# $X = [1_n \hspace{1mm} Z]$ with full column rank, then $(I_n - \frac{1}{n}J_n)Z$ is full-column rank

I'm trying to solve the following statement.

Consider a $$n \times p$$ matrix $$X = [1_n \hspace{1mm} Z]$$, where $$1_n = (1, \dots, 1) \in \mathbb R^n$$ and $$\text{rank}(X) = p$$. Let $$I_n$$ be a $$n \times n$$ identity matrix, and $$J_n$$ be a $$n \times n$$ matrix whose elements are all 1. Assuming $$n >\!\!> p$$, show that $$(I_n - \frac{1}{n}J_n)Z \in \mathbb R^{p-1}$$ is a full column rank.

Since $$I_n - \frac{1}{n}J_n$$ is not invertible, I can't use the property $$\text{rank}((I_n - \frac{1}{n}J_n)Z) = \text{rank}(Z)$$ directly. So, it would be grateful if I could get some hint regarding this approach.

Thank you in advance.

First, a $$m\times n$$ matrix $$A$$ is full-column rank if and only if for each vector $$\alpha\in \mathbb R^{n}$$ and $$\alpha\neq0$$, there is $$A\alpha\neq0$$.

Below, we use proof by contradiction and assume that $$X=\begin{bmatrix}1_n & Z\end{bmatrix}$$ is full-column rank but $$Y=(I_n-\frac1n J_n)Z$$ is not. Then throwgh above fact, there exists a vector $$\alpha\in \mathbb R^{n}$$ and $$\alpha\neq0$$, such that $$Y\alpha=0$$. Use the equation that $$J_n=1_n\cdot1_n^T$$, we have $$Y\alpha=\left(I_n-\frac1n1_n\cdot1_n^T\right)Z\alpha=Z\alpha-\frac1n1n1_n\cdot1_n^TZ\alpha=Z\alpha-\left(\frac1n1_n^TZ\alpha\right)1_n=0$$ in which $$\frac1n1_n^TZ\alpha$$ is a number. Rewrite it, we have $$\begin{bmatrix}1_n & Z\end{bmatrix}\begin{bmatrix}-\frac1n1_n^TZ\alpha\\\alpha\end{bmatrix}=0.$$ If we let $$\beta=\begin{bmatrix}-\frac1n1_n^TZ\alpha\\\alpha\end{bmatrix}$$, we will have $$X\beta=0$$ with $$\beta\neq0$$ and this is cotradictory to the face that $$X$$ is full-column rank.

So, we get the proof that $$Y$$ is full-column rank.

Let $$P=I_n - \frac 1n J_n$$ so that $$P$$ is the orthogonal projection on the orthogonal complement of $$1_n$$. Let $$P'$$ denote the restriction of $$P$$ to the subspace $$\operatorname{Im} Z$$.

Note that $$\operatorname{rank}(PZ) = \dim \operatorname{Im}(PZ) = \operatorname{rank} P'$$, and by the rank-nullity theorem, $$\operatorname{rank} P' = \dim \operatorname{Im}Z - \dim \ker P' = \operatorname{rank} Z - \dim\big((\ker P) \cap \operatorname{Im} Z \big)$$

Since $$X$$ has full column rank, $$\operatorname{rank} Z = (\operatorname{rank} X) -1 = p-1$$ and $$(\mathbb R 1_n) \cap \operatorname{Im} Z = \{0\}$$.
Since $$\ker P = \mathbb R 1_n$$, we obtain

$$\operatorname{rank}(PZ) = p -1,$$ thus $$PZ$$ has full column rank.