The question is related to the question Transition between matrices of full rank

Suppose we have in matrix space (I treat matrices here as vectors describing points in some $n \times n$ dimensional space) three real square matrices $A_1,A_2,A_3$ that all are of full rank and any matrix $P$, lying on the segment $A_1A_2$ or $A_2A_3$ or $A_3A_1$ between these matrices, is also of full rank.

(what is equivalent to the fact that this matrix $P = t_i{A_i}+{t_iA_j}$ where $t_i,t_j$ are positive and $t_i+t_j=1$)

  • Does it mean that any matrix $D$ in the interior of $\triangle ABC$ located in the two-dimensional plane determined by these matrices is also of full rank?
  • If not what condition should be stated additionally to satisfy non-singularity for all these internal matrices?

Matrix $D$ is treated as an internal point of $\triangle ABC$ if the equation $D= t_1A+t_2B+t_3C$ is satisfied for some positive $t_1$, $t_2$, $t_3$ constrained by the equation $t_1+t_2+t_3=1$.

  • $\begingroup$ 1) There is no reason that the full rank property is preserved ; there are many counter-examples. 2) little remark: It is not a hyperplane (dimension $n^2-1$) but a 3 dimensional subspace of the vector space of $n \times n$ matrices. $\endgroup$ – Jean Marie Mar 2 '17 at 12:25
  • $\begingroup$ @JeanMarie Ok just plane, two dimensional however, I suppose ( 3 points) $\endgroup$ – Widawensen Mar 2 '17 at 12:27
  • $\begingroup$ No: the set of matrices of the form $t_1A+t_2B+t_3C$ is 3-dimensional. (think to $A$, $B$,$C$ as there vectorized equivalent form as $1 \times n^2$ long vectors). $\endgroup$ – Jean Marie Mar 2 '17 at 12:55
  • $\begingroup$ @JeanMarie But they are additionally constrained by $t_1+t_2+t_3=1$ what gives I suppose two-dimensionality.. think of analogy with two points in 3d space - they determine one-dimensional line, it's not required that point $(0,0,0)$ belongs to that line.. $\endgroup$ – Widawensen Mar 2 '17 at 12:59
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    $\begingroup$ It might be notable that the matrices of deficient rank form an $(n^2-1)$-dimensional variety in $\Bbb R^{n \times n}$. $\endgroup$ – Ben Grossmann Mar 2 '17 at 13:42

The answer to this question is no. We can take an example directly from the complex numbers, effectively: consider $$ A_1 = I, \quad A_2 = \pmatrix{\cos 2\pi/3 & -\sin 2 \pi /3\\ \sin 2 \pi /3 & \cos 2 \pi /3}, \quad A_3 = \pmatrix{\cos 4\pi/3 & -\sin 4 \pi /3\\ \sin 4 \pi /3 & \cos 4 \pi /3} $$ where $I$ denotes the identity matrix. Verify that all matrices on the segment connecting $A_i,A_j$ are invertible (in particular, it is useful to note that $\det (\begin{smallmatrix} a&-b\\b&a \end{smallmatrix}) = a^2 + b^2$). However, we find that $$ \frac 13 (A_1 + A_2 + A_3) = 0 $$

  • $\begingroup$ ..hmm... so it seems that for 3 points situation wrt preserving non-singularity is much worse that for two points... it's hard to find criterion ...probably only some additional constraints (?) could guarantee that all matrices-"points in the region" would be non-singular.. $\endgroup$ – Widawensen Mar 2 '17 at 17:14
  • $\begingroup$ @Widawensen I don't think that we can be so lucky as to really leverage the two-matrix case. You should post something about the full problem $\endgroup$ – Ben Grossmann Mar 2 '17 at 17:21
  • $\begingroup$ the full problem is in avoiding singular matrices when we generate some real matrix or a set of matrices (for example from experimental data) and we want to know how they are distanced from singularity.. $\endgroup$ – Widawensen Mar 2 '17 at 17:28
  • $\begingroup$ @Widawensen that's not really enough information to go on. Anyway, I'm saying you might get better results if you made a new question about that directly, explaining what you're trying to accomplish $\endgroup$ – Ben Grossmann Mar 2 '17 at 17:35
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    $\begingroup$ Don't really have much besides what I gave you. A result you might find useful is as follows: the set of positive definite matrices is convex. So, if each of the $A_i$ is positive definite, then the matrices on the interior of their convex hull are also positive definite (and therefore have full rank). $\endgroup$ – Ben Grossmann Mar 2 '17 at 17:55

Another counter-example:

$$\pmatrix{2 & 2\\ 2 & 2}=\dfrac{1}{3}\left(\pmatrix{3 & 0\\0 & 3}+\pmatrix{3 & 3\\3 & 0}+\pmatrix{0 &3 \\3 & 3}\right)$$

is rank one, whereas the other matrices are rank-2.


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