1

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.

jason 1
  • 789

2 Answers2

2

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.

Lasswetz Patz
  • 339
1

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.

Gabriel Romon
  • 36,881