I'd like to add a few things to complement Arturo's and Fabian's answers.
If you take the outer product of a unit vector, $\hat{e}$ ($\lvert e \rangle$, Dirac notation), with its dual, $\hat{e}^\ast$ ($\langle e \rvert$), you get a matrix that projects vectors onto the space defined by the unit vector, i.e.
\begin{aligned}
\mathbf{P}_e &= \hat{e} \otimes \hat{e}^\ast \ &= \lvert e \rangle \langle e \rvert \ &= \left( \begin{array}{ccc} \lVert e_{1} \rVert^2 & e_{1} e_2^\ast & \ldots \\ e_2 e_1^\ast & \lVert e_{2} \rVert^2 & \ldots \\ \vdots & \vdots & \ddots \end{array} \right)
\end{aligned}
where
$ \mathbf{P}_e \vec{v} = (\vec{v} \cdot \hat{e}) \hat{e}. $
In other words, $\mathbf{P}_e$ is a projection operator. Using this, you can rewrite a matrix, $\mathbf{A}$, in terms of its eigenvalues, $\lambda_i$, and eigenvectors, $\lvert e_i \rangle$,
$ \mathbf{A} = \sum_i \lambda_i \lvert e_i \rangle \langle e_i \rvert$
which is called the spectral decomposition. From this, it is plain to see that any eigenvector, $\hat{e}_i$, with a zero eigenvalue does not contribute to the matrix, and for any vector component in one of those spaces, $\mathbf{P}_{e_i}\vec{v} = \vec{v}_i$, $\mathbf{A} \vec{v}_i = 0.$
This implies that the dimensionality of the space that $\mathbf{A}$ operates on is smaller than the dimension of the space used to describe it. In other words, $\mathbf{A}$ does not posses full rank, and is not invertible.