What is the intuition behind the projection can be computed from the outer product of two vectors. 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.
$$
    M \cdot \vec{u} = \vec{v'}
$$

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 $v$.
