Prove 'every Positive Definite Matrix is Symmetric BUT not the other way around' I'm studying Linear Algebra and I read that :
"Every Positive Definite Matrix is Symmetric BUT the vice versa is not correct..."
in the text it suggests using 'x= u +iv' and then prove it by calculating (x,Ax).
how can I prove it this way ?
Thank you.
 A: *

*Let us for simplicity assume that $V$ is a finite-dimensional vector space over a field $\mathbb{F}$ with an inner product $\langle \cdot, \cdot \rangle: V\times V\to \mathbb{F}$. Note that in the complex case the inner product is a sesquilinear form (as opposed to a bilinear form).


*Let us (for the purpose of this answer) define that an $\mathbb{F}$-linear map $A:V\to V$ is positive if $$\forall x\in V\backslash\{0\}: ~~ \langle x, Ax\rangle ~>~0.\tag{1}$$


*Also let us define that $A$ is symmetric if $$
\forall x,y\in V: ~~ \langle y, Ax\rangle ~=~\langle Ay, x\rangle.\tag{2}$$
In the complex case, the condition (2) is often called Hermiticity.


*$A$ positive does not imply that $A$ is symmetric, if the vector space $V$ is real $\mathbb{F}=\mathbb{R}$. Here is a 2D counterexample:
$$\begin{align} \langle x, y\rangle~=~&x^ty~=~x_1y_1+x_2y_2, \quad
 A~=~\begin{pmatrix} 1 & 2 \cr 0 & 2\end{pmatrix},\cr  \langle x, Ax\rangle~=~&(x_1+x_2)^2+ x_2^2.\end{align}\tag{3} $$


*Assume from now on that $V$ is a complex vector space $\mathbb{F}=\mathbb{C}$.


*Let us (for the purpose of this answer) define that $A$ is real if $$\forall x\in V: ~~ {\rm Im}\langle x, Ax\rangle ~=~0.\tag{4}$$


*$\fbox{$\text{Proposition: }  $A$ \text{ real } ~~\Leftrightarrow~~ $A$ \text{ symmetric.}$}$ The $\Leftarrow$ proof is trivial and left to the reader. The $\Rightarrow$ proof will be deduced in steps below. Assume from now on that $A$ is real.


*By replacing $x$ with $x+y$ in eq. (4), deduce that
$$ \forall x,y\in V: ~~ {\rm Im}\left(\langle y, Ax\rangle +\langle x, Ay\rangle\right)~\stackrel{(4)}{=}~0.\tag{5}$$


*By replacing $y$ with $iy$ in eq. (5), deduce that
$$ \forall x,y\in V: ~~ {\rm Re}\left(\langle y, Ax\rangle -\langle x, Ay\rangle\right)~\stackrel{(5)}{=}~0.\tag{6}$$


*By rewriting $$\langle x, Ay\rangle~=~ \overline{\langle Ay, x\rangle}\tag{7}$$
in eqs.  (5) & (6), we get
$$ \forall x,y\in V: ~~ {\rm Im}\left(\langle y, Ax\rangle - \langle Ay, x\rangle\right) ~=~{\rm Im}\left(\langle y, Ax\rangle + \overline{\langle Ay, x\rangle}\right) ~\stackrel{(5)+(7)}{=}~0,\tag{8}$$
and
$$ \forall x,y\in V: ~~ {\rm Re}\left(\langle y, Ax\rangle - \langle Ay, x\rangle \right)
~=~{\rm Re}\left(\langle y, Ax\rangle - \overline{\langle Ay, x\rangle}\right) ~\stackrel{(6)+(7)}{=}~0,\tag{9}$$
respectively. Now deduce eq. (2) from eqs. (8) & (9). End of proof.
