Calculating the rank and the order of block matrices I was trying to learn about matrix blocks and I came across with the following question while practicing: Let $A$ be a matrix of $4\times 5$ with rank of $3$ Consider the following two matrix:
$$B_{1}=\begin{pmatrix}0 & A\\
A^{t} & 0
\end{pmatrix},\,B_{2}=\begin{pmatrix}A & 0 & A\\
0 & A & 2A\\
A & 0 & A
\end{pmatrix}$$
How can I calculate the rank and the order of those matrices?
 A: I will consider in the sequel matrix $B_1$ only.
Reasoning on SVD (Singular Value Decomposition) is probably the most direct way.
Let us recall that $4 \times 5$ matrix $A$ can be decomposed in the following way (SVD), which can be seen at least approximatively as an extension of the diagonalization process to general matrices, not especially square) :  
$$A=U\Sigma V^T \ \ \ \text{or equivalently} \ \ \ A^T=V\Sigma^T U^T\tag{1}$$
where 


*

*$U$ is a $4 \times 4$ orthogonal matrix, 

*$\Sigma$ is $4 \times 5$ (same dimensions as $A$) and 

*$V$ is a $5 \times 5$ orthogonal matrix. 
The important thing is that $\Sigma$ begins like a diagonal matrix with $3$ non zero diagonal elements $\sigma_1,\sigma_2,\sigma_3$ in the following way :
$$\Sigma=\begin{pmatrix}\sigma_1 & 0 & 0 & 0 & 0\\
0 & \sigma_2 & 0 & 0 & 0\\
0 & 0 & \sigma_3 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{pmatrix}\tag{2},$$
Allowing to write
$$B_1=\begin{pmatrix}0 & A\\
A^{T} & 0
\end{pmatrix}=\underbrace{\begin{pmatrix}U & 0\\
0 & V
\end{pmatrix}}_{\Omega}\underbrace{\begin{pmatrix}0 & \Sigma\\
\Sigma^{T} & 0
\end{pmatrix}}_{\Lambda}\underbrace{\begin{pmatrix}U^T & 0\\
0 & V^{T}
\end{pmatrix}}_{\Omega^T=\Omega^{-1}}\tag{3}.$$
Therefore, $B_1$ is similar to $\Lambda$ which is clearly a rank-$6$ matrix. 
