Existence of $v\in\mathcal{L}(E)$ such as $u=u\circ v\circ u$ Let $E$ be a vector space of dimension $n$, $u\in\mathcal{L}(E)$. How can I show there exists $v\in\mathcal{L}(E)$ such as $u=u\circ v\circ u$ ?

$u(x)=0\Rightarrow u(v(u(x)))=0\Rightarrow u(v(0))=0\Rightarrow u(0)=0$ so there isn't any problem with $\ker$... what now ?
 A: Just take the Moore–Penrose pseudo inverse for the matrix $A$ of $u$, i.e., a matrix $A^+$ with $AA^+A=A$ etc, see point $1$ of the definition here: http://en.wikipedia.org/wiki/Moore-Penrose_pseudoinverse). The Moore–Penrose pseudoinverse exists and is unique.
A: Lauds and +1 to Dietrich Burde for giving the Moore-Penrose pseudo inverse as a solution to this problem, and providing a citing to support his assertions.
BUT . . .  if you want to see the actual mechanics and mechanism of a solution, try this:
Let $E$ be a vector space of finite dimension $n$ over any field $\Bbb F$.
If $u \in \mathcal L(E)$ is nonsingular, $u^{-1}$ exists and we can take $v = u^{-1}$.  Then
$uvu(x) = u(u^{-1}u(x)) = u(x) \tag{1}$
for all $x \in E$ and we are done.  If $u$ is singular, then since $\dim E = n < \infty$ there is a basis $X_1, X_2, \ldots X_m$ for $\ker u$ with $0 < m \le n$.  If $m = n$, then $\ker u = E$, i.e. $u = 0$; then we take $v = 0$ as well and it is easy to see that $u = uvu$; we are done.  So we assume $m < n$.  Again since $\dim E = n < \infty$, we can choose $l = n - m$ vectors $Y_1, Y_2, \ldots, Y_l$ which collectively extend the basis $\{ X_i \}$ of $\ker u$ to all of $E$; note we have $Y_j \notin \ker u$ for all $j$, $1 \le j \le l$, since any $Y_j \in \ker u$ can be expanded in terms of the $X_j$, but then $\dim(\text{span} (\{X_j\} \cup \{Y_i\})) < n$ and $\text{span} (\{X_j\} \cup \{Y_i\})$ cannot be a basis for all of $E$.  Furthermore, the vectors $u(Y_i)$, $1 \le i \le l$, are themselves linearly independent:  if
$\sum_1^l \alpha_i u(Y_i) = 0 \tag{2}$
for some $\alpha_1, \alpha_2, \ldots \alpha_l \in \Bbb F$, not all $\alpha_i = 0$, then by (2)
$u(\sum_1^l \alpha_i Y_i) = \sum_1^l \alpha_i u(Y_i) = 0, \tag{3}$
that is,
$\sum_1^l \alpha_i Y_i \in \ker u; \tag{4}$
but this implies
$\sum_1^l \alpha_i Y_i = \sum_1^m \beta_j X_j, \tag{5}$
$\beta_j \in \Bbb F$, $1 \le j \le m$, which is a linear relation between the independent vectors $X_j$, $Y_i$; hence we must have $\alpha_i = \beta_j = 0$ for all $i, j$.  This in turn contradicts the choice of the $\alpha_i$ as not all zero; thus we must have that the $u(Y_i)$ are independent vectors in $E$.  Then $F = \text{span} \{ u(Y_j), 1 \le j \le l \}$ is a subspace of $E$ of dimension $l = n - m$ with basis $\{u(Y_j), 1 \le j \le l\}$, and so once again we may extend the basis $\{ u(Y_i) \}$ of $F$ to all of $E$ by selecting $m = n - l$ linearly independent vectors $Z_j \notin F$, $1 \le j \le m$, so that the set $\{u(Y_j), 1 \le j \le l\} \cup \{Z_j, 1 \le j \le m\}$ then spans $E$.  We define $v$ in this basis by setting
$v(u(Y_i)) = Y_i, \; 1 \le i \le l, \tag{6}$
and
$v(Z_j) = 0, \; 1 \le j \le m, \tag{7}$
and extending $v$ to all of $E$ by linearity:
$v(\sum_1^l \alpha_i u(Y_i) + \sum_1^m \beta_j Z_j) = \sum_1^l \alpha_i v(u(Y_i)) + \sum_1^m \beta_j v(Z_j) = \sum_1^l \alpha_i Y_i, \tag{8}$
since $v(Z_j) = 0$ for all $j$.  Now, we have seen that any vector $x \in E$ may be uniquely written
$x = \sum_1^l y_i Y_i + \sum_1^m x_j X_j \tag{9}$
with the $x_j, y_i \in \Bbb F$, since $\{Y_i, 1 \le i \le l\} \cup \{X_j, 1 \le j \le m \}$ forms a basis for $E$.  Thus
$vu(x) = v(\sum_1^l y_i u(Y_i)) = \sum_1^l y_i vu(Y_i) = \sum_1^l y_i Y_i, \tag{10}$
by virtue of the fact that the $X_j \in \ker u$, whence
$uvu(x) = \sum_1^l y_i u(Y_i); \tag{11}$
but by (9) we see
$u(x) = \sum_1^l y_i u(Y_i), \tag{12}$
showing that
$u(x) = uvu(x) \tag{13}$
for all $x \in E$; thus $u = uvu$ in $\mathcal L(E)$, the requisite result.  And once again, we are done.  QED.
Note added Wednesday 18 June 2014 12:04 PM PST  The key idea in the above is to establish the subspace $\text{span} \{Y_i, 1 \le i \le l \} \subset E$ on which $u$ acts in an injectively, and then build $v$ from the isomorphic (under $u$) subspace $\text{span} \{u(Y_i), 1 \le i \le l \} = u(\text{span} \{Y_i, 1 \le i \le l \})$.  We are thus effectively working in the two bases $\{X_j, 1 \le j \le m\} \cup \{Y_i, 1 \le i \le l\}$ and $\{u(Y_j), 1 \le j \le l\} \cup \{Z_j, 1 \le j \le m\}$ of $E$ concurrently, an operation which I personally feel is clarified by the avoidance of a single basis and coordinate system for $E$.  The kernels $\ker u$ and $\ker v = \text{span} \{Z_k\, 1 \le k \le m \}$ of course "take care of themselves"; $v$ is an inverse of $u$ on $F = \text{span} \{ u(Y_j), 1 \le j \le l \}$.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
