Rank Deficient matrix The rank of a matrix A is the dimension of the vector space generated by its columns. It shows the dependent and independent column of matrix A. 
Suppose I have a nxn matrix A with rank n/2. As a specific example, let A =[3 1 3 1;1 4 1 4;3 1 3 1;1 4 1 4] with the two eigenvalues 4.7639, 9.23361 and the other two zero. Matrix A is 4x4 but its rank is 2. In other words, matrix A is actually 2D but it is represented in 4D.
My question is how to get actual 2D representation of matrix A (Eigenvalue in 2D and 4D should be same)? Since matrix A consist of four similar 2x2 matrices, can I take the 2x2 matrix as its representation in 2D(This matrix has eigenvalue half that of matrix A)? Or I have to do projection or something?
Please help!
 A: There’s a general result for quotient spaces that says that if you have two vector spaces $V_1, V_2$ with subspaces $W_1, W_2$, respectively, then a linear map $T:V_1\to V_2$, induces a map $\overline T:V_1/W_1\to V_2/W_2$ that’s consistent with $T$ in the sense that the diagram $$\require{AMScd}
\begin{CD}
V_1 @>T>> V_2 \\
@V\pi_1VV @VV\pi_2V \\
V_1/W_1 @>\overline T>> V_2/W_2
\end{CD}
$$ commutes ($\pi_1,\pi_2$ are the respective projections on the quotient spaces). What you’re basically asking about is a way to produce a matrix of $\overline T$ given a matrix of $T$. Note that while the map $\overline T$ is independent of the choice of basis for the spaces, the matrices of the two maps aren’t, so it doesn’t really make sense to speak of “the” matrix.  
In the case you’re looking at, $V_1$ and $V_2$ are both $\mathbb R^{2n}$ and $W_1=\ker A$, which is $n$-dimensional, making $V_1/W_1$ $n$-dimensional as well. We want $V_2/W_2$ to be isomorphic to $\operatorname{im}A$. A natural choice for $W_2$ is then $\ker A^*$, i.e., the kernel of the adjoint of $A$. Once you’ve chosen bases for these two quotient spaces, then the matrix $\overline A$ is simply the usual change-of-basis matrix. It shouldn’t be too hard to arrange for this matrix to have the same non-zero eigenvalues as $A$, but that only makes it unique up to similarity. You’ll need to impose some other conditions to come up with “the” smaller matrix.  
Note that this sort of construction has many interesting applications. For instance, when solving electrical networks (an application of algebraic topology) you have a space $C_1$ of branch currents and its dual, the space of branch voltages $C^1$. There is also a map $Z:C_1\to C^1$ that, with the obvious choice of bases, is represented by a diagonal matrix with the branch impedances along the diagonal. This matrix is $b\times b$, where $b$ is the number of branches in the graph. In the mesh-current method, you identify the subspace $Z_1$ of cycles in the graph and take $H_1$ as this subspace considered as a vector space in its own right rather than a subspace of $C_1$ (i.e., as $\mathbb R^m$ or $\mathbb C^m$, with $m=\dim Z_1$). A selection of a basis for $Z_1$ induces a map $\sigma:H_1\to Z_1$ that simply identifies the basis of $H_1$ with that of $Z_1$. With this map, you can then produce a smaller $m\times m$ matrix $\sigma^*Z\sigma$ that characterizes the impedances of the chosen meshes, which you can then use to solve for the mesh currents. This matrix is, of course, dependent on the choice of independent meshes, i.e., the choice of basis for $Z_1$.
A: The columns of the matrix in your example (because 2 eigenvalues are zero) form a 2 dimensional subspace of $\mathbb{R}^4$. This does not have anything to do with $\mathbb{R}^2$.
Consider e.g.
$$A = \begin{bmatrix}1&0&0&0\\2&1&0&0\\4&5&0&0\\5&6&0&0\end{bmatrix}$$
Then $rank\, A = 2$. The columns of $A$ form a 2 dimensional subspace of $\mathbb{R}^4$, that is $$U = span\,\{\begin{bmatrix}1\\2\\4\\5\end{bmatrix},\begin{bmatrix}0\\1\\5\\6\end{bmatrix}\} \subset \mathbb{R}^4$$
However, this has no direct connection to $\mathbb{R}^2$ (you can imagine $U$ as a plane in 4-dimensional space).
However in your case, If the matrix you call "2D representation of $A$" should just have the same eigenvalues as $A$, you can just take $$\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$$ where $\lambda_1,\lambda_2$ are the eigenvalues of A
