How many degrees of freedom does an $N\times N$ idempotent matrix have? Is there a way to generate them?

Question

If I have an $N\times N$ idempotent matrix, how many degrees of freedom does it have? (For example, a non-special $N\times N$ matrix would have $N^2$ degrees of freedom).

Also, would it be possible to create a generator function for these matrices? Assuming $N>1$, there must be less than $N\times N$ degrees of freedom for an idempotent matrix; let’s call that number $D$.

Would it be possible to create a formula that takes $D$ input numbers but whose domain covers the entire space of all $N\times N$ idempotent matrices? (It would take $D$ numerical arguments and output an $N\times N$ idempotent matrix)

Answer 1

An idempotent matrix is diagonalizable with eigenvalues that are all either $1$ or $0$. Conversely, any matrix of the form $BDB^{-1}$ with invertible $B$ and diagonal $D$ with entries all either $1$ or $0$ is idempotent.

Thus you can construct an idempotent matrix by

• choosing how many eigenvalues are $1$, and then
• choosing linearly independent eigenvectors (columns of $B$) to correspond to the eigenvalues.

Although all idempotent matrices can be obtained in this manner, there will be some redundancy in the above "generator," since, for instance, the order of the eigenvalues/eigenvectors can be shuffled without affecting the resulting idempotent matrix.

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I suppose a more abstract way to think about this is to find two subspaces $W_1$ and $W_0$ such that $W_1 \oplus W_0 = \mathbb{R}^n$ (i.e., the two subspaces span $\mathbb{R}^n$ and have intersection $\{0\}$), and define your idempotent matrix to have its $1$-eigenspace to be $W_1$ and its kernel to be $W_0$.

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Finally, the collection of idempotent matrices does not form a linear subspace of the vector space of $n \times n$ matrices, so I am not sure a simple notion of "degrees of freedom" can be obtained, although I may be wrong.

Answer 2

Let $f:P\in M_n\rightarrow P^2-P$ and $V$ be the variety $f^{-1}(0)$.

The derivative of $f$ in an idempotent matrix $P$ is:

$Df_P:H\in M_n\rightarrow PH+HP-H$, that is, $Df_P=P\otimes I_n+I_n\otimes P^T-I_n\otimes I_n$

when we stack the matrices row by row. $Df_P$ is the sum of three matrice that pairwise commute; moreover, since any idempotent $P$ is diagonalizable, these matrices are diagonalizable and, consequently, are simultaneously diagonalizable. Let $spectrum(P)=(\lambda_i)_i$; the previous result implies that $spectrum(Df_P)=(\lambda_i+\lambda_j-1)_{i,j}$.

The dimension of $V$ is the max of the $dim(\ker(Df_P))$ when $P$ varies through the idempotent matrices, that is, the max of the number of zero eigenvalues of $\ker(Df_P)$ when $P$ varies.

Since $\lambda_i+\lambda_j-1=0$ iff we consider a couple $0,1$ of eigenvalues, we can convince ourselves that this max is obtained when the $0$'s and the $1$'s are about equal in number in $spectrum(Df_P)$.

Case 1. $n=2p$ is even. We consider some $P$ s.t. $spectrum(P)$ has $p\times 0$ and $p\times 1$. The number of couples $(0,1)$ is $p^2$ and the number of couples $(1,0)$ is also $p^2$. Then the maximal local dimension is $2p^2$.

Case 2. $n=2p+1$ is odd. We consider some $P$ s.t. $spectrum(P)$ has $p\times 0$ and $p+1\times 1$. The number of couples $(0,1)$ is $p^2+p$ and the number of couples $(1,0)$ is also $p^2+p$. Then the maximal local dimension is $2p^2+2p$. In other words, $dim(V)$ is the integer part of $n^2/2$.

EDIT. Answer to @Jose Brox . We assume that the underlying field is $\mathbb{C}$. $V$ is not a pure variety -its local dimension is locally constant but not globally- but rather an algebraic set. Let $V_r$ be the set of idempotent matrices of rank $r$; then $V$ has $n+1$ connected components $V_0,\cdots,V_n$, each being a pure variety of dimension $2r(n-r)$ (cf. the above proof); in particular $V_0=\{0_n\},V_n=\{I_n\}$.

Another way to see that $dim(V_r)=2r(n-r)$ is as follows: to define an idempotent $P$ of rank $r$ is equivalent to choose two complementary vector spaces $W_0,W_1$ of dimension $n-r,r$: $W_0=\ker(P),W_1=im(P)$. Note that $W_0\in G_{n-r,n},W_1\in G_{r,n}$ where $G_{n-r,n},G_{r,n}$ are Grassmannian varieties of dimension$(n-r)r,r(n-r)$. Then $dim(V_r)=dim(G_{n-r,n}\times G_{r,n})=2r(n-r)$; indeed, if you are very unlucky, $W_0,W_1$ intersect; such a choice must satisfy a system of algebraic relations; then your choice is contained in a Zariski closed set of dimension $<2r(n-r)$.

Now, $dim(V_r)=2r(n-r)$ means that, for every $P_0\in V_r$, there are neighborhoods $R$ of $P_0$, $S$ of $0$ in $\mathbb{C}^{2r(n-r)}$ and a diffeomorphism $\Pi_{P_0}:R\rightarrow S$. Finally, $(\Pi_{P_0})^{-1}$ locally realizes a parametrization of $V_r$. That was the good news; the bad one, is that this result gives no explicit parameterization.

For example, consider the simple case when $n=2,r=1$; let $P_0=\begin{pmatrix}a_0&b_0\\c_0&1-a_0\end{pmatrix}\in V_1$ where $a_0(1-a_0)=b_0c_0$. If $b_0\not= 0$, then a local param. is $(a,b)\in \mathbb{C}\times \mathbb{C}^*\rightarrow \begin{pmatrix}a&b\\a(1-a)/b&1-a\end{pmatrix}$. We similarly treat the case $c_0\not=0$; yet, what is an admissible param. when $P_0=diag(0,1)$ or $P_0=diag(1,0)$ ?

Of course, $Q\in GL_n\rightarrow Qdiag(0_{n-r},I_r)Q^{-1}\in V_r$ is not an admissible param.!!