Find the projection from one vector $\vec{u}$ onto $\vec{v}$,
$$
\vec{v} = \left[ \begin{array}{c}
1 \\
2
\end{array} \right]
$$
we can find the projection matrix in following:
$$
M = \left[ \begin{array}{c}
1 \\
2
\end{array} \right]
\left[ \begin{array}{cc}
1 & 2
\end{array} \right] =
\begin{bmatrix}
1 & 2 \\
2 & 4
\end{bmatrix}
$$
If $$\vec{u} = \left[ \begin{array}{c}
2 \\
3
\end{array} \right] $$
then the projection from $\vec{u}$ onto $\vec{v}$ is following without any scale. e.g. normalize.
$$
M \cdot \vec{u} = \vec{v'}
$$
Best Answer
Notice that $M = vv^T$, so $Mu = vv^Tu$. Now, $v^Tu = v \cdot u$ is the dot product (or inner product for our usual Euclidean metric) between $u$ and $v$, a real number, so the result is always a multiple of $v$.