Real symmetric matrices are diagonalizable

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: \forall A symmetric,  \exists P orthogonal (P^tP=I) and D diagonal, such that A=P^{t}DP

To continue the proof I will use the following result:

Let let A be a real symmetric matrix and let V be a subspace of R^n and V^{\bot} its complement (R^n=V \oplus V^{\bot}). If \forall v \in V, Av \in V then: for w \in V^{\bot} \Rightarrow  Aw \in V^{\bot}.

Proof: I will use the dot product defined in the previous post. Given v \in V and w \in V^{\bot} , <Av, w> = <v, Aw> because A is real and symmetric. But <Av, w>=0 because Av \in V. Thus <v, Aw>=0, \forall v \in V. This means that Aw \in V^{\bot}.

Getting back to the problem, A has at least one eigenvalue. It results that exists X_1 and \lambda _1, such that AX_1=\lambda _1X_1.

If V_1 is the vector space generated by X_1, then the operator A is also symmetric when applied to the subspace V_1^{\bot} (this can be proven by changing the basis). This means that exists X_2 \in V_1^{\bot} such that AX_2 = \lambda_2X_2. Considering the vector space generated by X_1 and X_2, and by applying the operator A to its orthogonal we will get AX_3 = \lambda_3X_3. By induction we get: AX_i = \lambda_iX_i, i=1..n, where the vectors X_i are pair wise perpendicular: <X_i, X_j>=0, \forall i,j \in {1..n}.

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:

A[X_1 X_2 ... X_n]=[X_1 X_2 ... X_n]diag(\lambda_1, \lambda_2, ... ,\lambda_n)

Or A=P diag(\lambda_1, \lambda_2, ... ,\lambda_n) P^{-1}, where the columns of P are the vectors X_1, X_2, ...X_n. P is orthogonal because vectors X_i are pair wise perpendicular and unit vectors. This also means that P^{-1}=P^t.

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3 Responses to Real symmetric matrices are diagonalizable

  1. Pingback: Reynold’s Equation and Its Simplifications « SanBal

  2. Pingback: SVD – a simple proof | AlfaNotes

  3. Anonymous says:

    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)”

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