How many solutions can an onto matrix have and how many can one-to-one matrix have?
I think that Onto has one or infinity many and One-to-one can have only one unique solution but I'm not sure if this is correct. Also, must we have a solution for Onto and One-to-one or can we not have any solutions?
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$\begingroup$I will interpret your question in the following way:
Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ be a linear map. Then there is an $m\times n$ matrix $M$ such that $Mv = T(v)$ for each $v \in \mathbb{R}^n$.
Let $y \in \mathbb{R}^m$. We want to determine the number of solutions to $Mx = y$.
First, suppose $T$ is one-to-one. (Then we call $M$ a one-to-one matrix.) First we can prove there is at most one solution. Let $a, b \in \mathbb{R}^n$. Suppose $Ma = Mb = y$. Then $T(a) = Ma = Mb = T(b)$, so $a=b$ as $T$ is one-to-one. It is also possible that there are no solutions - we can use a small example: Consider the case where $T:\mathbb{R}\rightarrow\mathbb{R}^2$ is defined by $T(x) = \pmatrix{x\\x}$. (Then $M$ is given by $\pmatrix{1\\1}$.) Clearly this is one-to-one, but there are no solutions to the equation $Mx = \pmatrix{1\\0}$.
In summary, if $M$ is one-to-one, there is one solution or no solutions to the equation $Mx = y$ for a given $y$.
With similar reasoning, you can show that if $M$ is onto, there is one solution or infinitely many solutions to the equation $Mx = y$ for a given $y$. (I'll leave this as an exercise for you!)
Note that both of these results combine to give you: If $M$ represents a bijective linear transformation, then there is precisely one solution to the equation $Mx = y$.
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