Understanding Projector Matrices: A Math Problem with Complex Numbers

[math8]

Consider the matrix A=uv* where u and v lie in C^n (C:complex numbers). Under what condition on u and v is A a projector?

I know that a projector is a square matrix P for which P^2= P.
Now according to this definition, would that make uv*=Identity. and then u be the inverse of v*?

Also what is the inverse of a vector that belongs to C^n? I mean what is u^-1? Is u^-1 a vector such that uu^-1 = the (nxn) identity matrix?
In what set would such a vector (u^-1) lie in?

 

[lanedance]

Consider the matrix A=uv* where u and v lie in C^n (C:complex numbers). Under what condition on u and v is A a projector?

do you mean an outer product of the column vectors u,v given by
A = uv*
where v* is the complex conjugate transposed?

I know that a projector is a square matrix P for which P^2= P.
Now according to this definition, would that make uv*=Identity. and then u be the inverse of v*?

I haven't done a whole heap with projectors, but here's some ideas based on general linear algebra

does P²= P, imply P = I?
if the inverse exists
P^-1.(P.P) = I.P = P = P^-1(P) = I
what if the inverse does not exist?

Also what is the inverse of a vector that belongs to C^n? I mean what is u^-1? Is u^-1 a vector such that uu^-1 = the (nxn) identity matrix?
In what set would such a vector (u^-1) lie in?

I would try writing out the case for C², and you should be able to see a relation for u & v in terms of their inner product that is easily extendable to the Cⁿ,

 

[HallsofIvy]

Consider the matrix A=uv* where u and v lie in C^n (C:complex numbers). Under what condition on u and v is A a projector?

I know that a projector is a square matrix P for which P^2= P.
Now according to this definition, would that make uv*=Identity. and then u be the inverse of v*?

No, the fact that (uv*)^2= uv* does NOT make uv* the identity operator. For example, the operator P((x,y,z))= (0, y, z) is a projection because P^2((x, y, z))= P((0, y, z)= (0, y, z)= P but P does not map (x,y,z) to itself.

And, in any case, there is no such thing as an "inverse" vector. "Inverse" only applies to functions and operators, not vectors.

Also what is the inverse of a vector that belongs to C^n? I mean what is u^-1? Is u^-1 a vector such that uu^-1 = the (nxn) identity matrix?
In what set would such a vector (u^-1) lie in?

 

[lanedance]

After reading Halls's comment on vectors, consider the case when P is a square matrix defined by the outer product, ie if u,v are column vectors of length n, then P is the nxn matrix given by:

$$P = \textbf{u} \otimes \textbf{v} = \textbf{u} \overline{\textbf{v}}^T$$

in the case you give Halls (sorry for the poor matrix display)
P =
|100|
|010|
|000|
which is clearly not invertible

From the definition of a projector P^2=P. In the matrix format it is clear, that if P is invertible then P=I.

So perhaps in general projectors are not invertible except in the trivial case above. This seems to make sense with the example of projecting Rⁿ onto R^m, with m<n (and the 3-space onto a plane example given by Halls).

In any case, if you perform the outer product defined above for an arbitrary u,v. Then look at what is required to satisfy P²=P, it leads to a simple requirement for u,v in terms of their innner product

 

[HallsofIvy]

Yes, in general a projector "projects" a vector space onto a subspace of smaller dimension. The entire orthogonal subspace is the kernel of the projector and so the projector is not invertible.

But my remarks were really adressed, not to a projector, but to math8's talking about the inverse of vectors- which is not defined.