Suppose we are given some square $n$ by $n$ matrix $A$ which we think might be a standard matrix for a the orthogonal projection onto some subspace $V$. What are sufficient conditions for $A$ to be such a matrix?


$A^2=A$, because projecting onto a subspace twice results in the same image as only projecting once.

$E_{1}=\{x \in \mathbb{R}^n : Ax=x \}$ , the egienspace of $A$ corresponding to eigenvalue $1$ must contain $V$ since the projection of any vector in $V$ is itself. I think it also must be equal to $V$.

If that is the case, we must have $\text{Proj}_{E_1}e_1=Ae_1$ (the first column of A) and etc (likewise for second column, third).

But are these necessary conditions sufficient? If not, how to go about finding what conditions are sufficient?


If $A^2 = A$, then the only possible eigenvalues are $0$ and $1$, as $A$ is a root of the polynomial $(z - 0)(z - 1)$.

Suppose further that $A$ is diagonalisable. If $0$ is the only eigenvalue then the map is the $0$ map, hence a projection onto the trivial subspace. If $1$ is the only eigenvalue, then the map is the identity map, hence a projection onto the entire space.

Otherwise, $0$ and $1$ are both eigenvalues. It's not difficult to see, by considering a basis of eigenvalues, that this map is a (not necessarily orthogonal) projection onto $E_1$ via $E_0$.

If you want an orthogonal projection, you need these subspaces to be orthogonal. When you have diagonalisability, real eigenvalues, and all of your subspaces orthogonal, this makes your matrix self-adjoint (Hermitian, or symmetric in the real case). In fact, this is equivalent to the previous three properties.

So, in conclusion, $A$ being an orthogonal projection matrix is equivalent to $A^2 = A$ and $A^* = A$ ($A^T = A$ in the real case).


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