What is the interpretation of the difference? We have the linear maps \begin{equation*}f:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+x_2 \\ x_3 \\ x_1+x_2\end{pmatrix} \ \text{ and } \ h:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+ x_2 \\ x_3 \\ x_1-x_2\end{pmatrix}\end{equation*} 
We also have the vectors \begin{equation*}\vec{v}:=\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} \ \text{ und } \ \vec{w}:=\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\end{equation*} 
I want to check if $f(\vec{v}$ and $f(\vec{w})$ are linearly (in)dependent and then the same for $h(\vec{v})$ and $h(\vec{w})$ and then I want to interpret the difference. 
With $f$ we get \begin{equation*}f(\vec{v})=f\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \ \ \text{ and } \ \ f(\vec{w})=f\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}\end{equation*} 
The vectors $f(\vec{v})$ and $f(\vec{w})$ are linearly dependent, since $f(\vec{v})=0 \cdot f(\vec{w})$. 
With $h$ we get \begin{equation*}h(\vec{v})=h\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix} \ \ \text{ and } \ \ h(\vec{w})=h\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}\end{equation*} 
The vectors $h(\vec{v})$ and $h(\vec{w})$ are linearly independent, since the one vector has zeroes and the other one not, correct? 
What is the interpretation of the difference? 
 A: The matrices corresponding to the linear transformations $f,h$ are $$\begin{bmatrix}1&1&0\\0&0&1\\1&1&0\end{bmatrix},\begin{bmatrix}1&1&0\\0&0&1\\1&-1&0\end{bmatrix}$$
respectively. Obviously the determinant of the former is $0$ since there are two identical (and hence linearly dependent) rows, while you can show that the rows of the latter are linearly independent and hence the determinant is nonzero. What this means is that $h$ represents an injective (i.e. invertible) transformation while $f$ does not. So $f$ can and must take linearly independent vectors in its domain and "collapse" them to become dependent, since the dimension of its image is smaller than its domain, and conversely $h$ will take linearly independent vectors in its domain to linearly independent vectors in its codomain, which you can prove by considering the basis vectors.
A: Yes it is correct, the interpretation is that $v$ belongs to the nullspace of $f$ while by $h$ vectors $v$ and $w$ are mapped into two linearly independent vectors.
