I am working with $\Gamma$, a positive definite matrix of dimension 2, which is not necessarily symmetric. I assume that $\Gamma$ has real eigenvalues and that the eigenvalues are positive (since $\Gamma$ is positive definite).

Up to similarity, I think that there are two families of $\Gamma$:

  • \begin{pmatrix}\lambda_1 & 0\\0&\lambda_2\end{pmatrix} where $\lambda_1$ and $\lambda_2$ are positive scalars.
  • \begin{pmatrix}\lambda & 1\\0&\lambda\end{pmatrix} i.e. $\Gamma$ is similar to a Jordan canonical form where $\lambda$ is positive.

Do you think that I am correct or there exists another family for $\Gamma$ (up to similarity)?

Assuming everything is in $\mathbb{R}$.


You are wrong. $A>0$ does not imply that $P^{-1}AP>0$. Choose for example $A=diag(1,2)$ and $P^{-1}AP=\begin{pmatrix}-1&-6\\1&4\end{pmatrix}$.

In the sequel, I assume that the matrices are real.

On the other hand, for every $x\not= 0$, $x^TAx>0$ IFF for every $x\not= 0$, $x^TP^TAPx>0$ (where $P$ is invertible). Then we must consider congruence classes and not similarity classes.

There are at least $2$ congruence classes for your $>0$ matrices that have $>0$ eigenvalues. That of $I_2$ and for example that of $B=\begin{pmatrix}2&2\\3&4\end{pmatrix}$. The classification is known over $M_n(\mathbb{C})$ (Horn, Sergeichuk) but, over $M_n(\mathbb{R})$, I don't know.

EDIT. About the OP's question, the answer is YES, because any considered matrix (with $>0$ eigenvalues) in $M_n(\mathbb{R})$ is similar to one of the presented two models. Conversely, the first model is $>0$ and the second one is similar to $\begin{pmatrix}\lambda&a\\0&\lambda\end{pmatrix}$, where $4\lambda^2>a^2>0$, which is $>0$.

  • $\begingroup$ Thank you, but maybe my question is not clearly stated. I only deal with similarity, i.e. $P^{-1}AP$ is similar to $A$ but I do not require $P^{-1}AP$ positive definite. $\endgroup$ – TDT Mar 5 '18 at 11:15
  • $\begingroup$ @TDT , I understood that; and I explain above why that has not any interest. $\endgroup$ – loup blanc Mar 5 '18 at 14:35
  • $\begingroup$ I have thought that the result is still correct if I just assume $\Gamma$ has positive eigenvalues. Do you think so? $\endgroup$ – TDT Mar 6 '18 at 9:05
  • $\begingroup$ Yes, of course, and the change of basis matrices are over $\mathbb{R}$. $\endgroup$ – loup blanc Mar 6 '18 at 9:35

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