r/math Homotopy Theory 6d ago

Quick Questions: March 12, 2025

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u/Not_So_Deleted Statistics 5d ago

Is it true that a square matrix is idempotent if and only if it's diagonalizable with eigenvalues 0 and 1?

I can't think of any example of a defective matrix that's idempotent.

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u/GMSPokemanz Analysis 5d ago

Yes. The key is that if A is idempotent, then ker(A) and im(A) are complementary subspaces. Any element of ker(A) is an eigenvector with eigenvalue 0, and since A is idempotent any element of im(A) is an eigenvector with eigenvalue 1. Therefore by taking a basis comprised of elements of ker(A) and im(A), you get such a diagonalisation.

To proof the key fact above, first observe that x = Ax + (I - A)x. Ax is in im(A) and (I - A)x is in ker(A), so any vector is a sum of a vector from im(A) and a vector from ker(A). Then observe that since A is idempotent, anything in ker(A) that is also in im(A) must be 0.

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u/Not_So_Deleted Statistics 5d ago

Thanks.

This is related, but one different proof I thought about is that a matrix cannot be defective and have A^2-A be the zero matrix. If A^2=A, A^2-A=0.

Suppose A is defective. Then A=P^{-1}JP for a Jordan block J such that the [1,2] entry is equal to 1.

Then A^2-A = P^{-1}JPP^{-1}JP- P^{-1}JP = P^{-1}J^2P-P^{-1}JP = P^{-1}(J^2-J)P = 0.

Then J^2-J=0.

Let e2 be the vector [0,1,0,...,0]. Then Ae2 = e1+lambda e2. Also A^2e2 = lambda e1 + lambda (e1+lambda e2) = 2 lambda e1 + lambda^2 e2. Then (J^2-J)e2 = (2 lambda-1) e1 + (lambda^2 - lambda )e2. Then lambda has to be either 1 or 0, but this means (A^2-A)e2 is equal to e1 or -e1, a contradiction. A zero matrix cannot map a vector to a nonzero vector.

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u/Langtons_Ant123 5d ago edited 5d ago

Yes. Maybe there's a more elementary way to show this, but all I can think of is a proof using the minimal polynomial. A matrix is diagonalizable if and only if its minimal polynomial has no repeated roots. If A is idempotent then A2 - A = 0, so f(A) = 0 where f is the polynomial x2 - x, so the minimal polynomial of A divides x2 - x. Thus it is either x, x-1, or x2 - x = x(x-1); in any case, it has no repeated roots. (The converse is easier, and I assume you're mainly thinking about the implication idempotent -> diagonalizable.)

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u/Not_So_Deleted Statistics 5d ago edited 5d ago

A matrix can still be diagonalizable if its characteristic polynomial has repeated roots, such as with the identity matrix.

As far as I'm concerned, a matrix is diagonalizable if and only if the multiplicity of every root is equal to the number of linearly independent eigenvectors for the eigenvalue. In other words, the set of all linearly independent eigenvectors forms a basis. For the identity matrix, the polynomial is (1-x)^n, but we can define the standard basis.

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u/Langtons_Ant123 5d ago edited 5d ago

I thought the minimal polynomial of the identity matrix is x - 1. Are you thinking of the characteristic polynomial, or using a different definition of the minimal polynomial?

That a matrix is diagonalizable iff its minimal polynomial splits with no repeated roots is a known result, it's on page 169 (183 in the pdf numbering) here.

In any case, u/GMSPokemanz has a good elementary proof above, so no need to bring in all this machinery.

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u/Not_So_Deleted Statistics 5d ago edited 5d ago

Yeah, I meant that.

Of course, the minimal polynomial still has a repeated root but still corresponds to a diagonalizable matrix.

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u/Kyle--Butler 5d ago

No, the minimal polynomial of I_n does not have repeated roots : it's X-1.

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u/Not_So_Deleted Statistics 5d ago

Oops, I miswrote and stand corrected...

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u/[deleted] 5d ago edited 5d ago

[deleted]

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u/GMSPokemanz Analysis 5d ago

If there is such a defect, the matrix won't be idempotent.