definiteness of asymmetric matrices

Question

The english wikipedia page requires a symmetric matrix to determine definiteness. ( A is an n × n symmetric matrix ...)

The german Wikipedia page on the definiteness of matrices states that a $n\times n$ Matrix does not necessarily have to be symmetric to assert definitness: (translated into english): An arbitrary (possibly, symmetric or hermitic) $(n\times n)$-Matrix $A$ ...

The procedure to determine definiteness is the same both times: Simply evaluate the sign of $x^\top A x$ and do the usual case distinction.

So as an example: Let $A = \begin{pmatrix} \alpha & \beta\\ -\beta & \alpha \end{pmatrix},\quad \alpha, \beta \in \mathbb{R}$

$$ \begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} \alpha & \beta\\ -\beta & \alpha \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \alpha (x_1^2+x_2^2) \Rightarrow A \text{ is } \begin{cases} \text{positive definite} \quad \text{if} \quad \alpha > 0\\ \text{negative definite} \quad\! \text{if} \quad \alpha < 0 \end{cases}$$

So A is asymmetric (anti-symmetric), yet definite for all nonzero $\alpha, \beta$.

Why would a matrix have to be symmetric in order to tell whether it is definite ? Where does that restriction arise from ?

btw: the eigenvalue-method does not work for asymmetric matrices (complex eigenvalues), but the $x^\top A x$ does.

Answer 1

The "canonical" definition for (e.g.) positive definiteness of a symmetric matrix is $\text{Sp} (A) \subset \mathbb{R}^*_+$, so each eigenvalue must be strictly positive. It is equivalent to say that for each $x \in \mathbb{R}^n\setminus \{0\}, \, \langle x, Ax \rangle = x^T Ax>0$.

The reason here is that each symmetric matrix is diagonalisable in $\mathbb{R}$, and you can do that with an orthogonal change of basis, i.e. $\exists U \in O_n(\mathbb{R})$ such that $U^T A U$ is diagonal. You then just have $\langle x, Ax \rangle = \langle U U^Tx, A U U^Tx \rangle = \langle U^T x, D U^T x\rangle $ with $D$ diagonal (you can finish the proof alone from here).

The study of definite positive matrices is very practical and useful, hence the definition has purpose...

In summary, if we were to define some definite positivity on any matrix, the requirement would be to have a strictly positive spectrum (so if it has complex eigenvalue, forget about it). In this case, you could find another scalar product for which this matrix would be symmetric, so the problem would be exactly the same up to conjugation. But the equivalence (positive definite) iff ($\langle x,Ax\rangle_2, \, \forall x\neq 0$) would only be true for this new scalar product $\langle \ast, \ast \rangle_2$ which is different from $\ast^T \ast$.

Lastly, the word definite only means that it preserves non-nullity so a definite matrice is a matrice whose spectrum does not contain 0, thus an invertible matrix.

What I have written here can be adapted for all your matrices subspaces (non negative, negative, non-positive, hermitian...)

Maybe check some references about polar decomposition and the existence of a unique symmetric definite positive square root to see how we use symmetric matrices.