# Diagonalization of a symmetric matrix

I'm trying to prove that every symmetric matrix is diagonalizable. I know that there are already many answers regarding this question, but I just want to check out whether my approach is correct or not. After some searching, I found some clues to my problem (Here). Based on Tunococ's answer, I want to prove the following : For every symmetric matrix (with real entries), there exists an orthogonal matrix B and a diagonal matrix D such that $$B^{T}AB = D$$ Let's use induction on orders of $$A$$. For $$1 \times 1$$ matrix (which is symmetric), this is trivial. Suppose that this result is true for $$k \times k$$ symmetric matrix ($$k\geq1$$). Let $$A$$ be a $$(k+1) \times (k+1)$$ symmetric matrix. Let $$\lambda$$ be any eigenvalue of $$A$$ and $$v$$ be an associated eigenvector. Since an eigenvector multiplied by any nonzero scalar is still an eigenvector, we can assume that $$\left \| v \right \|=1$$. By Gram-Schmidt process, we can obtain an orthonormal basis $$\{v, w_2, \ldots, w_{k+1}\}$$ of $$\mathbb{R}^{k+1}$$. Let C be an orthogonal matrix defined by $$C = \begin{bmatrix} | & | & \ldots & |\\ v & w_2 & \ldots & w_{k+1} \\ | & | & \ldots & | \end{bmatrix}$$
Then, after some computation, we obtain $$C^{T}AC =\begin{bmatrix} \lambda & O_{1 \times k} \\ O_{k \times 1} & F \end{bmatrix}.$$ Here, $$F$$ is a $$k \times k$$ symmetric matrix. By the induction hypothesis, there exists an orthogonal matrix $$G$$ and a diagonal matrix $$H$$ such that $$G^{T}FG=H$$. Let $$J =\begin{bmatrix} 1 & O_{1 \times k} \\ O_{k \times 1} & G \end{bmatrix}.$$ Since $$G$$ is an orthogonal matrix, $$J$$ is also an orthogonal matrix. By computation, we obtain $$(CJ)^{T}A(CJ)=\begin{bmatrix} \lambda & O_{1 \times k} \\ O_{k \times 1} & H \end{bmatrix}.$$ Note that since $$C$$ and $$J$$ are orthogonal matrices, their product $$CJ$$ is also an orthogonal matrix.

Is this argument correct?

No, you need to explain why you can find an eigenpair $$(\lambda, v)$$.
• No, $1 + x^2$ has no real root. – user7440 Aug 14 at 16:20
• When A is symmetric, if there is an eigenvalue, then it is real. (using $x^* A x = (Ax)^* x$ ). – user7440 Aug 14 at 16:21
• I mean all the roots of the characteristic polynomial of a symmetric matrix are real numbers... So.... I think $1+x^2$ is not our candidate.... – Kim Aug 14 at 16:29