question about idempotent matrix
Let me assume that $M$ is an idempotent matrix ($M^2=M$) and $I-M$ is not zero ($I$ is identity matrix with the same dim as $M$).
If I multiply $I-M$ by $M$, $M-M^2=M-M=0$.
How could it be possible? multiplication with non-zero matrices yields zero matrix. Is there any one to help me out?
$\endgroup$3 Answers
$\begingroup$This means that the rows of $M$ are orthogonal to the columns of $I-M$.
A geometric interpretation of this is the following: Consider an orthogonal projection, i.e. $P$ with $P=P^2$ and $P=P^*$.
Then $P$ is a projection onto $\operatorname{ran} P$, and $Q=I-P$ is a projection onto $(\operatorname{ran} P)^\perp$.
The property of $PQ=0$ means that for any $x,y \in \mathbb R^n$ $Px$ and $Qy$ are orthogonal to each other, and we can not only write $x=Px+Qx$ but even $$x = Px \oplus Qx = Px \oplus (I-P)x$$.
$\endgroup$ $\begingroup$This can happen. Consider for instance the matrix $$M = \left[\begin{array}{cc} 1 & 0\\ 0 & 0 \\ \end{array}\right].$$
Then you can easily see that $(I - M)M = 0$ even thought $I - M \ne 0$ and $M \ne 0$. Remember that the set of the square matrices endowed with the sum and product is just a ring, not a field. (Note that $M$ and $I-M$ are not invertible!).
$\endgroup$ 3 $\begingroup$Let $A=\begin {pmatrix} 0&-1\\0&0\end {pmatrix}$ and $B=\begin{pmatrix} 1&1\\0&0\end {pmatrix}$. What is $AB$? Yes, it is possible.
$\endgroup$ 1
