Let $A,B\in M_n(\mathbb R)$ $\det(A)>0$ and $\det(B)<0$. For $t\in[0,1]$, consider $C(t)=tA+(1-t)B$.

Then $(a)$ $C(t)$ is invertible $\forall t\in[0,1]$.

$(b)$ $\exists t_{0}\in(0,1)$ s.t. $C(t_{0})$ is not invertible.

$(c)$ $C(t)$ is not invertible for each $\forall t\in[0,1]$.

$(d)$ $C(t)$ is invertible for only finitely many $t\in[0,1]$.

I got the answer that option $(b)$ is correct but I couldn't understand why $(c)$ and $(d)$ are incorrect. Please explain.

  • 1
    $\begingroup$ To show C is wrong, all you need is one example. Heck, all you need is to think about $C(0)$, or $C(1)$. $\endgroup$ – Gerry Myerson Feb 26 at 7:53
  • $\begingroup$ If B is correct then Amust be incorrect and then C must be correct. $C$ is convertible at $0$ and $1$ but by taking the example of an identity matrix and the transpose of that matrix, we can prove that B is correct. Then C and D must be correct but they ain't. And that is my question. $\endgroup$ – Huny Feb 26 at 8:05
  • $\begingroup$ Why would correctness of B imply correctness of C? Also, you can't show correctness of B by one example. You have to show that is true for any $A$,$B$ that meet the conditions. (Whereas you can show incorrectness of A, B, and D by providing a single counterexample for each one.) Can you clarify your example? The transpose of the identity is the identity, which doesn't have determinant $-1$. $\endgroup$ – Bungo Feb 26 at 8:09
  • $\begingroup$ Huh? C says $C(t)$ is not invertible etc., etc., so $C(0)$ being invertible is a counterexample, not a confirmation, of C, Huny. $\endgroup$ – Gerry Myerson Feb 26 at 8:11
  • $\begingroup$ The wording of C is a bit ambiguous; perhaps that's the issue? I assume it means to say that "for each $t \in [0,1]$, $C(t)$ is not invertible" as opposed to the possible alternative reading "it is not true that $C(t)$ is invertible for each $t \in [0,1]$". $\endgroup$ – Bungo Feb 26 at 8:15

Let $A=1$ and $B=-1$.


Clearly $C(0)$ is invertible and $C(t)$ is invertible as long as $t\ne \frac12$.

To construct counterexample for arbitrary $n$, let $A=\operatorname{diag}(1, I_{n-1}), B=\operatorname{diag}(-1, I_{n-1})$

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  • $\begingroup$ Thanks! Actually i also wanted to ask if Options c and d could be correct? $\endgroup$ – Huny Feb 26 at 8:09
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    $\begingroup$ note that determinant is a continuous function, by intermediate value theorem, no. $\endgroup$ – Siong Thye Goh Feb 26 at 8:11
  • $\begingroup$ Okay, got it! Thanks. $\endgroup$ – Huny Feb 28 at 7:59

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