# Question about idempotent matrix: how can M(I-M) equal zero?

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?

• This question belongs on MATH.se, as it is not about statistics. – Repmat Mar 31 '16 at 7:05
• An archetypical example, which will help you understand all idempotent matrices, is to consider $$M=\pmatrix{1&0\\0&0},$$ whence $$I-M=\pmatrix{0&0\\0&1}.$$ – 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.
• +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. – whuber Mar 31 '16 at 14:52
• 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. – whuber Mar 31 '16 at 21:43