Are These Projections Unitarily Equivalent? Consider the C${^*}$-algebra $A=C(X,M_{n})$, where $X$ is a compact Hausdorff space. Let $e_{1}=(1,0,\ldots,0)$ and let $E=e_{1}\otimes e_{1}$, where for $u,v\in \mathbb{C}^{n}$, $u\otimes v$ denoes the rank-one operator in $M_{n}$ given by $(u\otimes v)(z)=\langle z,v\rangle u$. 
Suppose $f\in A$ is such that $f(x)$ is a non-zero diagonal matrix with non-negative entries for all $x\in X$. Write $v_{x}$ to denote the vector in $\mathbb{C}^{n}$ whose entries are those along the diagonal of $f(x)$. Let $P\in A$ be the projection given by $P(x)=\|v_{x}\|^{-2}(v_{x}\otimes v_{x})$. I.e., for each $x$, $P(x)$ is the projection onto the subspace $\operatorname{span}\{v_{x}\}$. Note that $P$ is continuous since $v_{x}$ varies continuously in $x$. 
If we let $W(x)=\|v_{x}\|^{-1}(v_{x}\otimes e_{1})$, then $W\in A$ is a partial isometry such that $E=W^{*}W$ and $P=WW^{*}$. Thus, $E$ and $P$ are Murray-von Neumann equivalent projections in $A$ and, thus, $P$ is trivial. My question is:

Are $P$ and $E$ unitarily equivalent in $A$?

I know that for each $x\in X$, $E$ is unitarily equivalent to $P(x)$ under the unitary $U_{x}$, say, since $E$ and $P(x)$ are MvN equivalent, but I don't know whether or not $x\mapsto U_{x}$ is continuous.
EDIT: Maybe asking for unitary equivalence is too much. 

What about just approximately unitarily equivalent?

 A: In what follows, let me deal with column vectors, so that for $u, v \in \mathbb{C}^n$, the rank $1$-operator $u \otimes v$ is simply the matrix product $uv^\ast$.
Let $w_1 := (1,1,\dotsc,1)^T$, and complete $\{w_1\}$ to a basis $\{w_1,\dotsc,w_n\}$ of $\mathbb{R}A^n$ – in fact, you could just take $w_k = e_k$ for $2 \leq k \leq n$ – and observe that 
$$
 \forall x \in X, \quad v_x = f(x)w_1.
$$ 
Fix $\delta > 0$. For every $x \in X$, since $f(x;\delta) := f(x) + \delta I_n$ is invertible, we can perform Gram–Schmidt orthonormalisation on the basis $\{f(x;\delta)w_1,\dotsc,f(x;\delta)w_n\}$ for $\mathbb{R}^n$ to get an orthonormal basis $\{u_1(x;\epsilon),\dotsc,u_n(x;\epsilon)\}$ for $\mathbb{R}^n$ with $u_1(x;\delta) = \|v_x + \delta w_1\|^{-1}(v_x + \delta w_1)$; by taking a closer look at the Gram–Schmidt algorithm, one can prove by induction on $k$ that each $u_k : X \times (0,1) \to \mathbb{R}^n$ is continuous. At last, let 
$$
 U := ( u_1(\cdot) \vert \cdots \vert u_n (\cdot) ) \in C(X \times (0,1),O(n)) \subset U(C(X \times (0,1),M_n)).
$$
Then $U$ satisfies
$$
 U E U^\ast = U e_1 e_1^\ast U^\ast = (Ue_1)(Ue_1)^\ast = (\|v_x+\epsilon w_1\|^{-1}(v_x+\delta w_1))(\|v_x+\delta w_1\|^{-1}(v_x+\delta w_1))^\ast\\ = \|v_x + \delta w_1\|^{-2} (v_x + \delta w_1) (v_x + \delta w_1)^\ast \to_{\delta \to 0} \|v_x\|^{-2}v_x v_x^\ast = P;
$$
by existence of the above limit and continuity of $U$ in the variable $\delta$, we can now find $\delta > 0$ small enough, such that $\| U(\cdot,\delta) E U(\cdot,\delta)^\ast - P \| < 1/2$, but this is enough to conclude (see Rørdam–Larsen–Laustsen, An introduction to $K$-theory for $C^\ast$-algebras, section 2.2) that $E$ and $P$ are homotopy equivalent, and hence unitarily equivalent.
