Find an orthogonal projection given a matrix with identical rows Given a matrix A as seen below,
$$A=
        \begin{bmatrix}
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        \end{bmatrix}
$$
I am trying to find (1) a number $c$ so that $M=cA$ is an orthogonal projection onto a subspace $W$. (2) Additionally I will need to find a basis for $W$.
After finding $rref(A)$ it's obvious that the matrix $A$ as rank = 1 or that dim$(ColA)=1$. It's easy to find a basis for $W$, which is the one vector that forms a basis for $Col(A)$ = $(1,1,1,1,1,1)^T = v_1$. This solves (2).
Part (1) is what I am unsure of. Here is my approach: given the basis for $W$ found above, I will find an orthonormal basis for $W$ by normalizing the vector $v_1$ and forming $u_1$. I do this by calculating $u_1=\frac{1}{||v_1||}v_1=(\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{1}{\sqrt6})^T$.
Now to calculate the orthogonal projection onto subspace $W$ I will multiply $uu^T$ to get a 6x6 matrix with each entry of $\frac{1}{6}$. Pulling the scalar $\frac{1}{6}$ out of the calculated matrix I'll end up with $c=\frac{1}{6}$. Does this sound correct?
Work: 
$$ uu^T = 
\begin{bmatrix}
\frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\
\end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6}
\end{bmatrix}
$$
$$ M=
        \begin{bmatrix}
        \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\
\end{bmatrix}
$$
$$ = cA = \frac{1}{6}
\begin{bmatrix}
  1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 \\
\end{bmatrix}
$$
Giving us $c=\frac{1}{6}$. 

You can solve question 1 in a simpler way: $M$ the matrix of  a projection if and only if $M^2=M$, i.e. $c^2A^2=cA$, which implies $cA^2=A$. As you've calculated that $A^2=6A$, this means $c=\dfrac16$.
