@ Farid,

Thank you for the reply.

Sure, there are many results not listed with the question

such as the only nonsingular idempotent matrix is the identity, the literature is rich with such results that discuss the kernels, images, ranks, spectrums

of idempotent matrices, their additions as well as their multiplications.

I am interested in new forms of matrices that are idempotent, analog to that appears in (2) and (5).

Did you encounter such forms somewhere?

Best regards

Please see the attachment.

@ Wiwat,

Thank You for sharing your article.

I will back to you after reading the article.

Best regards

@Wiwat,

Thank you for the simple and readable article that includes a family of idempotent matrices that satisfy nice properties and relations.

Best wishes

Thank you Sir.

Dear Isam,

I just want to point these

in 4) the " A = A² if and only if rk(A) = tr(A " is sufficient, the rest is a consequence of it.

In 5) I want to reformulate your expression and so you see it is just the consequence in 1.

I-XX^+ , where X^+ is the Moore- Penrose inverse of X. (XX^+ is idempotent).

in 6) please add such that A is not of rank 1, otherwise k is an eigenvalue of A, which yields to k=O or 1

Best.

Dear Hanifa,

Thank you for your contribution. Notice that I start with:

[ Here are some well-known results about idempotent matrices].

The mentioned properties are well- known published results.(Not mine).

(*) < in 5) I want to reformulate your expression and so you see it is just the consequence in 1.

I-XX^+ , where X^+ is the Moore- Penrose inverse of X. (XX^+ is idempotent).>

regardless of the name:

M=I_{n} - [X](X^(T) [X])⁻¹[X]^(T) is idempotent, where

[X] is any n×k full rank matrix( not necessary a square matrix)

(*)< in 6) please add such that A is not of rank 1, otherwise k is an eigenvalue of A, which yields to k=O or 1>

In 6: If A²=kA, then (1/k)A is idempotent,( k is not 0 ).

Proof: ((1/k)A)2 =(1/k2)A2 =(1/k2)(kA)=(1/k)A

therefore (1/k)A is idempotent matrix and no further conditions are required.

Best wishes

Thank you dear Isam;

For 5) XX^+ where X any matrix not necessary square and not necessary of full rank. You can put X in full rank factorization and you have the expression in 5

For 6, yes you are right, I confused between idempotency of A and (1/k A)

Dear Hanifa,

In (5) X should be in full rank to make sure XT X is invertible and the formula is applicable.

check by examples.

Best regards

.

Dear Isam,

An idempotent is a projector, so what do you think in kernels lemma ( le lemme des noyaux)?

we can extract expressions:

1) Let u be an endomorphism ( so A is a square matrix of u) and let f(X) a polynomial such that f(u)=0 ( so we can take the characteristic polynomial of u).

2) put f(X)=g(X) h(X), such that gcd(g(X), h(X))=1, then there exist q(X) and l(X), such that g(X) q(X)+h(X) l(X)=1, thus we can find a projector (idempotent) p ' such that p=g(u)q(u), which let us build an idempotent from any square matrix.

For 5)

Let A=XY be full rank factorization of any matrix A, then AA^+=XY(Y^t(YY^t)^(-1))(X^t X)^(-1)X^t =X(X^t X)^(-1)X^t.

Conclusion:

For any full rank matrix X, the matrix X(X^t X)^(-1)X^t is idempotent, thus we do not need to say I_n-(X(X^t X)^(-1)X^t.) because it is consequence of result in 1.

That what I meant.

Best Regards

Dear Isam,

We know the necessary and sufficient conditions for a matrix to be idempotent, that is, a square matrix A is idempotent if and only if  ker(A) = Im(I - A).

Best Regards

Dear Hanifa,

In a previous post, you said that X is not necessarily a full rank. Now, in the conclusion of your last post, you considered X is full rank.

If either A or I - A is idempotent, then both are idempotent.

Now, you replaced idempotent by a projector.

All stated results are correct and selected from published papers.

Also, I know what Bezout's lemma is.

I appreciate if you can extract a new result concerning idempotent matrices using such lemma.

Thank you

Dear Wiwat,

Yes, you are right, if you back to the correspondence with prof. Fraid we agree that the literature is rich with results concerning kernels, images, etc.

The main goal of this question is how to build new forms of idempotent matrices and to express their entries explicitly.

Thank You

Dear Issam,

1) What is the difference between projectors and idempotents?

Which I know is that a projector is idempotent by definition and an idempotent matrix is a projector by results, because it is a matrix of index 1 (A²=A gives R(A²)=R(A) which also gives N(A)Direct sum R(A)=the entire space, where R(A) and N(A) designate the range and the null spaces of A. So A is a projector on its range along its kernel.

2) If we use the kernel lemma we get the same result because the minimal polynomial of A is X(X-1), so just we have A and I-A can not give much new results.

3) If you agree that idempotent matrices are projectors, then the field of generalized inverses may serve new constructions of idempotents

For example: Let A=XY be full rank factorization, such that A the group inverse A^# exists ( i.e. A is of index 1) then, we have the idempotents AA^#=XY(X(YX)^-2)Y and (A^#)A=X((YX)^-1)Y

Also for any matrix A, let A=UDV* be the SVD decomposition, then

AA^+ = U(DD^+)U* is idempotent.

Best Regards.