Projector of a euclidean space I have a Euclidean Space $V=R^3$ and a subspace E = {v} , v≠0 so to calculate the projection of $u \in V$.
My teacher wrote on the blackboard: P(u)=$\frac{<u,v>}{||v^2||}v$ so for this case would be that : $\frac{u^Tv}{||v||^2}v$  but my teacher wrote that as long as the dimension of E is one then P(u)=$\frac{1}{||v^2||}v v^Tu$ .I do not understand why he wrote that $v^T$  when $u$ was the only trasposed .
 A: In a real Euclidean space the inner product is symmetric, so
$$u^Tv=\langle u,v\rangle=\langle v,u\rangle =v^Tu\,.$$
Anyway, this can be directly seen as well, because $u^Tv$ is a scalar ($1\times 1$ matrix), its transpose is itself:
$$u^Tv=(u^Tv)^T=v^T(u^T)^T=v^Tu\,.$$
Also because this is a scalar, it doesn't matter whether we multiply $v$ by this from left or from right.
The comment that $E$ must be one dimensional means that $E$ is the subspace of $V$ where we want to project to, and here $E$ is the span (line) of $v$.
An additional suggestion is that it's easier to calculate with $v_1:=\frac v{\|v\|}$ which has norm $1$ in place of $v$: we simply get
$$P(u)=\langle u,v\rangle v\,.$$
By the way, a similar formula can be written when $E$ has any finite dimension, using an orthonormal basis of $E$.
A: For the inproduct of two vectors it doesn't matter which you transpose.
To show this let's compare the products $u^Tv$ and $v^Tu$:
The product $u^Tv$ evaluates to:
$$
\begin{pmatrix}
u_1 & u_2 & u_3
\end{pmatrix}
\begin{pmatrix}
v_1\\ v_2\\ v_3
\end{pmatrix}=u_1\cdot v_1 + u_2\cdot v_2 + u_3\cdot v_3
$$
The product $v^Tu$ evaluates to:
$$
\begin{pmatrix}
v_1 & v_2 & v_3
\end{pmatrix}
\begin{pmatrix}
u_1\\ u_2\\ u_3
\end{pmatrix}=u_1\cdot v_1 + u_2\cdot v_2 + u_3\cdot v_3
$$
Therefore we have that $u^Tv$ and $v^Tu$ are both equal to the inproduct $\langle u,v\rangle$.
