Let $A$ and $B$ be $n \times n$ real matrices such that det$(A) > 0$ and det$(B) < 0$. For $0≤ t ≤ 1$ let $C(t) = tA + (1-t)B$

Question

Let $A$ and $B$ be $n \times n$ real matrices such that det$(A) > 0$ and det$(B) < 0$. For $0≤ t ≤ 1$ let $C(t) = tA + (1-t)B$. Then there exists exactly one $t_{0}$ in $(0,1)$ such that $C(t_{0})$ is not invertible. (True/false)

I took two matrices satisfying the given conditions. For $t=0.5$, I'm getting the matrix $C$ invertible.

I want to know if there exists only one such $t_{0}$ or more$?$

i think that the thesis is false. Yes could be exist A, B such that exist more than one $t_0$ — tommaso faustini, Aug 28 '19 at 15:42
Hint: To show there is at least one such $t$ note that $\det(C(t))$ is a continuous function of $t$... — David C. Ullrich, Aug 28 '19 at 15:54

tommaso faustini · Accepted Answer · 2019-08-28T15:55:03.007

6

$A=\begin{pmatrix}1 & 0 & 0 \\ 0&1 &0 \\ 0 & 0 & 1\end{pmatrix}$ $B=\begin{pmatrix}-1 & 0 & 0 \\ 0&-1 &0 \\ 0 & 0 & -2\end{pmatrix}$

With $t_1=\frac{1}{2}$ and $t_2=\frac{2}{3}$ we have that $C$ is not invertible.

So the thesis seems false.

You can easily generalize this putting $1$ along the diagonal.

edited Aug 28 '19 at 15:55

answered Aug 28 '19 at 15:50

tommaso faustini

510

t should be in between 0 and 1. – Mathaddict Aug 28 '19 at 15:52
I'm sorry it was misstyping. @Mathsaddict – tommaso faustini Aug 28 '19 at 15:53
You can generalize for even $n\geq4$ as well, by just putting $1$ along the diagonal of $B$ after the initial $-1,-1,-2$. – Arthur Aug 28 '19 at 15:54
yes, thank you for the hint @Arthur – tommaso faustini Aug 28 '19 at 15:56
Thanks @tommaso One more thing I want to know, do there exist finitely many such $t_{0}$ or these may be infinte$?$ – Mathaddict Aug 28 '19 at 16:02
@Mathsaddict As regards diagonal matrices there are only finitely many such t. More generally it seems to me that the number of possible values of t is limited by n, and thus is finite,but this should be proved – tommaso faustini Aug 28 '19 at 16:08
1

The determinant is a degree-$n$ polynomial in $t$. Thus it can at most have $n$ roots. – Arthur Aug 28 '19 at 16:11

Let $A$ and $B$ be $n \times n$ real matrices such that det$(A) > 0$ and det$(B) < 0$. For $0≤ t ≤ 1$ let $C(t) = tA + (1-t)B$

1 Answers1