I am a bit confused about the following proof, and have a question about the last step:
First, assume there is more than one solution.
Suppose $(c_1, c_2, …, c_n), (d_1, d_2, …, d_n)$ are two different solutions.
Then $(d_1-c_1, …, d_n – c_n)$ is a homogenous system.
$(t(d_1-c_1), …, t(d_n – c_n))$ is also a solution to the homogenous system.
$(c_1+t(d_1-c_1), …, c_n+t(d_n – c_n))$ is a solution to the original system (infinite).
My confusion occurs at the last statement. I understand that $t$ can be any value, HOWEVER, $t(c-d)$ is equal to zero. So the last step simplifies to $(c_1, …, c_n)$ and we still have nothing!
Also, isn't the third step not proving anything either? All it's doing is basically showing that the only solution is $(0,…,0)$?
Exactly how does the last step prove that the system of linear equations has infinite solutions?
Best Answer
I'll do $2 \times 2$ here. Consider: $$\begin{cases} a_{11}x + a_{12}y = A \\ a_{21}x + a_{22}y = B\end{cases}$$ Suppose $(c_1, c_2)$ and $(d_1,d_2)$ are two different solutions. Let $t \in \Bbb R$. We have that $(c_1 + t(d_1 - c_1), c_2 + t(d_2 - c_2))$ is a solution. In details: \begin{align}a_{11}(c_1 + t(d_1 - c_1)) + a_{12}(c_2 + t(d_2 - c_2)) &= a_{11}c_1 + ta_{11}(d_1 - c_1) + a_{12}c_2 + ta_{12}(d_2 - c_2) \\ &=a_{11}c_1 + a_{12}c_2 - t(a_{11}c_1 + a_{12}c_2) + t(a_{11}d_1 + a_{12}c_2) \\ &= A - tA + tA \\ &= A\end{align}
In the same way you show that: $$a_{21}(c_1 + t(d_1 - c_1)) + a_{22}(c_2 + t(d_2 - c_2)) =B$$ Can you get a better picture of what's happening now?