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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