necessary condition for the sum of symmetric matrices to be an orthogonal projection matrix

Question

I am trying to prove a version of Cochran's theorem given here. But i am stuck in the following part of the theorem.

The Problem:-

Let, $A_1,A_2,A_3,\cdots\cdots, A_m$ be $n \times n$ symmetric matrices, with $rank(A_j)=n_j$ and $A_jA_k=0 ,\forall j\ne k$.

Let $A=\sum A_j$ be an orthogonal projection matrix.

Then, Prove that $A_j$ is an orthogonal projection matrix for all $j$.

What i have tried:

$A_i=A_i^T$ is already given and I was able to prove that $\sum A_i = \sum A_i^2$.
I can also prove that the range spaces of the $A_i$'s are disjoint. And this is where I am stuck.
I need to prove that $A_i^2=A_i$. How do i do that?

Answer 1

I think you already have most of the answer, and just one more small observation should finalize it for you. First, it is clear that proving the $m=2$ case also proves the general case, so I'll stick to that.

As you already noticed, we are able to prove that $A^2 + B^2 = A+B$. Now, remember that symmetric matrices have a nice property, namely that $\mathbb{R}^n = U \oplus V$, where $U$ and $V$ are the kernel and image of any fixed symmetric matrix $M$. Thus, in particular, we can take $M$ to be $A$ (or $B$). So in our setup, if we can prove that (1) $B^2u = Bu$ for any $u$ in $U$, AND that (2) $B^2v=v$ for any $v$ in $V$, we would be done (of course this also proves that $A^2=A$ because $A$ and $B$ are interchangeable).

(1) Let $u$ be any vector in $U$ (remember that is the kernel of $A$). Since we know that $A^2+B^2=A+B$, we get $$A^2u+B^2u = Au+Bu,$$ and using the fact that $Au=0$ by definition of $u$, we get $$0+B^2u = 0+Bu \implies B^2=B$$ on $U$.

(2) Notice that $BA=0$ is equivalent to saying that the image of $A$ is in the kernel of $B$, or equivalently $B=0$ on $V$. This obviously implies that $B^2=0$ on $V$, and thus $B^2 = B$ on $V$, which completes the proof.

As a summary, we have showed that

(i) $\mathbb{R}^n = U\oplus V$,

(ii) $B^2=B$ on $U$, and

(iiI) $B^2=B$ on $V$.

Hope this was clear, happy to clarify further if need be.

Answer 2

You can prove this by manipulating the trace. The problem statement implies for all $k$ that

$\sum_{j=1}^m A_j^k=A^k = A=\sum_{j=1}^m A_j$
$\implies \sum_{j=1}^m \text{trace}\big(A_j^k\big)= \text{trace}\big(A\big)= r =\sum_{j=1}^m \text{trace}\big(A_j\big)$

Now consider arbitrarily large even power $k$:
$A_i^k$ has real non-negative eigenvalues (i.e. the square of a real symmetric matrix must be PSD) and observe
$\Big\Vert A_i\Big\Vert_2^k = \Big\Vert A_i^k\Big\Vert_2\leq \text{trace}\big(A_i^k\big) \leq \sum_{j=1}^m \text{trace}\big(A_j^k\big)=r$
the boundedness of the operator 2 norm implies all eigenvalue of $A_i$ have modulus $\leq 1$. Further:

$\sum_{j=1}^m \text{trace}\big(A_j^4\big) \leq \sum_{j=1}^m \text{trace}\big(A_j^2\big)$
is met with equality hence each $A_i$ has no eigenvalues of modulus $\in (0,1)$

Thus all eigenvalues of $A_i$ are $\in\big\{-1,0,1\big\}$. Finally
$\sum_{j=1}^m \text{trace}\big(A_j\big)\leq \sum_{j=1}^m \text{trace}\big(A_j^2\big)$
and this is met with equality, hence there are no eigenvalues of $-1$.

I.e. each $A_i$ has eigenvalues only $\in \big\{0,1\big\}$ and being real symmetric this means it is an orthogonal projection