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I consider a collection of symmetric positive definite matrices of the same dimension. I've learned it's an open set but have no clue about the proof. Also, can the symmetry condition be dropped? Thanks.

Asaf Karagila
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  • no, you cannot drop symmetry. Symmetric matrices make a linear subspace of all matrices, of dimension strictly smaller than $n^2,$ unless $n=1. $ – Will Jagy Apr 07 '14 at 03:56
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    @WillJagy: Positive definiteness is defined for symmetric matrices- so trivially that condition can't be dropped. – voldemort Apr 07 '14 at 03:58
  • I think I remember that from the $LU$ decomposition that the set of positive definite matrices in the space of symmetric matrices is characterized by the conditions $\det(A_k)>0$ where $k=1,\dots,n$ and $A_k$ denotes the top left $k\times k$ submatrix of $A$. Probably this is in Strang's 'Linear algebra'. – Chertopkhanov on Malek Adel Aug 13 '21 at 23:53

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We have to show that if $Q$ is positive definite, then given a small perturbation $E$, the matrix $Q+E$ is still positive definite. For any particular $x$, it is clear that if $||E||$ is small enough (say, less than some $\epsilon_x$), then $x^T(Q+E)x > 0.$ Ideally, we would like to pick the smallest $\epsilon_x$, and we would be done. However, since there are (potentially) infinitely many $\epsilon_x's$, we cannot do that. In this sort of situation, you usually resort to a compactness argument.

In a compactness argument, you usually have two options: a direct proof (with open covers), or a proof by contradiction which involves sequential compactness. I will choose the second option.

Suppose, for a contradiction, that for every positive integer $n$, there exists a matrix $E_n$ with $||E_n|| < \frac 1n$ such that $Q + E_n$ is not PD. Therefore for each $n$, there exists an $x_n$ (which we can assume has norm $1$) such that $$x_n^T(Q+E_n)x_n \le 0.$$ Since the sequence $(x_n)$ is in a compact space (the unit circle) it has a convergent subsequence $x_{n_k}$ which converges to some $x^*$. Therefore $$x_{n_k}^T(Q+E_{n_k})x_{n_k} \le 0.$$

Taking the limit as $k \to \infty$, we get ${x^*}^TQx^* \le 0,$ which contradicts the assumption that $Q$ is P.D.

Ovi
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Hint: any symmetric matrix is diagonalizable with real diagonal entries. Since the matrix is also positive definite, the diagonal entries are positive. So, a little "perturbation" of the diagonal elements still gives you a positive definite matrix, and hence this set is open. Can you formalize this now?

Also, positive definiteness is defined for symmetric matrices, and hence symmetry condition can't be dropped :-P

However, in general its true that set of invertible elements is an open set (even in a Banach algebra). (See here).

voldemort
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Let $A$ be a positive definite matrix, and let $$ \delta = \inf x^T\! Ax > 0, $$ where the infimum is taken over all unit vectors $x$.

Then for any matrix $B$ and for any unit vector $x$, we have $$ |x^T\!Bx|\leq\|B\|\|x\|^2 = \|B\|, $$ so that $$ x^T\!(A+B)x = x^T\!Ax + x^T\!Bx \geq \delta - \|B\|. $$ This shows that if $B$ is small enough then $A+B$ is positive definite.

timur
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Thanks all. Now I see how to proceed using the symmetry assumption.

I was thinking to start from the definition that $P$ is positive definite (p. d.) if $x'Px>0$ for $ x\neq 0$. Suppose $P=\left(\begin{array}{r} 1 & 1\\ -1 & 1\\ \end{array}\right)$. Then $x'Px=x_1^2+x_2^2>0$ unless $x_1=x_2=0$.

Given an arbitrary $x\neq 0$, I consider the map $f(P): x'Px$ with domain as a set of p.d. matrices. The range of $f$ is $R^{+}$ and $f$ is continuous in its argument $P$. So can I say that the set of p. d. matrices is open? Something I missed? Or shouldn't I fix a non-zero $x$?

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