0
$\begingroup$

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?


Thoughts:

$A^2=A$, 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?

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.