how to show existence of such a matrix

Question

Given that $E$ is an $m_1\times n_1$ matrix with $m_1\le n_1$ and $C$ is a $p_1\times n_1$ matrix. We are given that rank of $\begin{bmatrix}E\\C\end{bmatrix}=n_1$, then we need to show that there exists a full column rank matrix $R$ which is of $n_1\times m_1$ such that

rank of $\begin{bmatrix}I_{n_1}-RE\\C\end{bmatrix}=rank(C)$

Answer 1

Let us translate the mentioned properties. By the rank-nullity theorem

$$\renewcommand{\arraystretch}{1.5} \begin{array}{rcl}\text{rank} \left[\begin{array}{c}E\\ C \end{array}\right] = {n}_{1}&\Longleftrightarrow &{n}_{1}-\text{rank} \left[\begin{array}{c}E\\ C \end{array}\right] = 0\\ &\Longleftrightarrow &\dim \text{ker} \left[\begin{array}{c}E\\ C \end{array}\right] = 0\\ &\Longleftrightarrow &\dim \left(\text{ker} \left(E\right) \cap \text{ker} \left(C\right)\right) = 0\\ &\Longleftrightarrow &\boxed{\text{ker} \left(E\right) \cap \text{ker} \left(C\right) = \left\{0\right\}} \end{array}$$

and

$$\renewcommand{\arraystretch}{1.5} \begin{array}{rcl}\text{rank} \left[\begin{array}{c}{I}_{{n}_{1}}-R E\\ C \end{array}\right] = \text{rank} \left(C\right)&\Longleftrightarrow &\dim \text{ker} \left[\begin{array}{c}{I}_{{n}_{1}}-R E\\ C \end{array}\right] = \dim \text{ker} \left(C\right)\\ &\Longleftrightarrow &\dim \left(\text{ker} \left({I}_{{n}_{1}}-R E\right) \cap \text{ker} \left(C\right)\right) = \dim \text{ker} \left(C\right)\\ &\Longleftrightarrow &\text{ker} \left({I}_{{n}_{1}}-R E\right) \cap \text{ker} \left(C\right) = \text{ker} \left(C\right)\\ &\Longleftrightarrow &\boxed{\text{ker} \left(C\right) \subset \text{ker} \left({I}_{{n}_{1}}-R E\right)} \end{array}$$

Decompose

$${\mathbb{R}}^{{n}_{1}} = \text{ker} \left(C\right) \oplus V \qquad {\mathbb{R}}^{{m}_{1}} = E \left(\text{ker} \left(C\right)\right) \oplus W$$

Remark that $E$ is one-to-one on $\text{ker} \left(C\right)$ because $\text{ker} \left(E\right) \cap \text{ker} \left(C\right) = \left\{0\right\}$. This implies that

$$\dim W = {m}_{1}-\dim \text{ker} \left(C\right) \leqslant {n}_{1}-\dim \text{ker} \left(C\right) = \dim V$$

We define $R \colon {\mathbb{R}}^{{m}_{1}} \rightarrow {\mathbb{R}}^{{n}_{1}}$ by choosing ${\left.R\right|}_{E \left(\text{ker} \left(C\right)\right)} = {\left({\left.E\right|}_{\text{ker} \left(C\right)}\right)}^{{-1}}$ and for ${\left.R\right|}_{W}$ we choose any one-to-one map from $W$ to $V$.

As $R$ is one to one, it follows that $\text{rank} \left(R\right) = {m}_{1}$, as required.

Here is a Scilab example that explain how to compute a matrix $R$

E = [1 1 1 1; 3 1 2 1; 1 2 4 8]
C = [1 2 3 1; 0 1 2 1; 2 1 0 -1]
m1 = size(E, 1)
n1 = size(E, 2)
K = kernel(C)
k = size(K, 2)
L = kernel(K')
M = E * K
N = kernel(M')
P = [K L(:,1:m1-k)]
R = clean(P * inv([M N]))
S = [eye(n1,n1)-R*E; C]
rank(S)

The example uses the Scilab function kernel() that computes a basis of the matrix's nullspace. The same function can be used to compute a complemement of a subspace by taking the nullspace of the transpose matrix (which is the orthogonal subspace).