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 .

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Cool! Helped a lot. Thanks.

Suggestion – you could elaborate on “A is also symmetric when applied to the subspace V_1^{\bot} (this can be proven by changing the basis)”