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I have read, even on this website, about connectedness of positive definite matrices. The proof is given in the form of convex equation. I wonder how to understand it intutively. e.g., as an analogy, a parabola is convex and continuous, so, its connectedness is obvious. How to understand positive definite matrices on this line. At first thought, I felt, in the vecor space, the next placed matrix might not be positive definite, as we need all leading principal minors to be positive. I felt, maybe this condition would be satisfied only after certain distance in the vector space. So, I felt these matrices would be disconnected. But of course I was wrong. I wonder how to build intuition for connectedness along these lines. Thanks.

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    $\begingroup$ The set of all $n\times n$ symmetric positive definite matrices, considered as a subspace of $R^{n^2}$ with the Euclidean norm, is path-connected, hence connected. Path-connectedness is easy to see if you work from the definition $x^TAx>0$ for $x\neq0$. $\endgroup$ Dec 25 '16 at 6:21
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    $\begingroup$ You can connect the current matrix to the "next" matrix by a straight line. By the convexity argument, all the matrices along this line are positive definite, too. $\endgroup$
    – bubba
    Dec 25 '16 at 7:44
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    $\begingroup$ You may have a wrong idea of convexity. The graph of $f(x)=x^2$ is not a convex set in the sense that is meant here. $\endgroup$
    – Carsten S
    Dec 25 '16 at 12:34
  • $\begingroup$ @Carsten S- Thanks. I indeed was mixing things up with a convex shape and convex set. Things have become clearer now. Thanks. $\endgroup$
    – aarbee
    Dec 26 '16 at 3:37
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The most straightforward intuition comes from the fact that the set $P$ of positive definite matrices is convex and convex sets are path connected by straight lines.

If $\langle x, A_k x \rangle > 0$ for all $x \neq 0$, then for $\lambda \in [0,1]$, we have $\langle x, (\lambda A_1 +(1-\lambda) A_2 )x \rangle = \lambda \langle x, A_1 x \rangle + (1-\lambda) \langle x, A_2 x \rangle > 0$, hence $\lambda A_1 +(1-\lambda) A_2$ is positive definite.

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A way is to take a positive definite matrix $M=PDP^{T}$ where $D$ is diagonal with all diagonal entries positive and $P$ is an orthogonal matrix of eigenvectors (use the Spectral Theorem). Now, consider the path $\mathcal{M}(t)$ given by

$$\mathcal{M}(t)=P\Big(tI+(1-t)D\Big)P^{T}$$

We have that $\mathcal{M}(0)=M$, $\mathcal{M}(1)=I$ and $\mathcal{M}(t)$ is positive definite for all $t\in[0,1]$. It follows that there is a path connecting each positive definite matrix to the identity, so they form a path-connected (and hence connected) set.

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