Curious Case of Idempotent Matrices – Seeking a Generalisation

idempotentslinear algebramatrices

Here's what I initially started with:

Find a 2×2 non zero matrix $A$, satisfying $A^2=A$, and $A\neq I$.

I understand that this is fairly easy, but please keep reading for something interesting coming up –

Let's bash it. Assume a matrix A = $\begin{bmatrix}a&b \\c&d\end{bmatrix}$. Putting $A^2 = A$ gives me the following system to solve:

  • $bc = a(1-a)$
  • $bc = d(1-d)$
  • $b(a+d)=b$
  • $c(a+d)=c$

Some conclusions:

  1. If $a+d = 1$, then $bc = ad$. That is, if we assume a certain value
    for a, we have d, and choosing a value for b, gets us c (or the
    other way round). So, knowing one of the tuples $(a,b), (a,c),
    (d,b),$     or $(d,c)$ determines the matrix $A$. On the other hand, if we choose $b$ and $c$ to begin with, we know $a$ and $d$ from the obvious quadratic equations that follow. Knowing the tuple $(b,c)$ also determines the matrix.

  2. If $a+d\neq 1$, then $(b,c)$ must be (0,0) for the last two
    equations to hold. Next, we're left with $a^2=a$ and $d^2=d$, which
    means $(a,d)$ is $(1,1)$ (we reject $(0,1)$ and $(1,0)$ since that'd
    mean $a+d=1$, and also $(0,0)$ since it'd result in a null matrix).
    This means that, if $a+d\neq1$, then $a+d=2$ with $a=d=1$, and
    $(b,c)=(0,0)$.
    We seem to have no power here (can't choose variables the way we did in the previous case), as $a+d\neq1$ alone determines the entire
    matrix.     Anyway, we shall ignore this for now, since we demand $A\neq I$

In conclusion, knowing one of the rows or columns determines the entire matrix. (the matrix not being null, or identity). Also, knowing the diagonal other than the main diagonal determines the entire matrix.

The observation here, is that in a $2$x$2$ matrix, which has $4$ entries, knowing any pair of entries other than the one along the main diagonal helps us determine other entries.

Why is it so? Could we have said this without going through such cumbersome algebraic weightlifting?

Does this generalise for $n$x$n$ idempotent matrices? That is, can we deduce something along the lines of:

  • Knowing any row or column determines the matrix OR
  • Knowing a certain minimum number of rows of columns (>1) determines the matrix OR
  • Knowing the diagonal other than the main diagonal determines the matrix OR

really anything along those lines. My gut came up with the above possibilities, if this apparently interesting pattern is to hold for matrices of higher order. I really feel there's something worth paying attention to, going on here.

I'd love it if y'all could share your thoughts on this, and help me identify a possible pattern. It'd be great to generalize this idea to higher order idempotent matrices, perhaps even others, if there's nothing special about $A^2=A$ here. I think this is a really important question, since it really boils down to, knowing a constraint in the form of a matrix, how many entries do I need to know, to determine the rest of the matrix? (uniquely determine, if that pleases you)

Hoping to find something amazing, wish y'all a great day!

Best Answer

$A\in M_n(\mathbb{R})$ is a projector (eventually non-orthogonal). The projectors are classified by their trace. Assume that $rank(A)=trace(A)=r\in (0,n)$.

$A$ is associated to a (unique) decomposition $\mathbb{R}^n=E\oplus F$ where $dim(E)=r,dim(F)=n-r$. The couple $(E,F)$ -and then $A$- depends on $r(n-r)+(n-r)r=2r(n-r)$ algebraically independent parameters.

Finally, if you choose skillfully $2r(n-r)$ entries of $A$, then there are only a finite number of possible values for the projector $A$.

In particular, if $r=1$ or $r=n-1$ (projection on a line or on a hyperplane), then it suffices to fix $2n-2$ entries of $A$ (but not just any).

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