On the image of some linear maps Let $V$ be an $n$-dimensional vector space with two endomorphisms $f,g:V\to V$. I have a linear map 
\begin{align}
\psi_{f,g}:V\times V&\to f(V)\times g(V)\subset V\times V\notag\\
(v,w)&\mapsto (f(w),g(v)).
\end{align}
I am interested in "how big" is the condition $f(w)=g(v)$, which means I want to compute the dimension of the vector space
$$K=\textrm{im }\psi_{f,g}\cap \Delta(V),$$ where $\Delta:V\to V\times V$ is the diagonal inclusion. For instance, I can say that if either $f$ or $g$ is invertible, then $K\cong \Delta(V)\cong V$. 
What can be said about arbitrary $f$ and $g$? Maybe the answer is some function of the ranks of $f$ and $g$, but I cannot figure how. Thanks for any help!
 A: This does not depend merely on the ranks of $f$ and $g$.
To compute the dimension of $K=Im(\psi_{f,g})\cap \Delta$ you can do the following:
First you find the cartesian equations of the image of $\psi_{f,g}$. This is easily done if you have the matrix expression of both $f$ and $g$ (because $\psi_{f,g}$ has a block diagonal matrix).
Then you form the system using the above equations and the equations of the diagonal. (If $V$ has coordinates $(x_1,\dots,x_n)$ and $V\times V$ has coordinates $(x_1,\dots,x_n,y_1,\dots,y_n)$, then the equations of the diagonal are $x_1=y_1, \dots, x_n=y_n$ and the matrix of such system is $\begin{pmatrix}I&-I\end{pmatrix}$)
Computing the rank $k$ of the matrix of the system you get that $\dim(K)=2n-k$.
In the particular case where $f$ is invertible, the cartesian equations of $Im(\psi_{f,g})$ involve only the $y_i$'s coordinates, therefore the matrix of the system is of the form  $\begin{pmatrix}0&M\end{pmatrix}$, then the system of $Im(\psi_{f,g})\cap \Delta$ is of the form $\begin{pmatrix}I&-I\\0&M\end{pmatrix}$ and its rank is $n+$ rank$(M)$, thus in this case we have $2n-k=2n-n-$rank$(M)=n-$rank$(M)=\dim(Im(g))$
