Let me assume that M is an idempotent matrix (MM=M) and (I-M) is not zero (I is identity matrix with the same dimensions as M). If I multiply (I-M) by M, (M-MM)=(M-M)=0. How could it be possible that multiplication of non-zero matrices yields the zero matrix?

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    $\begingroup$ This question belongs on MATH.se, as it is not about statistics. $\endgroup$ – Repmat Mar 31 '16 at 7:05
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    $\begingroup$ An archetypical example, which will help you understand all idempotent matrices, is to consider $$M=\pmatrix{1&0\\0&0},$$ whence $$I-M=\pmatrix{0&0\\0&1}.$$ $\endgroup$ – whuber Mar 31 '16 at 14:47

Have you come across the concept of the kernel or null-space? A matrix represents a linear map, and will map some vectors to zero (at the very least, the zero vector). More generally any vector such that $A \mathbf{x}=0$ lies in the null space of $A$.

Moreover, have you come across the idea that the matrix multiplication $AB$ can be considered as the application of the linear map represented by $A$ to the columns of $B$? If all columns of $B$ lie in the kernel of that map, then each gets mapped to the zero vector, and so $AB$ is the zero matrix. This is what is happening in your example.

If you are confused by how all columns of $B$ lie in this kernel, then it is worth thinking about how "big" the kernel is. It's actually a whole subspace, whose dimension is given by the rank-nullity theorem.

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    $\begingroup$ +1. It might be more consistent (and elegant) to stick with the interpretation of matrices as representing linear transformations rather than using a hybrid interpretation involving column vectors. Thus $AB$ represents the composition of two transformations (acting on column vectors). The product is zero if and only if the image of $B$ is a sub-vectorspace of the kernel of $A$. That should be easy to visualize. $\endgroup$ – whuber Mar 31 '16 at 14:52
  • $\begingroup$ @whuber Yes, I was toying with which way to go. I decided I wasn't entirely sure whether the OP would have come across the necessary vector space ideas to take that approach (the question is framed entirely in terms of matrices) so I was trying to add the minimum of new information: if we know are happy with the idea that a matrix can map a non-zero column vector to zero, and we can see that matrix multiplication can be seen as one matrix acting on the columns of another, then it shouldn't be so surprising that a matrix multiplication can fill all the columns with zeroes. $\endgroup$ – Silverfish Mar 31 '16 at 21:12
  • $\begingroup$ @whuber On the other hand I couldn't find a purely "matrixy" way to explain it - the idea that several quite different-looking column vectors can all end up mapped to zero seems worthy of explanation, but I couldn't see a nice way to explain that without venturing into the territory of the kernel of a transformation. If there's a nice way you can think of to handle that while talking only about "matrix arithmetic" rather than geometry, then I'd be interested to hear it. $\endgroup$ – Silverfish Mar 31 '16 at 21:18
  • $\begingroup$ I do intend to add a diagram at some point - the particular case of M and I-M has a particular geometric interpretation that's worth mentioning, even if the question is phrased more generally than that. $\endgroup$ – Silverfish Mar 31 '16 at 21:22
  • $\begingroup$ Pure "matrix arithmetic" arguments tend to be weak and cumbersome. In this case, since only a nontrivial example is needed, one might begin by contemplating a pair of nonzero $n$-vectors $x$ and $y$ for which $x^\prime y = 0$. These are easy to find (since the condition is geometrically interpretable in terms of orthogonality). Then it is immediate--using only matrix arithmetic--that the product $AB=0$ where the rows of $A$ all equal $x^\prime$ and the columns of $B$ all equal $y$. Obviously neither $A$ or $B$ is zero. $\endgroup$ – whuber Mar 31 '16 at 21:43

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