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Create a free Team What is Teams? Learn more. Why are real symmetric matrices diagonalizable? Asked 8 years, 2 months ago. Active 1 year, 4 months ago. Viewed 27k times. Marc van Leeuwen k 6 6 gold badges silver badges bronze badges. I would link to it, but I'm on my phone. Show 1 more comment. Active Oldest Votes. Ittay Weiss Ittay Weiss I'm not sure that part is as obvious to me as it should be. If the subspace is stable under the linear transformation given by the matrix for which there seems to be no reason here then one can choose a basis of the subspace and express the restriction of the linear transformation on that basis, giving a smaller square matrix.
However that matrix highly depends on the choice of basis. However what I said about restriction being not defined still applies. Also there is more wrong with this answer than would fit in my comment. Symmetry does not imply normality only works for real matrices.
And normality directly implies diagonalisable with orthogonal eigenspaces, not need to triangularise before applying that. Show 6 more comments. Rather consider orthogonality of distinct eigenspaces : that is easy to establish in general,, and once this is done it suffices to choose an orthonormal basis in each eigenspace which obviously can be done to get an orthonormal basis of eigenvectors.
Add a comment. Now our focus will be on proving the following fact. Marc van Leeuwen Marc van Leeuwen k 6 6 gold badges silver badges bronze badges. This avoids some technicalities. But as I said in the answer: I want to avoid the suggestion that the validity of the stated result somehow depends on that. But note that my first proof of the lemma does invoke FTA, so it is close but does not involve complex vector spaces.
Step 1: The eigen-vectors are orthogonal. Abramo Abramo 6, 2 2 gold badges 29 29 silver badges 50 50 bronze badges. Upcoming Events. Featured on Meta. Now live: A fully responsive profile. Indeed, the matrix:. So the semigroup generated by the perfect squares consists of just the perfect squares, which are not all the elements of the field, so the field can be ordered. However, the field need not be real-closed. Take a matrix over that field.
The only remaining case is if some are the same and some are distinct. If we can handle that case, then we can diagonalize any matrix. Proof: Consider the scheme of such orthogonal matrices. Each connected component of this scheme corresponds to a partition of the eigenvalues into blocks. Choose one. We want to lift this to a point on the whole ring. This is because the only way to keep a diagonal matrix block diagonal is to hit it with one of those.
This bit might not be entirely correct. Thus there is a lift and the matrix can be put in this form. Then we do an induction on dimension. This is a generalization of the idea of Will Sawin. The Stufe of such field should be infinite. So the base field should be a formally real field. A complete characterization is given in the following article a necessary and sufficient condition is that such field should be an intersection of real closed fields :.
MR D. Mornhinweg, D. Shapiro and K. Sign up to join this community. The best answers are voted up and rise to the top. Over which fields are symmetric matrices diagonalizable? Ask Question. Asked 8 years, 10 months ago. Active 5 years, 8 months ago.
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