Let \(P\in\mathbb R^{n\times n}\) satisfy \(P^3=P\). Let \(r=\operatorname{rank}P>0\). Show that there exist \(U,V\in\mathbb R^{n\times r}\) with \[ V^TU=I_r \] such that \[ P=USV^T, \] where \(S\) is an \(r\times r\) diagonal matrix whose diagonal entries are all \(\pm1\).
Solution
1. The matrix \(P\) is diagonalizable over \(\RR\), and everyeigenvalue of \(P\) belongs to\[ \{0,1,-1\}. \]
The relation \(P^3=P\) says that \(P\) is annihilated by \[ q(t)=t^3-t=t(t-1)(t+1). \] This polynomial splits over \(\RR\) into distinct linear factors. Hence the minimal polynomial of \(P\) also has no repeated root, so \(P\) is diagonalizable over \(\RR\). Every eigenvalue is a root of \(q\), giving the displayed set.
2. There is an invertible matrix \(T\in\RR^{n\times n}\) such that\[ P=TJT^{-1}, \qquad J= \begin{pmatrix} S&0\\ 0&0 \end{pmatrix}, \]where \(S\) is an \(r\times r\) diagonal matrix with diagonal entriesall equal to \(\pm1\).
By step 1, choose an eigenbasis for \(P\). Since \(r=\operatorname{rank}P\), exactly \(r\) diagonal entries in a diagonal form of \(P\) are nonzero. Reorder the eigenbasis so those entries come first. They are all \(\pm1\), giving the displayed block form.
3. Write\[ T= \begin{pmatrix} U&W \end{pmatrix} \]with \(U\in\RR^{n\times r}\) the first \(r\) columns of \(T\), and write\[ T^{-1} = \begin{pmatrix} V^T\\ Z^T \end{pmatrix} \]with \(V^T\) the first \(r\) rows of \(T^{-1}\).
2. In particular\(V\in\RR^{n\times r}\).
4. The matrices \(U\) and \(V\) satisfy\[ \boxed{V^TU=I_r}. \]
Since \(T^{-1}T=I_n\), the upper-left \(r\times r\) block of the product is \[ V^TU. \] The corresponding block of \(I_n\) is \(I_r\), proving the claim.
5. The same matrices satisfy\[ \boxed{P=USV^T}. \]
Using the block forms from steps <1>2–<1>3, \[ \begin{aligned} P &= \begin{pmatrix}U&W\end{pmatrix} \begin{pmatrix} S&0\\ 0&0 \end{pmatrix} \begin{pmatrix} V^T\\ Z^T \end{pmatrix} \\ &= USV^T. \end{aligned} \]
6. Q.E.D.
Steps 2, 4, and 5 give the required matrices and diagonal matrix \(S\).