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How do I construct matrices $P\in\mathbb{C}^{\ n\ x\ n}$ such that:

  1. They are not projectors (projector: $A=A^2$)
  2. $Col(P)\cap Ker(P) = \{0\}$
  3. $\forall x \in \mathbb{C}^n,\ x = y+z\quad \left(y\in Col(P),\ z\in Ker(P)\right)$
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2 Answers 2

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What you need is that your $P$ has no nilpotent part. That is, its Jordan form should be diagonal. In terms of blocks, your $P$ will be of the form $SXS^{-1}$ with $S$ invertible and $$ X=\begin{bmatrix}Y&0\\0&0\end{bmatrix}, $$ with $Y\in\mathbb C^{m\times m}$, $m\leq n$.

At its simplest form, for instance, here is an example of such $P$: $$ P=\begin{bmatrix}1&0&0\\ 0&2&0\\0&0&0\end{bmatrix}. $$ The column space is the span of the first two vectors in the canonical basis, while the kernel is the span of the third vector in the canonical basis.

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  • $\begingroup$ Would it be possible to do it without Jordan form? $\endgroup$
    – plr
    Commented Nov 14, 2016 at 18:39
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It was easier than I thought.

For example $T= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ has $T^2= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, so $T\neq T^2$. The $col(T)=\mathbb{C}^2$, and $ker(T)=0$. Then properties 2 and 3 are satisfied.

Another example $S= \begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix}$ has $S^2= \begin{bmatrix}4 & 0 \\ 0 & 0\end{bmatrix}$, so $S\neq S^2$. The $col(S)=\begin{bmatrix}1\\ 0 \end{bmatrix}$, and $ker(T)=\begin{bmatrix} 0 \\ 1\end{bmatrix}$; so properties 2 and 3 are also satisfied.

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