# Space of Defective Matrices

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## Summary:

What is the geometry/topology of the set of defective matrices

## Main Question or Discussion Point

Let me start by saying that my question will be somewhat vague by mathematical standards. I'm not a mathematician! I'm looking for some intuition about how defective matrices are distributed in the space of all matrices. I understand that they are rare and in some sense discontinuous - matrices switch from being non-normal to defective with an infinitesimal change of parameters. But for example, do they form some kind of manifold? Does the set of defective matrices have measure zero? Are there any interesting known properties about the topology or geometry of the set of defective matrices?

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I suspect I might be able to make some headway into my question using the Jordan normal form. We can use this as a constructive definition of the set of defective matrices. A defective matrix A must satisfy A = P*J*inv(P), where J is a Jordan matrix and P is an invertible matrix. Conversely for any invertible P and Jordan matrix J, such an A will be a defective matrix. In the most extreme case where the matrix has only one eigenvector, J has lambda on the diagonal, 1 on the superdiagonal and zeros elsewhere. We could then ask what the set of such A looks like for arbitrary (invertible) P. Then we could proceed to the more complicated cases where A has 2 eigenvectors, 3 eigenvectors, ... up to N-1 eigenvectors. I'm not sure I can even picture the constraints on A in the case of 1 eigenvector though!

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Infrared
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Does the set of defective matrices have measure zero?
Yes I think so, because a matrix can only be defective if its characteristic polynomial has a repeated root (this is a necessary, not sufficient condition). This has probability zero.

Are there any interesting known properties about the topology or geometry of the set of defective matrices?
I think the space of defective matrices is path-connected. Any Jordan block can be joined by a path of defective matrices to, say, the block with eigenvalue ##1## of the same size by continuously changing the eigenvalue. Do this for all blocks of a matrix in Jordan form simultaneously, and then continuously change the ##0## in the ##(i,i+1)## positions between blocks to a ##1## (this does not change that the matrix is still defective since if a matrix other than the identity has ##1## as its only eigenvalue, then it cannot be diagonalizable). This gives a path from any matrix in Jordan form to the Jordan block of the same size with eigenvalue ##1##. Since any defective matrix is of the form ##SJS^{-1}## where ##J## is a nontrivial Jordan form, this should prove path-connectedness.

Surprisingly, I think this space might actually be a manifold. I'll post an argument later.

I don't see how we could translate the existence qualifier on eigenvectors into an algebraic
A matrix fails to be diagonalizable if and only if its minimal polynomial has repeated roots.

fresh_42
fresh_42
Mentor
A matrix fails to be diagonalizable if and only if its minimal polynomial has repeated roots.
So the criterion could be that for ##\chi_\mathbb{C}(A)=\prod (x-\lambda_k)^{n_k}## we require ##\prod n_k \neq 1## which looks Zariski open and dense, and thus no variety!?

martinbn
... which looks Zariski open and dense, and thus no variety!?
Consider ##x\not =0## in the affine line. It is open and dense, yet it is a variety. It can be realized as the zero set of ##xy=1## in the plane. For the zeros of a polynomial you have ##p(x_1,x_2,...,x_n)y=1##. Same way as why the general linear group is an algebraic group.

fresh_42
Infrared
Gold Member
So the criterion could be that for ##\chi_\mathbb{C}(A)=\prod (x-\lambda_k)^{n_k}## we require ##\prod n_k \neq 1## which looks Zariski open and dense, and thus no variety!?
I think ##\chi## usually refers to characteristic polynomial, not minimal polynomial. Anyway, this set is way too small to be Zariski open- most matrices are diagonalizable. In particular, the space of matrices with distinct eigenvalues is disjoint from our space and is Zariski open (since the complement is the zero set of the discriminant of the characteristic polynomial), but two nonempty Zariski open sets in ##M_{n\times n}=\mathbb{R}^{n^2}## cannot be disjoint.

Let's think about the ##2\times 2## case more: To fix notation, let ##X## be the set of ##2\times 2## matrices which have a repeated eigenvalue, and let ##Y\subset X## be the subspace of non-diagonalizable (defective) matrices. Note that ##Y=X-\{cI:c\in\mathbb{R}\}##. This is because the only diagonalizable ##2\times 2## matrix with repeated eigenvalue ##\lambda## is ##\lambda I##. In particular, ##Y## is a open subset (even Zariski open) of ##X##.

Let ##A=\begin{pmatrix}a & b\\c & d\end{pmatrix}## and ##\chi_A(\lambda)=\lambda^2-(a+d)\lambda+(ad-bc)## be its characteristic polynomial. Then ##X## is the zero set of ##\text{disc}(\chi_A)=(a+d)^2-4(ad-bc)=(a-d)^2+4bc=0##.

Consider the map ##F(A)=(a-d)^2+4bc##. Note that ##\nabla F=(2(a-d),4c,4b,-2(a-d))## is zero if and only if ##a=d## and ##b=c=0##, i.e. ##A## is a scalar matrix. So ##X=F^{-1}(0)## is a manifold away from the scalar matrices. Since ##Y## is an open subset of ##X## not containing the scalar matrices, it should be a manifold.

I think the only part of this argument that doesn't generalize easily to higher dimensions is explicitly computing ##F## and ##\nabla F##.

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