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Prove or give a counterexample:If $ad-bc\neq0$ and $ax_n+by_n \rightarrow L$ and $cx_n+dy_n \rightarrow M$, then $x_n$ and $y_n$ both converges.

My proof:

I found a counterexample for this problem. If we let $x_n=\frac{n}{2}$ and $y_n=-n$, then we can have $a=2, b=1$. Then, $ax_n + by_n$ converges but $x_n$ and $y_n$ are divergent.

I am convinced my counterexample is valid, but the problem giving me $ad-bc \neq0$ makes me worried I am not doing this correctly. I feel like I have to use that inequality somehow.

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    For a counterexample, you need to specify $a,b,c,d$, not just $a,b$. – 2017-01-18
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    And then you need to check that, for some real numbers $L,M$ both of the limits $ax_n+by_n \rightarrow L$ and $cx_n+dy_n \rightarrow M$ hold. – 2017-01-18
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    Hint: $ad - bc \ne 0$ means that for every $r$ and $s$, the equations $au + bv = r; cu + dv = s$ have a unique solution $(u, v)$. In fact, you can write $u$ and $v$ in terms of $a, b, c, d, r, s$ in a pretty simple way. – 2017-01-18
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    @quasi Then can't I made $c=4$ and $d=2$? – 2017-01-18
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    @quasi I see, because for my case, the sequences will all converge to 0. – 2017-01-18
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    @John Huges Then I must replace $u$ and $v$ with sequences $x_n$ and $y_n$? But isn't your statement only valid for constants $u,v$? Would this also work for sequences as given in the problem? – 2017-01-18
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    @user3000482 If you take $c=4$ and $d=2$, then $ad-bc=0$. – 2017-01-19

2 Answers 2

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Let $n \in \{1 , 2 , \ldots\}$ and we consider $$ X_n = a x_n + b y_n \qquad Y_n = c x_n + d y_n\mbox{.} $$ We know that $\lim_{n \to \infty} X_n = L$ and $\lim_{n \to \infty} Y_n = M$, so $$ \lim_{n \to \infty} \left(d X_n - b Y_n\right) = d L - b M\mbox{,} \tag{1} $$ but obviously $$ d X_n - b Y_n = (a d - b c) x_n \quad \mbox{ for all } \quad n = 1 , 2 , \ldots \tag{2} $$ Then, from $(1)$ and $(2)$, you obtain that $$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{d X_n - b Y_n}{a d - b c} = \frac{d L - b M}{a d - b c} \in \mathbb{R} $$ because $a d - b c \neq 0$. Equally, $$ \lim_{n \to \infty} \left(c X_n - a Y_n\right) = c L - a M\mbox{,} \tag{3} $$ but obviously $$ c X_n - a Y_n = (a d - b c) y_n \quad \mbox{ for all } \quad n = 1 , 2 , \ldots \tag{4} $$ Then, from $(3)$ and $(4)$, you obtain that $$ \lim_{n \to \infty} y_n = \lim_{n \to \infty} \frac{c X_n - a Y_n}{a d - b c} = \frac{c L - a M}{a d - b c} \in \mathbb{R} $$ because $a d - b c \neq 0$. Finally, ${\{x_n\}}_{n = 1}^{\infty}$ and ${\{y_n\}}_{n = 1}^{\infty}$ both converge.

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    But aren't you making assumptions that $x_n$ and $y_n$ are convergent in the first sentence? – 2017-01-18
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    Now not, I only use that ${\{X_n\}}_{n = 1}^{\infty}$ and ${\{Y_n\}}_{n = 1}^{\infty}$ converge. – 2017-01-19
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$$\text {We have } \quad \binom {a\;\;b}{c\;\;d} \binom {x_n}{y_n}= \binom {L+e_n}{M+f_n}$$ where $e_n \to 0$ and $f_n\to 0$ as $n\to \infty.$ Since $ad-bc\ne 0$ we have $$\binom {x_n}{y_n}=(ad-bc)^{-1} \binom {\;\;\;d\;\;\;-b}{-c\;\;\;\;\;a} \binom {L+e_n}{M+f_n}=$$ $$=(ad-bc)^{-1}\binom {\;\;\;d\;\;\;-b}{-c\;\;\;\;a}\binom {L}{M}+(ad-bc)^{-1}\binom {\;\;\;d\;\;\;-b}{-c\;\;\;\;a}\binom {e_n}{f_n}.$$