It is well-known that the eigenvalues (in decreasing order) of a Hermitian matrix $A$ are Lipschitz continuous functions of $A$.
Do there exist orthonormal eigenvectors that vary in a Lipschitz continuous manner (as functions of A)?
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2 Answers
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Here is a counterexample: consider the real symmetric matrix
$$A=\begin{pmatrix}
\cos\phi&\sin\phi\\
\sin\phi&-\cos\phi
\end{pmatrix}.$$
The eigenvectors $v_\pm$ for the eigenvalues $\pm 1$ are
$$v_\pm=\left(\frac{\cos \phi\pm 1}{\sqrt{2\pm 2 \cos \phi}},\frac{\sin \phi}{\sqrt{2\pm 2 \cos \phi}}\right).$$
There is a discontuity in $v_-$ at $\phi=0\mod 2\pi$ and in $v_+$ at $\phi=\pi \mod 2\pi$.
What happens is that an eigenvector switches the sign of the eigenvalue as $\phi$ is advanced by $2\pi$, so if you consider an eigenvector for a particular eigenvalue it evolves discontinuously.
answered Jun 5, 2021 at 18:49
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2$\begingroup$ This example has a continuous choice of eigenvectors, $v_+=(\cos(\phi/2),\sin(\phi/2))$ and $v_+=(-\sin(\phi/2),\cos(\phi/2))$. $\endgroup$ Commented Jun 6, 2021 at 4:12
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2$\begingroup$ In this example, $v_+$, $v_-$ can be chosen continuously as function of $\phi$, but not as a continuous function of $A$. Indeed as a function of $A=A(\phi)$, we would get $v_+(A(\phi+2\pi))=-v_+(A(\phi))$. So there is no continuous determination, but there is continuous determination along paths (which more generally holds when multiplicities of eigenvalues are fixed). $\endgroup$– YCorCommented Jun 6, 2021 at 6:18
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1$\begingroup$ @MichaelMontgomery still this answers the question, since the question was to find a determination as a function of $A$. But your other example exhibits another, local, obstruction. $\endgroup$– YCorCommented Jun 6, 2021 at 6:21
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This cannot happen in general. When the number of unique eigenvalues changes we can have a discontinuity in the eigenspaces. For example, let $$f(t)=\left\lbrace\begin{array}{cc} 0 & \text{for }t<0 \\ t & \text{for }0\leq t \leq 1 \\ 1 & \text{for } t>1 \end{array}\right.$$ and define the matrix $$A(t)=\left(\begin{array}{cc} 1 & f(t-1) \\ f(t-1) & f(t) \end{array}\right).$$ $A(t)$ has eigenvalues $t,1$ and eigenvectors $(0,1),(1,0)$ for $t \in [0,1]$. For $t \in [1,2]$ $A(t)$ has eigenvalues $t,2-t$ and eigenvectors $(1,-1),(1,1)$. There is no way for this continuous family of Hermitian matrices to have a continuous choice of eigenvectors as the eigenspaces make a 45° jump at $t=1$.
