can a quadratic form be represented by a non-symmetric matrix? If so, hwo to determine if its postive definite.

Question

holsteremission · Answer

Sure. Here's an example:
$$x^2-xy-y^2=\begin{bmatrix}x\y\end{bmatrix}^\intercal\begin{bmatrix}1&1\-2&-1\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}$$The polynomial is a quadratic form because it's homogeneous and of order $2$:
$$\begin{align*}
f(x,y)=x^2-xy-y^2\implies f(tx,ty)&=(tx)^2-(tx)(ty)-(ty)^2\[1ex]
&=t^2x^2-t^2xy-t^2y^2\[1ex]
&=t^2\left(x^2-xy-y^2\right)\[1ex]
&=t^2f(x,y)
\end{align*}$$You can check for positive definiteness by checking the eigenvalues of the matrix. They all have to be positive (and thus real - you'll notice that this particular matrix has complex eigenvalues).

caozeyuan · Answer

but I was told that eigenvalue check only applies to symmetical ones? so its actually generally true for non symmetrical matrices as well?

holsteremission · Answer

I suppose the standard definition of positive definiteness only applies to Hermitian matrices ($\mathbf A=\mathbf A^*$, with $\mathbf A^*$ the conjugate transpose), or, in the real case, symmetric ones ($\mathbf A=\mathbf A^\intercal$).

And apparently, the notion of a "quadratic form" depends on the context. I'm interpreting it as referring to a homogenous second order polynomial (see the Wiki page), something of the form $ax^2+bxy+cy^2$.

In linear algebra, it seems, a quadratic form is more specifically the product $f(\mathbf x)=\mathbf x^\intercal \mathbf{Ax}$, where $\mathbf A$ is taken to be a symmetric real matrix. Then $\mathbf A$ is positive definite iff $f(\mathbf x)>0$ for any $\mathbf x$. So in this sense, there's no meaning of positive definiteness for the example I used above.

You can extend the definition of positive definiteness for asymmetric matrices, though. It involves decomposing $\mathbf A$ into a linear combination of what are called its symmetric and antisymmetric parts,
$$\mathbf A=\color{red}{\mathbf A_S}+\color{blue}{\mathbf A_A}=\color{red}{\frac{\mathbf A+\mathbf A^\intercal}{2}}+\color{blue}{\frac{\mathbf A-\mathbf A^\intercal}{2}}$$Then $\mathbf A$ can be defined to be positive definite iff $\mathbf A_S$ is too.

$\mathbf A_S$ is of course symmetric, since
$${\mathbf A_S}^\intercal=\left(\frac{\mathbf A+\mathbf A^\intercal}{2}\right)^\intercal=\frac{\mathbf A^\intercal+\left(\mathbf A^\intercal\right)^\intercal}{2}=\mathbf A_S$$and in fact generates the same quadratic form. So we're really not changing much as far as the definition of positive definiteness goes.