# Does orthogonal eigenvectors imply symmetric matrix?

If an $$n \times n$$ matrix $$\mathbf A$$ is diagonalizable, and orthogonal eigenvectors of $$\mathbf A$$ form a basis of $$R^n$$, then is $$\mathbf A$$ symmetric?

Here is what I tried:

Suppose {$${\vec{a_1},\vec{a_2},...,\vec{a_n}}$$} is a set of eigenvectors of $$A$$ which forms a basis of $$R^n$$. Suppose $$P$$ is the matrix with columns $${\vec{a_1},\vec{a_2},...,\vec{a_n}}$$, and $$D$$ is the diagonal matrix with eigenvalue corresponding to $$\vec{a_i}$$ on the i-th entry.

Then $$A = PDP^{-1}$$ and $$A^{T} = {P^{-1}}^{T}D^{T}P^{T}$$. $$D$$ is diagonal, so $$D^{T} = D$$. But I got stuck here.

• Yes. .... can you show us your effort? What did you try? – user376343 Feb 12 '19 at 22:41

$$A=P^tDP$$, where $$D$$ is diagonal and $$P^{-1}=P^t$$. So, $$A^t=(P^tDP)^t=((P^t)(DP))^t=(DP)^t(P^t)^t=(P^tD^t)P=P^tDP=A$$.
• @raeyu: This is equivalent to having the columns of $P$ being the orthonormal basis of eigenvectors assumed in the Question. – hardmath Feb 13 '19 at 0:12
Suppose $$\{ x_1,x_2,\cdots,x_N \}$$ is an orthonormal basis of $$\mathbb{R}^N$$, and suppose $$A$$ is a real $$N\times N$$ matrix such that $$Ax_n = \lambda_n x_n$$ for real $$\lambda_n$$ and all $$1 \le n \le N$$. Then \begin{align} Ax & = A\sum_{n=1}^{N}\langle x,x_n\rangle x_n \\ &= \sum_{n=1}^{N}\langle x,x_n\rangle \lambda_n x_n. \end{align}
Therefore, for all $$x,y\in\mathbb{R}^{N}$$, \begin{align} \langle Ax,y\rangle& =\left\langle\sum_{n=1}^{N}\langle x,x_n\rangle\lambda_n x_n,y\right\rangle \\ &=\sum_{n=1}^{N}\lambda_n\langle x,x_n\rangle\langle x_n,y\rangle \\ &=\left\langle x,\sum_{n=1}^{N}\lambda_n\langle y,x_n\rangle x_n\right\rangle = \langle x,Ay\rangle. \end{align} In particular, $$\langle Ae_n,e_m\rangle = \langle e_n,Ae_m\rangle$$ for the standard basis $$\{ e_n \}$$.