# 4 questions on invertible matrices

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.

• To show C is wrong, all you need is one example. Heck, all you need is to think about $C(0)$, or $C(1)$. – Gerry Myerson Feb 26 at 7:53
• 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. – Huny Feb 26 at 8:05
• 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 $-1$. – Bungo Feb 26 at 8:09
• Huh? C says $C(t)$ is not invertible etc., etc., so $C(0)$ being invertible is a counterexample, not a confirmation, of C, Huny. – Gerry Myerson Feb 26 at 8:11
• 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 $t \in [0,1]$". – Bungo Feb 26 at 8:15

Let $$A=1$$ and $$B=-1$$.
$$C(t)=tA+(1-t)B=t-(1-t)=2t-1$$
Clearly $$C(0)$$ is invertible and $$C(t)$$ is invertible as long as $$t\ne \frac12$$.
To construct counterexample for arbitrary $$n$$, let $$A=\operatorname{diag}(1, I_{n-1}), B=\operatorname{diag}(-1, I_{n-1})$$