Kernel and image of a linear map (with parameter) Let $T: \mathbb{R^3} \to \mathbb{R^4}$ such that
$f(1,1,0) = (1,h,1,0)$
$f(0,2,0) = (1,h,1,0)$
$f(0,1,-1) = (h,2,1,1)$
I have to determine the kernel and the image of $T$ for $h \in \mathbb{R}$. I've done some work, but I must have got wrong somehow, because I get


*

*$Im(T)=<(1,h,1,0), (h,2,1,1)>$

*$Ker(T)=<(-1,1,0)>$
regardlessly of $h$. Can you show me how to solve this correctly?
 A: Your results are correct.
Note that your expression for the image does contain $h$, so there's no reason to be unhappy with that.
That the kernel is the same no matter what $h$ is is just a fact about this particular family of $T$s. Note that the matrix representation of $T$ is
$$\begin{align}\begin{bmatrix} 1 & 1 & h \\ h & h & 2 \\ 1 & 1& 1 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & -1 \end{bmatrix}^{-1} &=
\begin{bmatrix} 1 & 1 & h \\ h & h & 2 \\ 1 & 1& 1 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ -\frac12 & \frac12 & \frac12 \\ 0 & 0 & -1 \end{bmatrix} \\[1ex]
&= \begin{bmatrix} 1 & h \\ h & 2 \\ 1& 1 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ -\frac12 & \frac12 & \frac12 \\ 0 & 0 & -1 \end{bmatrix} \\[1ex]
&= \begin{bmatrix} 1 & h \\ h & 2 \\ 1& 1 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} \frac12 & \frac12 & \frac12 \\ 0 & 0 & -1 \end{bmatrix}
\end{align}$$ 
and in the last form both factors have full rank, so the image depends only on the left fractor and the kernel only on the right factor (which happens to contain no $h$s).
