Setting up the standard matrix for orthogonal projection 
For a subspace $V$ of $\mathbb{R}^4$, you are given these three ordered bases:
$A= (\mathbf{a}=(1,-2,-1,3), \mathbf{b}=(1,3,-2,-4))$
$B= (\mathbf{a}=(1,-2,-1,3), \mathbf{c}=(2,1,-3,-1))$
$C= (\mathbf{b}=(1,3,-2,4), \mathbf{d}=(0,5,-1,-7))$
Let $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ be the orthogonal projection of $\mathbb{R}^4$ onto $V$.
$a)$ Explain why $T$ is or is not invertible.
$b)$ Find the matrix $[T]$ which represents $T$ with respect to the standard coordinates.

I really need some guidance with this... Why isn't the matrix invertible? Is it because the matrix $T$ is not a square matrix?
For part $b)$, how do I do it? I tried to use the projection formula $[T]=A(A^TA)^{-1}A^T$ but the matrix isn't invertible...
How do I approach this? I'm very much stumped on this.
 A: Recall that a linear transformation from $A^m\to B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $n\times m$. Thus, $T:\Bbb R^4\to\Bbb R^4$ is a linear map having a $4\times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $\ker(T)$ other than $(0,0,0,0)$, that is, $\dim(\ker(T))>0$. For example, $\mathbf u=(7,1,5,0)$ has $\bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $\text{rank}([T])<4$.
For part $(b)$, let $\mathbf v=(x,y,z,w)\in\Bbb R^4$. The basis $B=\{\mathbf a,\mathbf c\}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $\mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$\displaystyle\frac{\langle\mathbf v,\mathbf a\rangle}{\langle\mathbf a,\mathbf a\rangle}\cdot\mathbf a+\frac{\langle\mathbf v,\mathbf c\rangle}{\langle\mathbf c,\mathbf c\rangle}\cdot\mathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $\mathbf v=(x,y,z,w)$ as 
$\displaystyle=\Big[\frac{x-2y-z+3w}{15}\Big]\cdot(1,-2,-1,3)+\Big[\frac{2x+y-3z-w}{15}\Big]\cdot(2,1,-3,-1)\\\displaystyle=\Big(\frac{5x-7z+w}{15},\frac{5y-z-7w}{15},\frac{-7x-y+10z}{15},\frac{x-7y+10w}{15}\Big)$
Thus, the matrix of $T,[T]=\begin{bmatrix}1/3&0&-7/15&1/15\\0&1/3&-1/15&-7/15\\-7/15&-1/15&2/3&0\\1/15&-7/15&0&2/3\end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$. 
