Standard matrix, 1-to-1, onto questions I was sincerely hoping someone could explain to me how for the function found below I would determine its standard matrix and whether or not the function is 1-to-1 and onto. 

The linear function $T:\mathbb{R}^2 \to \mathbb{R}^3$ is given by $$ T(x,y) = \begin{pmatrix} x-y \\ 5x+3y \\ 2x+4y \end{pmatrix} $$



*

*I believe here the standard matrix would be: $\begin{bmatrix} 1 & -1 \\ 5 & 3 \\ 2 & 4\end{bmatrix}$, because I think multiplying that matrix with $(x,y)^T$ would result in that function. I'm not sure about this though.  

*As for the 1-to-1 question; I know that that function is 1-to-1 if each $x \in \mathbb{R}^2$ is related to a different $y \in \mathbb{R}^3$. But how do I test and prove this?  

*As for onto: A function is onto when its image equals its co-domain, so here that would be if $T(x)=y$? But yet again, how would I test/prove this on this function? 


Some very basic questions that I tried googling, but the explanations I found so far did not help much unfortunately. For example, I found explanations for functions like $f(x,y) = x-y$, but none like the one I have here. I also did a lot of looking in the slides from the school course, but that didn't help either. 
 A: Let $T(\mathbf x) = A\mathbf x$, where $A$ is a matrix and $\mathbf x\in\mathbb R^2$ is a vector.

1-to-1: From the equation that $A\mathbf x = A\mathbf y$, try to prove that $\mathbf x = \mathbf y$; if true, then $T$ is 1-to-1.
From $A\mathbf x = A\mathbf y$,
$$A(\mathbf x - \mathbf y) = \mathbf 0,$$
which means it is equivalent to proving whether only $\mathbf v = \mathbf 0$ satisfies $A\mathbf v = \mathbf 0$.
On the other hand, if you find $\mathbf v\ne \mathbf 0$ that satisfies $A\mathbf v = \mathbf 0$, then both $\mathbf v$ and $\mathbf 0$ are counter examples that show why $T$ is not 1-to-1.

Onto: Generally, verify that $A\mathbf x = \mathbf b$ is always consistent for all $\mathbf b\in\mathbb R^3$; if true, then some $\mathbf x$ will map to $\mathbf b$ and so $T$ is onto.
For $\mathbb R^2\to \mathbb R^3$, it is certain that the image of $T$ will not span the whole $\mathbb R^3$, so $T$ is not onto.

Finally, about finding matrix A,
$$T\left(1,0\right) = \begin{bmatrix}1\\5\\2\end{bmatrix}, \quad T\left(0,1\right) = \begin{bmatrix}-1\\3\\4\end{bmatrix}$$
And so for general input $x,y$:
$$\begin{align*}
T\left(x,y\right) &= xT\left(1,0\right) + yT\left(0,1\right)\\
&= x\begin{bmatrix}1\\5\\2\end{bmatrix} + y\begin{bmatrix}-1\\3\\4\end{bmatrix}\\
&= \begin{bmatrix}1&-1\\5&3\\2&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}
\end{align*}$$
A: Suppose
$$T(x,y)=\begin{pmatrix}x-y\\5x+3y\\2x+4y\end{pmatrix}=\begin{pmatrix}a-b\\5a+3b\\2a+4b\end{pmatrix}=T(a,b)\iff \begin{cases}x-y=a-b\\5x+3y=5a+3b\\2x+4y=2a+4b\end{cases}$$
Taking, for example, the first and third equations, you get:
$$\begin{cases}x-a=y-b\\
x-a=2y-2b\end{cases}\stackrel{\text{substract}}\implies\;0=y-b\implies y=b\;\implies x=a$$
and from here $\;T\;$ is $\;1-1\;$ .
Now, is it true that any vector in $\;\Bbb R^3\;$ can be expressed as  $\;\begin{pmatrix}x-y\\5x+3y\\2x+4y\end{pmatrix}\;$ , for some $\,\binom xy\in\Bbb R^2\;$ ?
Clearly not, since if $\;x-y=0\;$ then $\;x=y\;$ , but then 
$$\;5x+3y=5x+3x=8x\;,\;\;2x+4y=2x+4x=6x\;$$ , and thus for example
$$\begin{pmatrix}0\\0\\1\end{pmatrix}\neq\begin{pmatrix}x-y\\5x+3y\\2x+4y\end{pmatrix}\;\;\;\text{for any}\;\;\binom xy\in\Bbb R^2$$
so $\;T\;$ cannot be onto.
When you'll learn some more of this stuff you'll see that your question is almost trivial, as linear maps (matrix is one of this) cannot increase dimension, so a linear map $\;\Bbb R^2\to\Bbb R^3\;$ cannot possibly be surjective. About injectivity this follows at once from rank, kernel and stuff that you shall study later.
