# Finding Toeplitz matrix nearest to a given matrix

For an arbitrary $N\times N$ Hermitian matrix $A$, I want to derive a Toeplitz matrix from $A$ such that the eigenvectors of both matrices have minimal change.

Specifically I want find the Toeplitz matrix such that the $L^2$ norm between the eigenvectors of the Toeplitz matrix and eigenvectors of the matrix $A$ is minimal. Is there any alternative method other than searching numerically for the matrix? What is the computational cost of such such search?

I am aware of some work done related to perturbations of Toeplitz matrices, in addition eigenvectors of banded toeplitz matrix is studied, but the matrix I want in my application is not banded. I would appreciate any suggestion.

Edit: Is the problem tractable/solvable/realistic if we are given a sequence of matrices $A^n$ instead of $A$?

• One way is to solve $\min \|A-X\|$ such that $X \in \mathcal{T}$, where $\mathcal{T}$ is the set of Toeplitz matrices (this is a linear structure, so this particular problem can be solved using an SDP solver). May 6, 2016 at 1:38
• What do you mean exactly by "the L^2 norm between the eigenvectors of the Toeplitz matrix and eigenvectors of the matrix $A$"? Can you turn that into a formula? Keep in mind that the eigenvectors are not uniquely defined. So, for instance, one could claim that the solution to your problem is always the identity matrix $I$, which has the exact same eigenvectors as $A$, so the distance is 0. (That's a valid choice for the eigenvectors of $I$, right?) Mar 15, 2018 at 10:56
• @Suvrit Note that if you change norm the problem $\min_{X\in\mathcal{T}} \|A-X\|_F$ in the Frobenius norm has a trivial solution instead --- no need to use SDP. Mar 15, 2018 at 13:11

The set of $$n \times n$$ symmetric Toeplitz matrices is

$$\left\{ x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n \mid x_1, x_2, \dots, x_n \in \mathbb R \right\}$$

where $$\mathrm M_1, \mathrm M_2, \dots, \mathrm M_n$$ are $$n \times n$$ symmetric Toeplitz basis matrices. Let $$\mathrm M_1 = \mathrm I_n$$ correspond to the main diagonal, whereas the remaining basis matrices correspond to super and sub diagonals.

Let $$\mathrm M : \mathbb R^n \to \mbox{Sym}_n (\mathbb R)$$ be defined by

$$\mathrm M (\mathrm x) := x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n$$

### Spectral norm

To complement Suvrit's comment, using the spectral norm, we obtain the following unconstrained optimization problem in $$\mathrm x \in \mathbb R^n$$

$$\begin{array}{ll} \text{minimize} & \| \mathrm M (\mathrm x) - \mathrm A \|_2\end{array}$$

which can be rewritten as the following semidefinite program (SDP) in $$\mathrm x \in \mathbb R^n$$ and $$t \geq 0$$

$$\boxed{\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & - t \,\mathrm I_n \preceq - \mathrm A + x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n \preceq t \,\mathrm I_n\end{array}}$$

whose solution can be found numerically.

### Frobenius norm

To complement Federico's comment, using the squared Frobenius norm, we obtain the following unconstrained quadratic program (QP) in $$\mathrm x \in \mathbb R^n$$

$$\begin{array}{ll} \text{minimize} & \| \mathrm M (\mathrm x) - \mathrm A \|_{\text{F}}^2\end{array}$$

where the objective function is

$$\begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix}^\top \begin{bmatrix} \langle \mathrm M_1, \mathrm M_1 \rangle & \langle \mathrm M_1, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_1, \mathrm M_n \rangle\\ \langle \mathrm M_2, \mathrm M_1 \rangle & \langle \mathrm M_2, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_2, \mathrm M_n \rangle\\ \vdots & \vdots & \ddots & \vdots\\ \langle \mathrm M_n, \mathrm M_1 \rangle & \langle \mathrm M_n, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_n, \mathrm M_n \rangle\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} - 2 \,\begin{bmatrix} \langle \mathrm A, \mathrm M_1 \rangle\\ \langle \mathrm A, \mathrm M_2 \rangle\\ \vdots \\ \langle \mathrm A, \mathrm M_n \rangle\end{bmatrix}^\top \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} + \| \mathrm A \|_{\text{F}}^2$$

where $$\langle \mathrm M_i, \mathrm M_j \rangle$$ denotes the Frobenius inner product of (symmetric) basis matrices $$\mathrm M_i$$ and $$\mathrm M_j$$.

Computing the gradient of the objective function and finding where it does vanish, we obtain the following linear system

$$\begin{bmatrix} \langle \mathrm M_1, \mathrm M_1 \rangle & \langle \mathrm M_1, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_1, \mathrm M_n \rangle\\ \langle \mathrm M_2, \mathrm M_1 \rangle & \langle \mathrm M_2, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_2, \mathrm M_n \rangle\\ \vdots & \vdots & \ddots & \vdots\\ \langle \mathrm M_n, \mathrm M_1 \rangle & \langle \mathrm M_n, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_n, \mathrm M_n \rangle\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} = \,\begin{bmatrix} \langle \mathrm A, \mathrm M_1 \rangle\\ \langle \mathrm A, \mathrm M_2 \rangle\\ \vdots \\ \langle \mathrm A, \mathrm M_n \rangle\end{bmatrix}$$

Fortunately, the basis matrices are orthogonal and, thus, the matrix above is diagonal. Hence,

$$\begin{bmatrix} n & 0 & \cdots & 0\\ 0 & 2(n-1) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} = \begin{bmatrix} \sum_{i=1}^n a_{i,i}\\ 2 \sum_{i=1}^{n-1} a_{i,i+1}\\ \vdots \\ 2 a_{n,n}\end{bmatrix}$$

and, thus, the solutions $$x_1, x_2, \dots, x_n$$ are the arithmetic means of the $$n$$ (distinct) diagonals of $$\rm A$$

$$\begin{array}{rl} x_1 &= \dfrac{\langle \mathrm A, \mathrm M_1 \rangle}{\langle \mathrm M_1, \mathrm M_1 \rangle} = \dfrac 1n \displaystyle\sum_{i=1}^n a_{i,i}\\ x_2 &= \dfrac{\langle \mathrm A, \mathrm M_2 \rangle}{\langle \mathrm M_2, \mathrm M_2 \rangle} = \dfrac 1{n-1} \displaystyle\sum_{i=1}^{n-1} a_{i,i+1}\\ x_3 &= \dfrac{\langle \mathrm A, \mathrm M_3 \rangle}{\langle \mathrm M_3, \mathrm M_3 \rangle} = \dfrac 1{n-2} \displaystyle\sum_{i=1}^{n-2} a_{i,i+2}\\ & \qquad\qquad\quad\vdots\\ x_n &= \dfrac{\langle \mathrm A, \mathrm M_n \rangle}{\langle \mathrm M_n, \mathrm M_n \rangle} = \displaystyle\sum_{i=1}^1 a_{i,i+n-1} = a_{1,n}\end{array}$$

Lastly, the Toeplitz matrix nearest to the given symmetric matrix $$\rm A$$ is

$$\boxed{\hat{\mathrm X} := \left( \dfrac 1n \displaystyle\sum_{i=1}^n a_{i,i} \right) \mathrm I_n + \left( \dfrac 1{n-1} \displaystyle\sum_{i=1}^{n-1} a_{i,i+1} \right) \mathrm M_2 + \left( \dfrac 1{n-2} \displaystyle\sum_{i=1}^{n-2} a_{i,i+2} \right) \mathrm M_3 + \cdots + a_{1,n} \mathrm M_n}$$

• I don't think you understood my approach. Note that, for $n=3$, $$\begin{bmatrix} x_1 & x_2 & x_3\\ x_2 & x_1 & x_2\\ x_3 & x_2 & x_1\end{bmatrix} = x_1 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1\end{bmatrix} + x_2 \begin{bmatrix} 0 & 1 & 0\\ 1 & 0& 1\\ 0 & 1 & 0\end{bmatrix} + x_3 \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{bmatrix}$$ Mar 15, 2018 at 9:30
• @FedericoPoloni Indeed. I didn't know how to answer the original question, so I took the liberty of minimizing the spectral norm of the difference instead. Mar 15, 2018 at 11:29
• @FedericoPoloni I used your suggestion and considered also the case where the squared Frobenius norm is minimized. The solution is indeed trivial. Updated my answer. Mar 15, 2018 at 14:08
• @RodrigodeAzevedo Yes, in my first read, I did not understand. Now, I am clear. I have implemented the Toeplitz matrix in this way (your way) few years back for my work, but never knew the method was optimal. Therefore, I was looking for an alternative way to get a Toeplitz matrix. In any case, I am happy to see that the solution can be derived as a minimal norm, so full credit to you. Thanks. Mar 15, 2018 at 20:27
• @RodrigodeAzevedo Yes, it is purely Engineering application. It is related to the MUSIC algorithm. If your google MUSIC algorithm you should get the idea. It relates to target direction in radar. Mar 15, 2018 at 21:43