There are some equivalent statements in the classical stability theory of linear homogeneous differential equations $ \dot{x} = Ax, x \in \mathbb{R}^n $ such as:

  1. All eigenvalues of $A$ have negative real parts
  2. For any symmetric, positive definite matrix $Q$ there is a unique symmetric, positive definite solution $P$ to the Lyapunov equation $A^T P + PA = -Q$
  3. The equilibrium of $\dot{x} = Ax$ is exponentially stable

All these statements in some way or another use the matrix exponential $e^{At}$. In particular, the solution to the Lyapunov equation can be found by $P=\int \limits_{0}^{\infty} e^{A^Tt}Qe^{At} dt$. In turn, solutions to the ODE have the from $e^{At}x_0$.

All proofs I know that show equivalence of these three statements use the Jordan normal form of $A$. In fact, the matrix exponential itself is notoriously hard to handle in practice, and one tries to represent it in some suitable form via the said normal form, or by partial fraction decomposition of $sI - A$ and inverse Laplace transforming it.

My question is: do there exist proofs of equivalence of the said three statements (or some of them) that do not use the Jordan normal form and possibly do not explicitly use eigenvectors?

  • 3
    $\begingroup$ I am confused by your motivation. What are you really trying to do? There are perfectly good ways to compute (numerically) matrix exponentials and solutions of linear ODES without Jordan forms. On the other hand, Jordan forms are more convenient for proofs. But the two tasks are fundamentally different and you shouldn't think of handling them both with the same apparatus. $\endgroup$ Dec 26, 2016 at 11:38
  • $\begingroup$ I am away from my department, so don't have access to the library. But I remember that the 2nd volume of Gantmacher's book is fully dedeicated to these stability issues. Did you check it ? $\endgroup$ Dec 28, 2016 at 7:52
  • $\begingroup$ @DenisSerre As far as I remember, he addressed the problem with matrix exponential and eigenvalues in Vol. 1. And yes, it used the Jordan normal form $\endgroup$
    – Rubi Shnol
    Dec 28, 2016 at 9:19

2 Answers 2


I do not know if this is what you are after. The following shows that the Jordan form can be replaced by the Schur form. This feels a bit like cheating. Is this enough? The trickier parts are the statements about the real parts of the eigenvalues. It is of course a bit harder to say useful things about eigenvalues when we are not allowed to speak about eigenvectors. If computational considerations are a concern then this might be admissible as the Schur form is obtained via orthogonal/unitary transformations. On the other hand in computing the Schur form you compute at least one eigenvector.

It is bit more convenient to argue over $\mathbb{C}$. Real version of what is below are easy to come by.

Note 1: I tried my best to avoid eigenvectors. To me this feels artificial. The best proof I know for (3) $\Rightarrow$ (1) is: ``Assume $\lambda$ is an eigenvalue of $A$ with nonnegative real part. Let $x$ be an eigenvector. Then $e^{At}x = e^{\lambda t}x$ does not converge to zero, contradicting exponential stability.'' Compared with this the proof below is a bit over the top.

Note 2: If computational approaches to stability are what you are after and you do not like eigenvectors, why do you not simply dispense with (1) and live with the equivalence of (2) and (3). The solution of (2) asks for solving a system of linear equations. This can frequently be done and is often much easier than computing the eigenvalues anyway.

Finally, the answer: Most of this is only a sketch. I hope this will be sufficient. For the equivalence (2) $\Leftrightarrow$ (3) no information about the eigenstructure of $A$ is needed.

We call the equilibrium position $\hat x:=0$ of $\dot x = Ax$ exponentially stable, if there are $M,\beta>0$ such that for all $t\geq0$ we have $$ \|e^{At}\| \leq M e^{-\beta t}.$$ By equivalence of norms on finite-dimensional vector spaces, we may use any norm on $\mathbb{C}^{n \times n}$ for this. I will allow myself the (common) abuse of terminology of calling the system exponentially stable, if the equilibrium position $\hat x=0$ is.

(2) $\Rightarrow$ (3): Let $P$ be the solution corresponding to $Q=I$. Consider the Lyapunov function $V(x) = x^T P x$ and compute that the derivative along solutions is $$ \dot V (x(t)) = - \|x(t)\|^2. $$ From this conclude exponential stability.

(3) $\Rightarrow$ (2): Given $Q$ positive definite, define (as you say) $$ P := \int_0^\infty e^{A^*t}Q e^{At} dt .$$ Check that this is well-defined by exponential stability, Hermitian and positive definite and that the integral evaluates to $-Q$.

(1) $\Rightarrow$ (3): We will use the following small

Lemma: If $M\in \mathbb{C}^{n \times n}$ defines an exponentially stable system and $\lambda \in \mathbb{C}$ with $\mathrm{Re} \lambda <0$ then for all $K \in \mathbb{C}^{n \times 1}$ the matrix $$ \begin{bmatrix} M & K \\ 0 & \lambda \end{bmatrix} $$ defines an exponentially stable system.

Proof: Follows by a computation using the variation-of-constants formula. $\square$

With this lemma at hand, the argument is now as follows. Assume all eigenvalues of $A$ have negative real part. Let a Schur form of $A$ be given by $$ UAU^* = \begin{bmatrix} \lambda_1 & * & \ldots& * \\ 0 & \lambda_2 & *&*\\ & \ddots& \ddots & * \\ 0 &&0& \lambda_n \end{bmatrix}$$ The $1\times 1$ submatrix $\lambda_1$ clearly defines an exponentially stable system. The exponential stability of the upper left $k \times k$ submatrices then follows from the lemma. By continuing until $n$ we obtain exponential stability of $UAU^*$ and then also of $A$.

(3) $\Rightarrow$ (1): Again let $UAU^*$ be a Schur form of $A$. The diagonal entries of $U e^{At}U^*$ are of the form $e^{\lambda_j t}$, where $\lambda_j, j=1,\ldots,n$ are the eigenvalues of $A$. As $e^{At}$ goes to zero exponentially as $t\to\infty$, it follows that $e^{\lambda_j t}\to 0$ exponentially for all $j$. Thus all eigenvalues have negative real part.


You can use a weaker version of Jordan normal form, namely an upper triangularization. There is a very straightforward conceptual proof that every square matrix over an algebraically closed field $k$ admits an upper triangularization: take $T : V \to V$ to be a linear operator where $V$ is a vector space over $k$, so $T$ admits an eigenvector $v_1$. Now consider the induced linear map

$$T : V/\text{span}(v_1) \to V/\text{span}(v_1).$$

This new linear map also admits an eigenvector; pick some lift $v_2 \in V$ of it. Continuing in this way, we construct a basis $v_1, \dots v_n$ of $V$ with the property that

$$T v_i \in \text{span}(v_1, \dots v_i).$$

This is precisely a basis with respect to which $T$ is upper triangular. This allows us to write down a matrix for $T$ which is the sum of a diagonal matrix (whose entries are the eigenvalues of $T$) and a nilpotent matrix (the rest of the triangle). From here it is straightforward to bound the coefficients of the matrix exponential $e^T$ in the case that $k = \mathbb{C}$.

I agree with Federico Poloni's comment that this discussion has little relevance to the question of how difficult it is to compute the matrix exponential in practice.

  • $\begingroup$ The Schur form used in Fabian Wirth's answer is precisely a version over $\mathbb{C}$ of the upper triangularization that you mention. So your answers are related, although with different language. $\endgroup$ Dec 30, 2016 at 9:50

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