Do real eigenvalues $\implies$ symmetric matrix? And why is a positive definite matrix symmetric?

Question

Proof:

$Av$ = $\lambda v$

$\implies \bar{v}^{T}Av = \lambda \bar{v}^{T} v$ ------(1)

And,

$Av$ = $\lambda v \implies \bar{A}\bar{v}$=$\bar{\lambda}\bar{v} \implies \bar{v}^{T}\bar{A}^{T}=\bar{\lambda}\bar{v}^{T}$

$\implies \bar{v}^{T}\bar{A}^{T} v = \bar{\lambda}\bar{v}^{T} v$ ------(2)

Since $A$ has real eigenvalues, $\lambda = \bar{\lambda} \implies \bar{v}^{T}\bar{A}^{T} v = \lambda\bar{v}^{T} v$ ------(3)

Now, Assuming A is real ($A=\bar{A}$), and comparing equation (1) and (3):

$\bar{v}^{T}A^{T} v = \bar{v}^{T}Av$

Does that mean $A=A^{T}$?

And hence can I infer that a positive definite matrix (which of course has all eigenvalues real and positive) is symmetric? Is that the right reason behind it? If not, what makes a positive definite matrix symmetric?

EDIT: This question is not the same as "Prove that the eigenvalues of a real symmetric matrix are real.", but it actually asks about whether or not the converse it true.

Answer 1

This is not true. The following matrix $$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$has real eigenvalues but is asymmetric. Also it is positive definite but this too does not imply symmetry.

Answer 2

The question has been correctly answered by Mostafa Ayaz, but, just in case you suspect that the problem arises because his example isn't diagonalizable, here's a diagonalizable example: $$ \begin{pmatrix} 2&-1\\0&1 \end{pmatrix}. $$ The eigenvalues are $1$ and $2$, with eigenvectors $\binom11$ and $\binom 10$, respectively.

More generally, notice that, for a symmetric matrix, not only are the eigenvalues real but the eigenvectors for different eigenvalues are orthogonal. So you can produce many counterexamples like mine by just choosing some different real numbers to serve as eigenvalues and some non-orthogonal vectors to serve as eigenvectors.