TITLE: Finding the nearest matrix with real eigenvalues
QUESTION [17 upvotes]: In this thread on MATLAB Central, I found a discussion on finding the nearest matrix with real eigenvalues. The first hypothesis was to simply truncate the complex part of the eigenvalues. So, given matrix $A$, the closest matrix to $A$ in some norm would be 
$$A' = V \; \mathrm{real}(D)\; V^{-1}$$
where $A= V D V^{-1}$ is the eigendecomposition of $A$ (assuming $A$ is diagonalizable). This has been found to be false, however, with the counterexample
$$A = \begin{bmatrix} 
     1& 1& 0 \\
    -1& 0& 0 \\
     0& 0& 0 \\
\end{bmatrix}$$ 
The procedure above would produce 
$$A'=\begin{bmatrix} 
     0.5& 0& 0 \\
     0& 0.5& 0 \\
     0& 0& 0 \\
\end{bmatrix}$$ 
but there is another matrix 
$$A''=\begin{bmatrix} 
     0.9& 0& 0 \\
    -2& -0.1& 0 \\
     0& 0& 0 \\
\end{bmatrix}$$
that has real eigenvalues and is closest to $A$ both in spectral and Frobenius norm.
Now, let $ A = U T U' $ be the real Schur decomposition of $A$, with $U$ orthogonal and $T$ quasi-upper triangular (there are 2-by-2 blocks on the diagonal corresponding to complex eigenvalues). Consider $ A_s = U \; \mathrm{triu}(T) \; U' $ where triu returns the upper triangle part.  By numerical simulations $A_s$ it is always closer to random $A$ than $A'$ and on the counterexample it is also closest to $A$ than $A''$ (in Frobenius and spectral norm).
My conjecture is that $A_s$ is the closest matrix with real eigenvalues to $A$ in Frobenius (or spectral) norm.
Unfortunately I could not prove or disprove this conjecture. Can anybody?
A starting point can be 
$$||A-A_s||_F = ||T-\mathrm{triu}(T)||_F$$

REPLY [4 votes]: Vanni Noferini and I have published an arxiv e-print, Nearest $\Omega$-stable matrix via Riemannian optimization, in which we address also this problem (in Section 7.5).
Quick summary of our findings:

There does not seem to be a simple closed-form solution. In particular, you can't get the solution by just truncating the Schur form or the eigendecomposition of $A$.
We suggest an interesting  reformulation of the problem. Namely, let $\operatorname{triu}$ be the operator that returns the upper triangular part of a matrix (diagonal included)
$$
\operatorname{triu}(A)_{ij} =
\begin{cases}
A_{ij} & i\leq j,\\
0 & i > j
\end{cases}
$$
and $\operatorname{tril}(A) = A - \operatorname{triu}(A)$ the one that returns its lower triangular part (diagonal excluded). Then the nearest matrix to $A$ with real eigenvalues (returned already in Schur decomposition) is $B = Q\operatorname{triu}(Q^*AQ) Q^*$, where the orthogonal matrix $Q$ solves
$$
Q = \arg\min_{Q \in O_n} \|\operatorname{tril}(Q^*AQ)\|_F^2.
$$
In this reformulation, the objective function is a nice simple smooth function: it just the sum of squares of the strictly lower triangular entries of $Q^*AQ$. In addition, this formulation has a smaller dimension (the manifold of orthogonal matrices has dimension $\frac{n(n-1)}{2}$ rather than $n^2$) and simpler constraints ("$Q$ is orthogonal" is easier to deal with numerically than "all eigenvalues of $B$ are real").
Unfortunately this problem (in both formulations, the original one and the alternative formulation) is non-convex and has multiple local minimizers. In similar problems, the number of minimizers grows exponentially with the dimension $n$, and we believe that this holds also for this problem.
However, code for optimization on matrix manifolds can be used to compute a (local) minimizer numerically in a very fast way, at least for small matrices (up to $n\approx 50$). Our Matlab code is available. 
None of the minimizers suggested by the previous answers for the $3\times 3$ and $4\times 4$ examples in this thread are global minima; we can beat them all. For matrices that are so small, repeating the minimization procedure with various starting points one gets only few local minima, so numerical evidence suggests that the matrices that we compute in Section 7.5 are global minimizers. For instance, for the matrix
$$
A = \begin{bmatrix}
    1 & 1 & 0\\
    -1 & 0 & 0\\
    0 & 0 & 0
\end{bmatrix}
$$
in the original question, our method computes the minimizer
$$
B \approx \begin{bmatrix}
    1.2110 &   0.7722 &  -0.0962\\
   -0.7722 &  -0.2416 &  -0.0962\\
   -0.0962 &   0.0962 &   0.0306\\
\end{bmatrix},
$$
with $\|A-B\|_F = 0.4946$.
It is not hard to prove that for each (local) minimizer $B$, $\operatorname{Tr} B = \operatorname{Tr} A$; otherwise one could construct a closer matrix $B + (\operatorname{Tr} A - \operatorname{Tr} B)\frac{1}{n}I$. 
Matrices with real distinct eigenvalues have a neighbourhood formed of matrices with real distinct eigenvalues: hence a (local) minimizer $B$ cannot have distinct eigenvalues. Numerically, often the minimizers have eigenvalues with high multiplicities. For instance, in the example above $B$ has a triple eigenvalue in $1/3$.
It does not seem easy to extend the results to other matrix norms.
We started our research by thinking about this problem, but then we were happily surprised to find out that the technique can be applied also to a harder problem that had already been studied in the literature on numerical linear algebra and control theory, that of finding the nearest Hurwitz stable matrix. And, more generally, it can be extended to solve numerically the problem of finding
$$
B = \arg \min_{S_\Omega} \|B-A\|_F,
$$
where $S_\Omega$ is the set of all (real or complex) matrices with all eigenvalues in a given closed set $\Omega$. Another nice example of how procrastinating and thinking about Mathoverflow questions can lead to useful research sometimes. :)