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Let $A$,$B$ $ \ $ be two $n \times n$ real matrices of full rank
and
$C$ a matrix generated with segment formula $C= (1-t)A+ tB$, where $0 <t <1 $
(so in some sense $C$ is "between" $A$ and $B$ as an internal point of segment $AB$).

  • Is there a method of checking whether any $C$ is also a full rank matrix $ - $ other one than just writing determinant $\det(C(t))$ and checking whether $det(C(t)) \neq 0$ for all $t$?
  • Maybe if general case is too hard to tackle some method for orthogonal matrices exists?
Widawensen
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1 Answers1

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For $t \ne 1$ you have $$ \frac{1}{1-t} C B^{-1} = A B^{-1} - \frac{t}{t-1} I, $$ where $I$ is an appropriate identity matrix.

So $C$ is not full rank precisely when $\dfrac{t}{t-1}$ hits one of the eigenvalues of $A B^{-1}$.

Andreas Caranti
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  • so it is sufficient to use only eigenvalues of $AB^{-1}$ ?... and that means for example that for orthogonal matrices the only possibility of loosing full rank is for $t=1/2$ ? – Widawensen Mar 01 '17 at 13:08
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    I would say so, yes. – Andreas Caranti Mar 01 '17 at 13:10
  • Andreas, as I understand your great answer the formula provides events of loosing the rank also for any real $t $ not just from interval $(0;1)$ and additionally: the formula can be used starting from eigenvalues also for detection all loosing rank matrices C (with the calculation of $t$) lying on the "line" through $A$ and $B$ ? – Widawensen Mar 01 '17 at 14:57
  • @Widawensen. thx. Yes, I believe this to be the case. – Andreas Caranti Mar 01 '17 at 16:18
  • The 'symmetric' complement to the Andreas' answer - the condition is equivalent to: $\dfrac{t-1}{t}=1-\dfrac{1}{t}$ hits one of the eigenvalues of $BA^{-1}$. On the whole line $AB$ consisting mainly of non-singular matrices it seems there is maximum only $n$ points representing singular matrices.. – Widawensen Mar 01 '17 at 21:24
  • The other conclusion from your formula Andreas is that for $0<t<1$ we have $ 1-(1/t) <0 $ so if happens that $BA^{-1}$ has only positive eigenvalues then for sure $C$ is a matrix of full rank.. Is it a proper conclusion? – Widawensen Mar 02 '17 at 10:18
  • @Widawensen, sorry, I am taken by other things, will possibly consider later. – Andreas Caranti Mar 02 '17 at 11:07
  • I agree with your conclusion. – Andreas Caranti Mar 02 '17 at 11:30