Is there any non-trivial decomposition of matrices that involve idempotents that provide the shortest sum description of given matrix?
Decomposition theorems of matrices involving idempotents?
1 Answers
Yes, the spectral decomposition of a symmetric matrix (or a Hermitian matrix). To be more precise, let $A$ be symmetric. Then $$A=\sum_{\lambda\in \sigma(A)}\lambda P_\lambda,$$ where $\sigma(A)$ is the spectrum of $A$, i.e. the eigenvalues of $A$ and $P_{\lambda}$ is the orthogonal projection onto the corresponding eigenspace. Obviously $P_{\lambda}^2=P_{\lambda}$. Hence $P_{\lambda}$ is idempotent.
The above decompostion holds for normal matrices as pointed out in the comments.
Edit: Let's look at the $2\times 2$ matrices. Any idempotent can be written as $\begin{pmatrix} a& b\\ c&1-a \end{pmatrix}$ with $a^2+bc=a$. Notice that the trace of such a matrix is $1$. Given any matrix $X=\begin{pmatrix} x& y\\ z&t \end{pmatrix}$, is it possible to write $X=\lambda A+\mu B$ where $A$ and $B$ are idempotent? If the answer is no, is it possible to write $X=\sum_{i=1}^n\lambda_iA_i$ with the $A_i$ idempotent? These are interesting first questions to look at. I don't know the answer, and perhaps the non-linear condition $a^2+bc=a$ for idempotent matrices might be very difficult to work with.
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0what is $\lambda\sigma(A)$? what makes it fail for skew symmetric? – 2017-01-19
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0More generally, the spectral theorem holds for normal matrices. – 2017-01-19
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0This description needs $n$ summations in worst case. Could there be a shorter non-trivial one? – 2017-01-19
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0Interesting question, I'm not convinced that it is even possible to write any matrix as a linear combination of idempotent ones, I'm thinking about it. – 2017-01-19
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0@Mathematician42 Any diagonalizable matrix can be written as a linear combination of at most $n$ commuting idempotents (the product of any two yields $0$). Any matrix can be written as a linear combination of at most $n$ idempotents, but we can't always choose them so that they commute. The number of idempotents needed is at most the rank of the matrix. – 2017-01-19