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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?

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  • $\begingroup$ "Positive real part" - they are real, since $x^TAx=\lambda |x|^2>0$ implies $\lambda$ is real. $\endgroup$
    – A.Γ.
    Jul 26, 2018 at 15:21
  • $\begingroup$ @A.Γ. $x^TAx > 0$ only has to hold for real $x$ in this context. If $\lambda$ is a complex eigenvalue, then there is no real vector $x$ which is an eigenvector. $\endgroup$ Jul 26, 2018 at 15:31
  • $\begingroup$ @Omnomnomnom Indeed, I see it now. $\endgroup$
    – A.Γ.
    Jul 26, 2018 at 23:53
  • $\begingroup$ @RideTheWavelet Again, this is not true since the similar matrix may have non-real eigenvalues while still being positive definite (as in the definition above). $\endgroup$
    – smalldog
    Jul 27, 2018 at 7:04
  • $\begingroup$ Thanks Nao, I think instinct took over when I saw the phrase "positive definite." $\endgroup$ Jul 27, 2018 at 7:39

2 Answers 2

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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 $\ker(\mathcal{A})=\ker(\mathcal{B})=span(e_1)$,$\ker(\mathcal{A}^2)=\ker(\mathcal{B}^2)=span(e_1,e_2)$,$\ker(\mathcal{A}^3)=\ker(\mathcal{B}^3)=span(e_1,e_2,e_3)$.

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  • $\begingroup$ Could you give a reference to why the Jordan blocks are given by (i) or (ii), and why similarity with an epsilon perturbation holds? I'm not sure I understand the argument. $\endgroup$
    – smalldog
    Jul 31, 2018 at 9:06
  • $\begingroup$ Firstly, don't say thanks for the answer. Secondly for i), see wiki and for ii), its your business (it seems to me that you have not done much work so far). $\endgroup$
    – user91684
    Jul 31, 2018 at 10:05
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    $\begingroup$ I had the impression that saying "thanks" was not recommended, as suggested by virtually all comments I've seen so far. I also think it's a little rude and I usually can't help it, as for the answer above. Am I in the wrong impression? In any case: thanks. You are equally welcome to be a little more polite, as I am doing my best in understanding the argument and working through this problem myself. Wikipedia is the first place I looked, but the article on Jordan normal form does not seem to invoke matrices like $U$ in your case (ii). $\endgroup$
    – smalldog
    Aug 1, 2018 at 8:14
  • $\begingroup$ I just read the @RideTheWavelet 's answer. I wrote too fast. He is right; it is necessary to use an orthonormal change of basis. In a moment, I'll delete my post. $\endgroup$
    – user91684
    Aug 3, 2018 at 15:29
  • $\begingroup$ Why is it necessary? Where is the flaw in your answer? $\endgroup$
    – smalldog
    Aug 3, 2018 at 16:00
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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).

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  • $\begingroup$ Thanks for your answer. I don't think this answers the question, since there may be another matrix similar to $A$ which is positive definite, even if it fails for $J$. Also, the assumption that $O$ is orthogonal seems superficial since $A$ is not assumed positive definite, so equivalence between positive definiteness of $A$ and $J$ does not help...? $\endgroup$
    – smalldog
    Jul 27, 2018 at 8:34

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