagonal matrices always commute
V*TVV*T*V = V*T*VV*TV
because VV* = 1
V*TT*V = V*T*TV
==> TT* = T*T
If the vector space is Complex then the orthonormal basis is the same as eigenvector of the Matrix?
I was following the proof of theorem 7.27 until I got to the words, "Thus the equation above implies that m > 0..." I can follow before and after this point, but I don't understand what implies that m > 0.
This is a question about something in the book, not about something in the videos. Thus this venue is not the appropriate place to ask such questions.
I do not possess the subscript of capital M, so this shall be proceeded by replacing M with n.
if m=0,
0 = c(T²-b₁T+c₁)...(T²-bₙT+cₙ)v
T is self adjoint,
so T²-bⱼT+cⱼ is invertible ∀j∈[1,n], j∈ℕ and c≠0
whence v=0, contradicting the fact that v≠0.
te amo porra
I wish he would reply to comments. They are good videos but I haven’t seen him answering anyone’s questions.
Edward Hartz he does