How do i read if a transformation is bijective, surjective or injective based on reduced row echelon form? i am looking for a easy way to read if a transformation is bijective, surjective or injective based solely on the reduced echelon form? someone told me that this was possible, but i cannot figure it out.
I have the following transformation:
$$
T(X) = \left(
  \begin{array}{c}
      x_{1} \\
      x_{2} \\
      x_{3}
  \end{array}
       \right) =  \left(
  \begin{array}{c}
      -x_{1}-3x_{2}+x_{3} \\
      3x_{1}+4x_{2}+2x_{3} \\
      x_{2}-x_{3}
  \end{array}
  \right) for x = \left( 
      \begin{array}{c}
   x_{1}\\
   x_{2}\\
   x_{3}
      \end{array}
  \right) \in \mathbb{R}^{3}
$$
Then i have the reduced echelon form:
    $$
 \left( 
     \begin{array}{ccc|c}    
   1 & 0 &  2 \\
   0 & 1 & -1 \\
   0 &  0 & 0 \\
     \end{array}
 \right)
 $$
I have done a lot of calculations so i have the null-space aswell, the column space, i added a null vector to my reduced echelon form and solved for all sorts of x's.. but i am not sure what kind of information you need to decide whether something is injective, surjective or bijective. 
 A: Is the null space of the matrix trivial(consisting only of the zero vector)? If yes, the linear transformation is injective. Otherwise it is not injective.
Next for a linear transformation $T:\mathbb{R}^n\to \mathbb{R}^m$, we have that
$$
n=\dim(\ker T)+\dim \,(\text{im}\, T)\tag{1}
$$
Based on the null space you can determine $\dim \,(\text{im}\, T)$ from (1). Then T is surjective iff $\dim \,(\text{im}\, T)=m$
A: Recall that by definition, the transformation is injective if $T(X)=T(X')$ implies $X=X'$. Now, check that this is not true in the case $X=(2,-1,0)$ and $X'=(0,0,1)$. Looking at the reduced echelon form, can you deduce how I came up with those vectors?
For surjectivity, you want the matrix to have full rank, as somebody else has already said. The reason for this is that the rank is the dimension of the image of the transformation, so if the rank is equal to $3$, then the dimension of the image is $3$, so it must give us all of $\Bbb R^3$. So this transformation is definitely not surjective either. To find an explicit example of a vector not lying in the range, take the images of $(1,0,0)$ and $(0,1,0)$, which are $Y_1=(-1,3,0)$ and $Y_2=(3,4,1)$, and find some vector not lying in the span of these vectors. For instance, you can take $Y$ to be the cross product $Y=Y_1\times Y_2$, and check that $T(X)=Y$ has no solutions.
It's hard at first to see what row echelon form is supposed to be telling you, but essentially the most important thing is that in our case, the reduced echelon form is telling you that we have the relation $2T(1,0,0)-T(0,1,0)=T(0,0,1)$ between the images of the standard basis vectors (and you can check for yourself that this is actually true).
A: Hints: Fill in details in the following
Your linear map is represented by a square matrix, and in fact $\;T:\Bbb R^3\to\Bbb R^3\;$ . Now
(1) $\;T\;$ is injective iff it is surjective iff it is bijective
(2) $\;T\;$ is injective iff its matrix (wrt any basis) has full rank, and in this case: if its matrix has rank three.
(3) After being fully reduced by rows, a matrix has rank $\;r=\,$ the number of non-zero rows in the reduced form
