## 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 $w \in V^{\bot}$ , $ = $ because $A$ is real and symmetric. But $=0$ because $Av \in V$. Thus $=0$, $\forall v \in V$. 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: $=0, \forall i,j \in {1..n}$.

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 $P^{-1}=P^t$.