This is a page from the book 'Linear algebra done right'. The assumption that V is finite-dimensional is not stated. This proof seems to imply that every operator on a finite-dimensional vector space can be diagonalised. Is that true? I recall being told that diagonalising linear maps is not easy, so I wonder if I am misinterpreting this.
Best Answer
No, general linear transformations can be written in Jordan normal form which is the best thing we can say about diagonalization in general.
However, for any self adjoint transformation $U$ ($U^*=U$) there's an orthonormal basis where its matrix is diagonal. (That's basically the Spectral theorem.)
In particular, this holds for the positive semidefinite $\sqrt{T^*T}$.