Prove 'every Positive Definite Matrix is Symmetric BUT not the other way around'

Question

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.

Answer 1

1. 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).

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

3. 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.

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

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

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

7. $\fbox{$\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.

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

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

10. 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.