Diagonalization of symmetric matrices
We are looking for conditions under which a matrix in some orthonormal basis takes a diagonal form.
Definition 1. We say that is diagonalizable if there exists an orthonormal basis such that is diagonal: (all matrices are real).
The usual way of solving a mathematical problem is this. Look for necessary conditions. Then try to find weak sufficient conditions. If they coincide with the necessary ones, you have an exact solution to the problem.
Exercise 1 (necessary condition) Suppose is diagonalizable by an orthogonal matrix. Then it is symmetric.
Proof. A diagonal matrix is obviously symmetric. Hence, from and using Exercise 1 on the orthogonal matrix inverse
Exercise 2. If is symmetric and a subspace is invariant with respect to , then is also an invariant subspace of .
Proof. Let We need to show that Take any Since is invariant, we have Hence, by Exercise 1 Since is arbitrary, it follows that
Exercise 3 (sufficient condition) If is symmetric, then it is diagonalizable by an orthogonal matrix.
Proof. Let be symmetric. By Exercise 3 has at least one eigenvector Denote the set of vectors orthogonal to It has dimension and it is invariant by Exercise 2. By Exercise 4 has an eigenvector in
Let be the set of vectors from which are orthogonal to As above, , is invariant and in there is an eigenvector of .
Continuing in this way, eventually we obtain a system Since is orthogonal to and all of belong to , is orthogonal to all of . Similarly, at each step is orthogonal to all of Thus, the system is orthogonal. Its elements can be scaled to satisfy for all
The resulting system will be an orthonormal basis. Its completeness follows from the fact that there are elements in this system and they are linearly independent.
Denoting the eigenvalue corresponding to we have This means that is of diagonal form in the basis The transition matrix from the basis consisting of unit vectors to is orthogonal. If then the same is the vector of coefficients of in the basis By Exercise 3,
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