Definiteness of a nonsymmetric matrix

[Mathsadness]

TL;DR

How do we find what type of definiteness a nonsymmetric matrix has? All the definitions in my book refer to symmetric ones as they have nice properties.

There is no specific example but my attempt at one would be to make the non-symmetric matrix symmetric. Then we would be able the usual formulas as designed for symmetric matrices. Is this how it works?

Alternatively, do I just calculate the Eigen values without making it symmetric? I don't think that would be correct as that test is used for symmetric matrices. Any input would be appreciated.

 

[fresh_42]

Asymmetry results in independent mixed terms which can be used to prove indefiniteness.

 

[Mathsadness]

Asymmetry results in independent mixed terms which can be used to prove indefiniteness.

Can you please elaborate a bit? Wouldn't the process I mentioned above apply i.e making the matrix symmetric first and then use the usual procedure?

 

[fresh_42]

Let us consider $Q=\begin{bmatrix}1&b\\c&1\end{bmatrix}$. The corresponding quadratic form is thus
$$
\vec{x}^\tau\cdot Q \cdot \vec{x}=(x,y)\;Q\;\begin{pmatrix}x\\y\end{pmatrix}= x^2+cxy+bxy+y^2=\left(x+y\right)^2+(b+c-2)xy
$$
Depending on the actual values of $b$ and $c$ you can easily make $(b+c-2)xy$ greater or less than $-(x+y)^2$ for some points $(x,y)^\tau$. Hence it cannot always be positive or always be negative, hence indefinite.

For $b=c$ we get $\vec{x}^\tau\cdot Q \cdot \vec{x}=(x+ by)^2+(1-b^2)y^2$ which is at least positive definite if $|b|<1\,$.

 

[StoneTemplePython]

I'm not sure where Fresh was going with this. Since OP is evidently working over reals, the standard approach is to write

$A = \frac{1}{2}\big(A+A^T\big)+\frac{1}{2}\big(A-A^T\big) = S + Z$
which decomposes $A$ into symmetric and skew parts. Now for any $\mathbf x\in \mathbb R^n$ a defining characteristic of a real skew symmetric matrix $Z$ is $\mathbf x^TZ\mathbf x =0$ (check the transpose and confirm this for yourself).

Thus for any $\mathbf x\in \mathbb R^n$
$\mathbf x^TA\mathbf x= \mathbf x^T\big(S + Z\big)\mathbf x = \mathbf x^TS\mathbf x + \mathbf x^TZ\mathbf x= \mathbf x^TS\mathbf x + 0 = \mathbf x^T\Big(\frac{1}{2}\big(A+A^T\big)\Big)\mathbf x$

thus to check for definiteness of $A$ you can apply the 'usual techniques' in your book, to $\frac{1}{2}\big(A+A^T\big)$