Let $A\in M(n,\mathbb{R})$ be an invertible matrix. Consider the (real) eigenvalues $\lambda_1,\cdots,\lambda_n$, in increasing order, of the positive-definite symmetric matrix $A^t A$. We shall denote the eigenvalues as $\lambda_i(A)$.

Question What can be said about the differentiability of the functions $\lambda_i:GL(n,\mathbb{R}) \to \mathbb{R}$?

[We may assume that the domain is $GL^+(n,\mathbb{R})$ for differentiability/smoothness.]

Any reference for this or relevant results would be appreciated.

  • $\begingroup$ After writing my answer below, I noticed that similar questions have been asked on MO: link, link. $\endgroup$ Apr 28, 2020 at 0:04

6 Answers 6


In the open subset of $M_n(\mathbb{R})$ where the $\lambda_i$ are distinct, they are $C^{\infty}$ functions: this follows from the implicit function theorem.

On the other hand, when some eigenvalue has multiplicity $>1$ you don't get more than continuity. For example if $A=\begin{pmatrix} 0 & 1\\ 1 & t \end{pmatrix}$ the largest $\lambda_i$ is $\dfrac{1}{2}\left(t^2+2 +|t|\sqrt{t^2+4}\right)$, which is not differentiable (as a function of $t$) at $t=0$.

  • $\begingroup$ Note that the question is about the eigenvalues of $A^\dagger A$ not about the eigenvalues of $A$ itself. But basically the same example works for that case too. $\endgroup$ Apr 27, 2020 at 5:55
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    $\begingroup$ I am dicussing the eigenvalues of $A^{\dagger}A$, not of $A$ (I am using the notation of the OP). The eigenvalues of $A$ are distinct for $t=0$. $\endgroup$
    – abx
    Apr 27, 2020 at 5:56
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    $\begingroup$ In the simple $A$ given as example, it appears that if we disregard $\lambda_i$ and use a new numeration $\mu_1,\mu_2$ of the eigenvalues that is wrt. the sign of the square root of the discriminant, not wrt. the relative size of the roots, then you can have that both $\mu_1$ and $\mu_2$ are smooth functions of $t$. Possibly see this Wolfram Alpha illustration. Under what conditions is it possible to pick $\mu_i$ with $i=1\ldots n$ "smart" so they are smooth? $\endgroup$ Apr 28, 2020 at 11:08
  • $\begingroup$ Sorry, I should have read the answers by Denis Serre and Marc Nardmann before writing this comment. They address it. $\endgroup$ Apr 28, 2020 at 11:28

The keyword is the Cartan decomposition in the theory of symmetric spaces.

In short, when an eigenvalue is simple (its multiplicity is $1$) it is locally an analytic function. But when the eigenspace is degenerate (the multiplicity is greater than $1$), the eigenvalue function is not differentiable. The problem is essentially one of choosing branches: if you try to deform the identity matrix, there is no consistent way to say which of the resulting distinct eigenvalues after deformation is the eigenvalue that you should have kept track of.

Let $K = \mathrm{O}(n)$, and let $A$ be the group of diagonal matrices with positive entries. You then have $G=KAK$ and if $g=k_1 a k_2$ then the eigenvalues of $g^\dagger g$ are exactly the squares of the eigenvalues of $a$. The problem is that the decomposition is not unique: you can conjugate $a$ by a permutation matrix, and there will be problems when $a$ is fixed by a permutation matrix.

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    $\begingroup$ If I follow what you are saying, then does it follow that $a$ (in the decomposition of $g$) is unique up to conjugation by permutation matrices? In that case, although each $\lambda_i$ may not be well-defined, any smooth function $f$ of $\lambda_i$'s which is invariant under permutations will be smooth. $\endgroup$ Apr 27, 2020 at 5:47
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    $\begingroup$ Yes. A smooth symmetric function will be smooth. The best way to think about it for analytic functions is as follows: the elementary symmetric polynomials are polyomials in the traces of powers of the matrix (that's the Newton identities), and the Taylor expansion of an analytic symmetric function is a sum of elementary symmetric polyomials. $\endgroup$ Apr 27, 2020 at 5:54

As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.

The first one is that each $\lambda_j$ is a Lipschitz function. This statement is still valid if you replace ${\bf Sym}_n({\mathbb R})$ by a subspace $E\subset{\bf M}_n({\mathbb R})$ with the property that the eigenvalues are always real.

The second one is that if $t\mapsto A(t)$ is a smooth curve in ${\bf Sym}_n({\mathbb R})$, then there is a labelling of the eigenvalues $t\in{\cal V}\rightarrow(\mu_1(t),\ldots,\mu_n(t))$ such that each $\mu_j$ is smooth. Mind that this labelling does not respect the order between eigenvalues when the multiplicities vary. Mind also that this becomes false if we replace a curve by a surface.

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    $\begingroup$ A smooth eigenvector labelling (where "smooth" means $C^\infty$, not just $C^1$) does not exist in the generality you claim (and, as far as I can tell, this is not claimed in Kato's book); see the references in my answer below. $\endgroup$ Apr 27, 2020 at 18:38
  • $\begingroup$ @MarcNardmann. Thank you for the precision. I am stuck at home, without access to our library. $\endgroup$ Apr 27, 2020 at 20:06

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2(t) = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

  • If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
  • If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
  • If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
  • Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
  • Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.
  • $\begingroup$ Very clear answer (+1). Do all the bullet points still hold for hermitian matrices? $\endgroup$
    – lcv
    May 1, 2020 at 21:13
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    $\begingroup$ @lcv: Yes. I just cited a few theorems that can be stated succinctly. Many of these results hold in greater generality, for instance for normal matrices or even unbounded normal operators on Hilbert spaces. The reference in Ben McKay's answer contains much more information. $\endgroup$ May 2, 2020 at 3:21
  • $\begingroup$ Your last bullet is false. Your condition about $\lambda_i,\lambda_j$ is used to avoid $\exp(-1/t^2)$-type functions. Then there is a numbering of the eigenvalues $(λi)i≤n$ s.t. the associated functions are $C^{\infty}$. However, the natural ordering of the eigenvalues is not necessarily met; consider, for example, $B(t)=diag(t+2,2t+2)$ when $t$ goes through $0$. See my post in math.stackexchange.com/questions/3601351/… $\endgroup$
    – loup blanc
    May 3, 2020 at 9:49
  • $\begingroup$ I do not understand your objection; I rather suspect that we mean the same thing. The last bullet point is correct. There might be a misunderstanding: Of course I do not claim that the ordered tuple $(\lambda_1,\dots,\lambda_n)$ is $C^\infty$. I claim that one can choose another tuple of eigenvalue functions which is $C^\infty$. I mention the ordered tuple only to state the sufficient condition under which this other choice is possible. $\endgroup$ May 4, 2020 at 3:40
  • $\begingroup$ OK I agree with your comment. $\endgroup$
    – loup blanc
    May 7, 2020 at 10:16

For the best known positive results under mild hypotheses, you might want to look at

Armin Rainer, Perturbation theory for normal operators, Trans. A.M.S., Volume 365, Number 10, October 2013, Pages 5545–5577


In this paper by Xuwen Zhu it is shown that, after resolution by radial blow-ups, the eigenvalues can be made to be smooth: https://arxiv.org/abs/1504.07581.


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