Dimension of image and kernel? If W is a subspace of $\mathbb R^4$ given by $W=\{(x,y,z,w):x+z+w=0,y+z+w=0\}$ . Then what will be the dimension of image and kernel of W. 
I just learned that dimension of $W=\text{number of independent variables}-\text{number of constraints}=4-2=2$. I amcurious to know abou the dimension of kernel and image.
 A: You need a linear function for a kernel and an image. A set doesn't have one. If you consider $A=\begin{pmatrix}1 & 0 & 1& 1\\0&1&1&1\end{pmatrix}$ then it induces a linear map $L:\mathbb R^4\to\mathbb R^2$ defined by $$
L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=A\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}x+z+w\\y+z+w\end{pmatrix}.
$$
For this linear map we have a kernel
\begin{align}
ker(L)&=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in \mathbb R^4~:~L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\right\}=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in\mathbb R^4~:~\begin{pmatrix}x+z+w\\y+z+w\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\right\}\\
=&=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in\mathbb R^4~:~x+z+w=0,y+z+w=0\right\}
\end{align}
You see, this is your set $W$. Now consider $L$. You might can use the following formula
$$
\dim(W)=\dim(ker(L))=\dim(domain(L))-\text{rank}(L)=\dim(\mathbb R^4)-\text{rank}(A)=4-2=2.
$$
The rank of $A$ is the maximal linear amount of linear independent rows.
And the image of $L$ is given by
$$
image(L)=\left\{\begin{pmatrix}a\\b\end{pmatrix}\in\mathbb R^2~:~\text{there exists }\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\text{ such that }L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}\right\}.
$$
The dimension of $image(L)$ is such the rank of $A$ and therefore it can't be greater than the dimenion of the codomain. Here it is $2$ and since $\dim(\mathbb R^2)=2$ you get $image(L)=\mathbb R^2$.
