Triangle made from full rank matrices

Question

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 $t_1+t_2+t_3=1$.

@JeanMarie Ok just plane, two dimensional however, I suppose ( 3 points) — Widawensen, Mar 02 '17 at 12:27
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). — Jean Marie, Mar 02 '17 at 12:55
@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.. — Widawensen, Mar 02 '17 at 12:59
@JeanMarie In this question I'm simply interested whether results from the linked previous question can be somehow extended into more complicated case ( two-dimensional plane instead of one-dimensional line) — Widawensen, Mar 02 '17 at 13:13
It might be notable that the matrices of deficient rank form an $(n^2-1)$-dimensional variety in $\Bbb R^{n \times n}$. — Ben Grossmann, Mar 02 '17 at 13:42
@Omnomnomnom Thank you for this extending knowledge comment.. — Widawensen, Mar 02 '17 at 17:18

Ben Grossmann · Accepted Answer · 2017-03-02T13:42:53.737

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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 $$ \frac 13 (A_1 + A_2 + A_3) = 0 $$

edited Mar 02 '17 at 13:42

answered Mar 02 '17 at 13:37

Ben Grossmann

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..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.. – Widawensen Mar 02 '17 at 17:14
@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 – Ben Grossmann Mar 02 '17 at 17:21
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.. – Widawensen Mar 02 '17 at 17:28
@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 – Ben Grossmann Mar 02 '17 at 17:35
o.k. I'll think it over and post some day a problem about singularity. For now I just wanted to grasp some intuition about this variety which was mentioned by you... Maybe you know some additional links, all new pointers could be helpful.. – Widawensen Mar 02 '17 at 17:40
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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). – Ben Grossmann Mar 02 '17 at 17:55

score 1 · Answer 2 · answered Mar 02 '17 at 13:49

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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.

answered Mar 02 '17 at 13:49

Jean Marie

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Triangle made from full rank matrices

2 Answers2