1
$\begingroup$

Let $A$ be an $m \times n$ matrix and $\operatorname{rank}(A) =r$. Let $P$ be an $m \times (m-r)$ matrix, such that $\mathbb{C}^m= \mathcal{R}(A) \oplus \mathcal{R}(P)$, then $P$ is of full column rank.

$\mathcal{R}(A)$ : Range of $A$ in $\mathbb{C}^m$

I'm trying to prove that with such conditions on $P$, $P$ is of full column rank.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

$$\begin{cases}\operatorname{rank}P=\dim\mathcal R(P) \\m=\dim\mathcal R(A)+\dim\mathcal R(p)=r+\dim\mathcal R(P)\\ \operatorname{rank}P\le \min(m,m-r)=m-r\end{cases}$$

What does this yield, in light of the definition of "being full-rank"?

$\endgroup$
1
  • $\begingroup$ that means P is of full column rank $\endgroup$
    – Amanda
    Aug 3, 2016 at 7:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .