Existence of a similar positive definite matrix

Question

Assume $A$ is a real (non-symmetric) positive definite matrix, that is, $$ x^T A x > 0 $$ for all real non-zero real $x$. It is easy to prove that the eigenvalues of $A$ have positive real part.

Conversely, assume $A$ is a (non-symmetric) real matrix, whose eigenvalues have positive real part. It does not necessarily follow that $A$ is positive definite. Nonetheless, does there exist a similar matrix $B$ which is positive definite? More precisely, does there exist an invertible matrix $P$ such that $P^{-1} A P$ is positive definite?

Answer 1

Here's a special case in which a complete answer may be given, and hopefully this leads to better understanding in the general case. We know that there is a real Jordan canonical form available for any real matrix. Suppose that the matrix leading to the Jordan canonical form of $A$ is real orthogonal. Then $A=O^{T}JO,$ where $J$ is the real Jordan canonical form, which means that $x^{T}Ax=(Ox)^{T}J(Ox)$ for all $x\in\mathbb{R}^{n}.$ Thus, the positive definiteness of $A$ is equivalent to that of $J$ in this case. Now if we inspect the Jordan blocks for some eigenvalue, we have four cases:

1. $[\lambda]$, $\lambda\in\mathbb{R}$, $\lambda>0;$
2. $\begin{bmatrix}\lambda&1&\cdots&0\\0&\lambda&\cdots&0\\ \vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&\lambda\end{bmatrix}$, $\lambda\in\mathbb{R},\lambda>0$ ($1$s only on the first superdiagonal);
3. $\begin{bmatrix}a&b\\-b&a\end{bmatrix}$, $a,b\in\mathbb{R},$ $a>0$;
4. $\begin{bmatrix}\begin{bmatrix}a&b\\-b&a\end{bmatrix}&I&\cdots&0\\0&\begin{bmatrix}a&b\\-b&a\end{bmatrix}&\cdots&0\\ \vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&\begin{bmatrix}a&b\\-b&a\end{bmatrix}\end{bmatrix}$, $a,b\in\mathbb{R}$, $a>0$ ($I$s only on the first superdiagonal).

It should be clear that $J$ is positive definite if and only if all blocks are of types (1) and (3).

Answer 2

The fact that $U\in M_n(\mathbb{R})$ satisfies for every $x$ s.t. $||x||=1$, $x^TUx>0$, is denoted by $U>0$; if moreover $U$ is symmetric, then the notation is $U\succ 0$.

Recall $(*)$: $U>0$ IFF $U+U^T\succ 0$.

$\textbf{Proposition}.$ Let $A\in M_n(\mathbb{R})$ s.t. for every $\lambda\in spectrum(A)$, $Re(\lambda)>0$. Then there is $B\in M_n(\mathbb{R})$ s.t. $A,B$ are similar over $\mathbb{R}$ and $B>0$.

$\textbf{Proof}.$ Let $spectrum(A)=(\nu_j),(\lambda_j \pm i\mu_j)_j$ where $\nu_j>0,\lambda_j>0,\mu_j\not= 0$. $A$ is similar over $\mathbb{R}$ to a block diagonal matrix where each block is

i) either in the form $\nu_j I_r+J_r$ where $J_r$ is a $r\times r$ nilpotent Jordan matrix

ii) or in the form (for $r=6$) $\begin{pmatrix}U&I_2&0_2\\0_2&U&I_2\\0_2&0_2&U\end{pmatrix}$ where $U=\begin{pmatrix}\lambda_j&\mu_j\\-\mu_j&\lambda_j\end{pmatrix}$.

Then (cf. $(*)$), it suffices to prove the required result for the two above matrices.

Case i). $\nu I+J$ is similar to $B=\nu I+\epsilon J$ ($\epsilon >0$) and $B+B^T=2\nu I+\epsilon (J+J^T)\succ 0$ (as $2\nu I$) for $\epsilon$ small enough.

Case ii). Our matrix $\tilde{A}$ (with $\lambda +i\mu$ and $r=6$)

$(**)$ is similar to $B=\begin{pmatrix}U&\epsilon I_2&0_2\\0_2&U&\epsilon I_2\\0_2&0_2&U\end{pmatrix}$

and $B+B^T=diag(2\lambda I_2,2\lambda I_2,2\lambda I_2)+\epsilon K$ which is $\succ 0$ (as $diag(2\lambda I_2,2\lambda I_2,2\lambda I_2)$ for $\epsilon$ small enough. $\square$

EDIT. About $(**)$. Let $\mathcal{A}=\tilde{A}-(\lambda+i\mu)I_6$, $\mathcal{B}=B-(\lambda+i\mu)I_6$, $e_1=[-i,1,0,0,0,0]^T,e_2=[0,0,-i,1,0,0]^T,e_3=[0,0,0,0,-i,1]^T$. Then $\ker(\mathcal{A})=\ker(\mathcal{B})=span(e_1)$,$\ker(\mathcal{A}^2)=\ker(\mathcal{B}^2)=span(e_1,e_2)$,$\ker(\mathcal{A}^3)=\ker(\mathcal{B}^3)=span(e_1,e_2,e_3)$.