This article involves advanced linear algebra knowledge but it definitely worth understanding it. The previous post contains a proof that a real symmetric matrix has real eigenvalues. Additionally the real symmetric matrices are diagonalizable by an orthogonal matrix. This means: symmetric, orthogonal () and diagonal, such that
To continue the proof I will use the following result:
Let let be a real symmetric matrix and let be a subspace of and its complement (). If , then: for .
Proof: I will use the dot product defined in the previous post. Given and , because is real and symmetric. But because . Thus , . This means that .
Getting back to the problem, has at least one eigenvalue. It results that exists and , such that .
If is the vector space generated by , then the operator is also symmetric when applied to the subspace (this can be proven by changing the basis). This means that exists such that . Considering the vector space generated by and , and by applying the operator to its orthogonal we will get . By induction we get: , where the vectors are pair wise perpendicular: .
Additionally the vectors can be divided by their norm to make them unit vectors. In a matrix form the relations above can be written as:
Or , where the columns of are the vectors . P is orthogonal because vectors are pair wise perpendicular and unit vectors. This also means that .