Given a two positive matrices $A,B$. For simplicity, let's assume that $Tr A=Tr B=1$. Assume that $\|A-B\|_1\leq\varepsilon$, for some small $\varepsilon>0$, where $\|\cdot\|_1$ is the $l_1$-norm, namely the sum of all its singular values.

My question is whether there is a generic transformation to move $A$ to $B$ for arbitrary such $A$ and $B$. The following are some simple observations. There are two simple transformations moving $A$ to its neighbours.

(1) Find an eigensystem (eigenvalues with corresponding eigenvectors) of $A$ and replace the eigenvalues $(\lambda_1(A),\cdots,\lambda_k(A))$ by a new sequence $(\theta_1,\cdots,\theta_k)\in[0,1]^k$ where $\sum_i\theta_i=1$ and $\sum_i|\lambda_i(A)-\theta_i|\leq\varepsilon$. Here we need the eigensystem of $A$ because the eigenvalues may not be unique.

(2) Choose a unitary matrix $U$ satisfying $\|U-I\|\leq\varepsilon$, where $\|\cdot\|$ is the spectral norm and replace $A$ by $UAU^*$.

It is easy to see neither single transformation (1) nor single transformation (2) is enough to cover all the neighbors.

Consider a naive example

$$ A=\left(\begin{array}{cc} \frac{1+\varepsilon}{2} & 0 \\ 0 & \frac{1-\varepsilon}{2} \end{array}\right) $$


$$B=(\frac{1}{2}+\varepsilon) \left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)+(\frac{1}{2}-\varepsilon) \left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array}\right).$$

$\|A-B\|_1\leq\varepsilon$ because $\|A-I/2\|_1\leq\varepsilon/2$ and $\|B-I/2\|_1\leq\varepsilon/2$, where $I$ is the $2\times 2$ identity matrix. Transformation (2) cannot bring $A$ to $B$ because the eigenvectors of $A$ and $B$ are far from each other. Also transformation (1) is not enough because the eigenvectors are different. But we can first apply transformmation (1) moving $A$ to $I/2$ and then apply transformation (2) moving $I/2$ to $B$ (There is freedom to choose the eigenvectors for $I/2$).

A more specific question. Can we move $A$ to all its $\varepsilon$-close (in $\|\cdot\|_1$-distance) neighbours by the composition of constant number of transformation (1) and (2)? Even more, can we do it with one transformation (1) and one transformation (2)?

I am not sure the question is research-level. Do not hesitate to close if it is not. Thank you.


What do you need your transformation for? One possible transformation would be $f(t)=A^{1/2}(A^{-1/2}BA^{-1/2})^tA^{1/2}.$ It satisfies $f(0)=A, f(1)=B$ and is the geodesic w.r.t. to the metric $d(A,B)=\|\log(A^{-1/2}BA^{-1/2})\|_{tr}$ where $tr$ denotes the trace norm, i.e. $\|A\|_{tr}=\sqrt{trace(AA')}$, see also "The Riemannian Geometry of the Space of Positive-Definite Matrices and Its Application to the Regularization".

Regarding the transformations (1) and (2): First note that the operations permute. Second note that if all eigenvalues have multiplicity one then the function which assigns the eigenvalues and the functions which assigns the eigenvectors are continuous. Hence in this case the conjecture holds true. If a function has multiplicity more than 1 it gets more complicated...

  • 1
    $\begingroup$ or even simpler transformation: $f(\lambda)=\lambda A+(1-\lambda)B$ $\endgroup$
    – user35593
    Apr 21 '14 at 8:25

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