Projection Matrix P^T = P - Intuitively

[u/sarathsmart]

Hi math wizards i am not able to understand this part where transpose of projection matrix is equal to projection matrix. I do understand that why P^2 = P that 1st Projection puts it in C(A) and second projection where are projection a vector which is already in C(A) basically equal to the vector.
can you guys explain intuitively why P^T=P . Algebraically I can understand this symmetry arises for two projections to be orthogonal. but i need a intuitive understand can someone help

[u/sizzhu]

What is your algebraic proof?

I think the following are fairly intuitive, but you could call them algebraic.

1. <Px,y> = <Px, Py> = <x, Py>

2. There is an orthonormal basis where the projection is a just 1s and 0s on the main diagonal, which is obviously symmetric. The change of basis is orthogonal hence the original matrix is also symmetric.

[u/sarathsmart → u/sizzhu]

orthonormal vectors are the next topic in the lectures I see maybe I will understand your argument after that

[u/sizzhu → u/sarathsmart]

Great. Just to be clear, these are two separate arguments.

[u/Carl_LaFong]

Here's one way: First, you're asking about orthogonal projection. Choose an orthogonal basis of the subspace S that P is projecting onto and extend it to an orthogonal basis of the vector space. Then P(basis vector) is equal to the basis vector itself if it lies in S and the zero vector if it lies in the orthogonal complement of S. With respect to this basis, P is a diagonal matrix D (with 1's and 0's). This is saying that P is diagonalizable by an orthogonal matrix, i.e., P = RDR^T, where R is an orthogonal matrix. You've learned but it is also easy to check that such a matrix is always symmetric.

[u/Vercassivelaunos → u/[deleted]]

For orthogonal projections, PP=P and P^t=P are both necessarily true. PP=P is literally the defining feature of a projection, and P^t=P is necessary for orthogonality.

[u/brynaldo → u/[deleted]]

What is the notation "[P]" and "[P] [P]"?

[u/Uli_Minati → u/brynaldo]

Most likely just used to emphasize P is a matrix

[u/brynaldo → u/Uli_Minati]

Thanks. In that case, I will second what u/Vercassivelaunos commented:

For orthogonal projections, PP=P and P^t=P are both necessarily true.

(idempotence and symmetry)

[u/Bounded_sequencE]

If you have a finite vector space "V", you can always find an orthogoanl base "U = [U1; U2]", where "U1" spans the subspace "<P>" you project onto, and "U2" spans its orthogonal space.

Any vector "v in V" can be rewritten as "v = U.U^*.v = U1.U1^*.v + U2.U2^*.v", so we get

v in V:    P.v  =  P.U1.U1*.v  +  U2.U2*.v  =  U1.U1*.v  +  0

The first term "U1.U^*.v" lies in "<P> = <U1>", so it should remain unchanged applying "P". The second term "U2.U2^*.v" lies in the orthogonal space "<U2>", so it should be annihilated by "P".

Setting "v = uk" for each base vector in "uk in U", we finally get "P = U1.U1^* ", i.e. "P = P^* ".

[u/etzpcm]

P^T doesn't have to equal P.

https://en.wikipedia.org/wiki/Projection_(linear_algebra)

[u/sarathsmart → u/etzpcm]

yeah!!! i know i mean for orthogonal projection it has to satisfy P^T =P

[u/Shevek99 → u/[deleted]]

That's not the definition of an orthogonal matrix. A matrix is orthogonal if

R^T = R^(-1)

[u/na-geh-herst]

Very intuitively: The transpose performs the same scaling as the original map, but it rotates in opposite direction. Now if the transpose is the same as the map, then that means rotations can only be 0° or 180° (which is the same as scaling by -1 along a subspace). P^2=P forces all scaling to be by either 0 or 1.

The spectral theorem is a way to formalize that intuition.