| Sbornik: Mathematics |
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This article is cited in 2 scientific papers (total in 2 papers) A well-posed setting of the problem of solving systems of linear algebraic equations E. E. Tyrtyshnikovab a Marchuk Institute of Numerical Mathematics of the Russian Academy of Sciences, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. Introduction and aims of the studyIt is known that an arbitrary system of linear algebraic equations $Ax=b$ has a uniquely defined normal pseudosolution. When the natural scalar product is used, it is obtained by the least square method as a vector of minimum length among all vectors minimizing the residual length. If the matrix $A$ is of incomplete rank, then the problem of seeking a normal pseudosolution is ill posed, since there is no continuous dependence on the entries of $A$. It is quite a common situation when instead of the system $Ax=b$ one proposes to solve an approximate system $\widetilde{A} \widetilde{x}=\widetilde{b}$, where
$$
\begin{equation*}
\| \widetilde{A} - A \| \leqslant \mu\quad\text{and} \quad \| \widetilde{b} - b \| \leqslant \delta.
\end{equation*}
\notag
$$
In this case one must understand that small values of the errors $\mu$ and $\delta$ do not guarantee any closeness of the normal pseudosolutions of the original and approximate systems. Using Tikhonov’s regularization method (see [1]–[4]) one considers regularized systems
$$
\begin{equation*}
(\widetilde{A}^* \widetilde{A} + \alpha I) x_\alpha=\widetilde{A}^* \widetilde{b};
\end{equation*}
\notag
$$
it claims that the parameter can be chosen so that the vector $x_\alpha$ converges to the normal pseudosolution of the system $Ax=b$ as $\mu, \delta \to 0$. Nevertheless, we cannot but think about the physical meaning attached to the search for an object that can vary arbitrarily strongly under small perturbations of the original data. Apparently, problems of this kind do not take account of some important additional information which could help one to restate the original problem so that it becomes well posed. Then solution methods could be sought for this new setting of the problem. It can be noted that a necessary attribute of regularization methods is the involvement of some additional information. However, in a huge number of works devoted to the development and applications of regularization methods, the new setting using additional information is usually not explicitly stated. In this regard, the work [5] by Tikhonov is of exceptional interest. It is proposed there to include the accuracy parameters $\mu$ and $\delta$ mentioned above in the statement of the problem of solving systems of linear algebraic equations. It is proposed to treat a whole class of systems equivalent in accuracy as input data. A separate class is specified by a matrix $A_0$, a right-hand side $b_0$ and two accuracy parameters $\mu$ and $\delta$. All systems of the form $Ax=b$ are under consideration, where
$$
\begin{equation*}
\| A - A_0 \| \leqslant \mu\quad\text{and} \quad \| b - b_0 \| \leqslant \delta.
\end{equation*}
\notag
$$
As a normal solution for $\Sigma$ it is proposed to take a vector of minimum length among all solutions of all consistent systems in this class. The main theorem in [5] claims that a normal solution for a class containing at least one consistent system exists and is unique, provided that the Frobenius (Euclidean) norms are used. A method for the calculation of this solution is also proposed, and some relationship with the idea of regularization is discussed. Nevertheless, [5] contains no theorem stating that the Tikhonov normal solution is well defined. The aim of this paper is to study the well-posedness of Tikhonov’s setting. The main new result is the theorem on the continuous dependence of the Tikhonov normal solution on the original data $A_0$, $b_0$, $\mu$ and $\delta$. In addition, the possibility of some generalizations is investigated, when norms different from the Frobenius ones are used. § 2. Uniqueness of Tikhonov’s solutionAssume that the spaces $\mathbb{C}^n$ and $\mathbb{C}^m$ are endowed with vector norms and that the norms of $m \times n$ matrices obey the inequality $\|Ax\| \leqslant \|A\| \,\|x\|$ for any vector $x$ and any matrix $A$. The spaces $\mathbb{C}^n$ and $\mathbb{C}^m$ are called the basic space and the image space, respectively. The class of systems
$$
\begin{equation*}
\Sigma=\Sigma (A_0, b_0, \mu, \delta ), \qquad A_0 \in \mathbb{C}^{m \times n}, \quad b_0 \in \mathbb{C}^m, \quad \mu, \nu > 0,
\end{equation*}
\notag
$$
consists of all systems of the form $Ax=b$, where
$$
\begin{equation*}
\|A-A_0\| \leqslant \mu\quad\text{and} \quad \|b-b_0\| \leqslant \delta.
\end{equation*}
\notag
$$
Systems can be consistent and inconsistent alike. We do not assume that the individual system $A_0x=b_0$ is consistent, but at least one system in this class must be so. Among the vectors solving consistent systems in the class $\Sigma$ we choose the ones of minimum norm. We call them Tikhonov solutions. We denote the set of Tikhonov solutions by
$$
\begin{equation*}
N_0=\bigl\{ \arg\min \|x\| \colon \|Ax-b_0\| \leqslant \delta, \, \|A-A_0\| \leqslant \mu \bigr\}.
\end{equation*}
\notag
$$
It follows directly from compactness arguments that $N_0 \ne \varnothing$. Following [5] we also introduce the important sets
$$
\begin{equation*}
N_1=\bigl\{ \arg\min \|x\| \colon \|Ax-b_0\|=\delta, \, \|A-A_0\|=\mu \bigr\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
N_2=\bigl\{ \arg\min \|x\| \colon \|A_0x-b_0\|=\mu \|x\| + \delta \bigr\}.
\end{equation*}
\notag
$$
Proof. Let $z \in N_0$. Proving by contradiction, we assume that $\|Az-b_0\| < \delta$. We set $z_\varepsilon=(1-\varepsilon) z$. Then for all small $\varepsilon > 0$ we obtain the inequalities which contradict the fact that $z$ has the minimum norm among all solutions of consistent systems in $\Sigma$. Thus, $\|Az-b_0\|=\delta$.
Now we assume that $\|A - A_0\| < \mu$. This inequality remains valid for all sufficiently small perturbations $F$ of the matrix $A$. We choose a perturbation so that To do this it suffices to take any matrix $F_0$ such that and set This contradicts the minimality of the norm of $z$ again. Thus we have proved that for any vector $z \in N_0$. Therefore, $N_0=N_1$. The lemma is proved.Note that Lemma 2.1 does not use any restrictions on norms, and thus is extends slightly a similar result stated in [5] for the Frobenius (Euclidean) norms. However, to deduce the equality we must introduce some restrictions.A norm on the $m \times n$ matrices is said to be supercompatible with the vector norms in the spaces of columns of sizes $m$ and $n$ if $\bullet $ the inequality $\|Ax\| \leqslant \|A\|\,\|x\|$ holds for any matrix $A$ of size $m \times n$ and any vector $x$ of size $n$ and, in addition, $\bullet $ for any nonzero vectors $x$ and $y$ of size $n$ and $m$, respectively, there exists a matrix $A$ such that $y=Ax$ and $\|y\|=\|A\| \,\|x\|$. If the Hölder 2-norms are used as vector norms, then the Frobenius (Euclidean) norm and the spectral norm have the property of supercompatibility. In this case, given two vectors $x$ and $y$, we can choose $A$ in the form
$$
\begin{equation*}
A=\frac{yx^*}{\|x\|_2^2}, \qquad \|A\|_2=\|A\|_F=\frac{\|y\|_2}{\|x\|_2}.
\end{equation*}
\notag
$$
Lemma 2.2. If the matrix norm is supercompatible with the vector norms in the basic space and the image space, then $N_0=N_1=N_2$. Proof. Let $z \in N_0$ and set $r_0=b_0 - A_0 z$. Then we certainly have $\|r_0\| - \mu \|z\| > 0$. Proving by contradiction we assume that $z \ne 0$ and $\| r_0 \| / \|z\| \leqslant \mu$. We choose $F$ from the equation By supercompatibility there exists a matrix $F$ such that We have obtained a contradiction with the equality $N_0=N_1$ established above.
Hence $\alpha=\mu \|z\| / \|r_0\|$ lies in the interval $0 < \alpha < 1$. We choose a matrix $F$ satisfying the equalities
$$
\begin{equation*}
Fz=\alpha r_0, \qquad \|F\|=\frac{\|\alpha r_0\|}{\|z\|}=\mu.
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\|b_0-(A_0+F)z\|=\|(1-\alpha) r_0\|=\|r_0\| - \alpha \|r_0\|=\|r_0\| - \mu \|z\|.
\end{equation*}
\notag
$$
Hence, by Lemma 2.1 Thus, the Tikhonov solution $z$ satisfies the equation
It remains to note that any vector $z$ satisfying this equation is a solution of some consistent system in the class $\Sigma$. In fact, choosing $F$ in the same way as above and setting $A=A_0 + F$, we infer that $\|b_0 - Az\|=\delta=\|b_0-A_0z\| - \mu \|z\|$. The lemma is proved. Clearly, all Tikhonov solutions have the same (minimum) norm. We denote it by $\nu$. Then Lemma 2.2 implies that Tikhonov solutions have the smallest norm among the vectors $x$ satisfying
$$
\begin{equation*}
\|b_0 - A_0 x\|=\rho, \quad\text{where } \rho=\mu \nu + \delta.
\end{equation*}
\notag
$$
They are obviously also vectors of minimum norm among the vectors in the closed convex set
$$
\begin{equation*}
M_\rho=\{ x \colon \|b_0 - A_0 x\| \leqslant \rho \}.
\end{equation*}
\notag
$$
If the norm $\|x\|$ is strictly convex, then this vector is uniquely defined. So we have proved the following theorem. Theorem 2.1. If the matrix norm is supercompatible with the vector norms and the vector norm in the basic space is strictly convex, then the Tikhonov solution is unique. § 3. Continuity of the Tikhonov solutionWe let $R_0$ denote the minimum of the norm of the residual $b_0-A_0x$ on the basic space. For $\rho \geqslant R_0$ we consider the function
$$
\begin{equation*}
f(\rho) :=\min_{\|b_0-A_0x\| \leqslant \rho} \|x\|=\min_{\|b_0-A_0x\|=\rho} \|x\|.
\end{equation*}
\notag
$$
We obviously have $f(\rho)=0$ for $\rho \geqslant R_1=\|b_0\|.$ Lemma 3.1. The function $f(\rho)$ is continuous and strictly decreasing on the interval $R_0 \leqslant \rho \leqslant R_1$. Proof. Monotonicity is verified in an evident way. To establish continuity from the right we consider a sequence of points Let $f(\rho_k)=\|x_k\|$ and $\|b_0-A_0x_k\|=\rho_k$. The sequence of vectors $x_k$ lies in the ball of radius $f(\rho)$; therefore, it contains a convergent subsequence $x_{k_l} \to x$ $\Longrightarrow$ $\|x_{k_l}\| \to \|x\|$. If $\rho_k \to \rho$, then $\|b_0 - A_0 x\|=\rho$ $\Longrightarrow$ $\lim_{k \to \infty} f(\rho_k)=\|x\|=f(\rho)$.
To prove continuity from the left, in view of monotonicity it suffices to construct a special sequence of points such that $\rho'_k \to \rho$ and $f(\rho'_k) \to f(\rho)$ as $k\to\infty$. Assume that $x$ and $x_0$ are vectors of minimum norm in the sets $M_\rho$ and $M_{\rho'}$:
$$
\begin{equation*}
\|b_0-A_0x\|=\rho, \qquad \|x\|=f(\rho), \qquad \|b_0-A_0x_0\|=\rho_0, \qquad \|x_0\|=f(\rho_0).
\end{equation*}
\notag
$$
For any convex combination $tx_0+(1-t)x$, where $0 < t < 1$, we obtain It follows that in any neighbourhood of $x$ there is a point $x'$ such that Hence we can choose a sequence of vectors $x_k \to x$, $k\to\infty$, such that the number sequence $\rho'_k :=\|b_0-A_0x_k\|$ increases monotonically and converges to $\rho$. It is clear that $f(\rho'_k) \leqslant \|x_k\| \to \|x\|=f(\rho)$. Let $f(\rho'_k)=\|x'_k\|$ and $\|b_0-A_0x'_k\|=\rho'_k$. If we suppose that then $x$ is not a vector of minimum norm in $M_\rho$. Thus, we have obtained the required special sequence. The lemma is proved. We let $z(\rho)$ denote a function that identifies a vector of minimum norm in the set $M_\rho$. Lemma 3.2. If there is a unique vector $z(\rho)$ of minimum norm in each set $M_\rho$, then the function $z(\rho)$ is continuous in $\rho$. Proof. Consider an arbitrary sequence $\rho_k \to \rho$, $k\to\infty$. The corresponding sequence of points $z_k=z(\rho_k)$ is bounded and thus contains convergent subsequences. Let $z_{k_l}$ be any of these subsequences, and let $z_{k_l} \to z'$ as $l\to\infty$. Clearly, $\|b_0-A_0z'\|=\rho$. As we have already shown that the function $f(\rho)=\|z(\rho)\|$ is continuous, we have $\|z_{k_l}\| \to \|z\|$ as $l\to\infty$. Therefore, $\|z'\|=\|z\|$; by uniqueness $z'=z$.
Thus, any convergent subsequence $z_{k_l}$ has the same limit $z$. Hence any convergent subsequence of the bounded number sequence $\|z_k-z\|$ converges to zero, so that $\|z_k-z\| \to 0$ as $k\to\infty$, and therefore $z_k \to z$ as $k\to\infty$. The lemma is proved. According to the definitions, the quantities $R_0$ and $f(\rho)$ and the set $M_\rho$ depend on the matrix $A_0$ and the vector $b_0$. Therefore, we write
$$
\begin{equation*}
R_0=R_0(A_0,b_0), \qquad f(\rho)=f(\rho,A_0,b_0)\quad\text{and} \quad M_\rho=M_\rho(A_0,b_0).
\end{equation*}
\notag
$$
In what follows we assume that for any $A_0$, $b_0$, and $\rho \geqslant R_0(A_0,b_0)$ there exists a unique vector of minimum norm in the set $M_\rho(A_0,b_0)$. We denote it by $z(\rho,A_0,b_0)$. Lemma 3.3. For fixed $\rho > R_0(A_0, b_0)$ the function is continuous with respect to $A$, $b$ in a neighbourhood of ($A_0$, $b_0$). Proof. It suffices to prove continuity at the point ($A_0$, $b_0$). We let $z_0$ denote the normal pseudosolution of the system $A_0x=b_0$. It is obvious that $z_0=z(\rho_0,A_0,b_0)$, where therefore, $\|b-Az_0\| < \rho$ for all $A$ and $b$ that are sufficiently close to $A_0$ and $b_0$. Thus, for $\|A-A_0\| \leqslant \varepsilon$ and $\|b-b_0\| \leqslant \varepsilon$, where $\varepsilon>0$ is sufficiently small, we see that
$$
\begin{equation*}
\|z(\rho,A,b)\| \leqslant \|z_0\|\quad\text{and} \quad f(\rho,A,b)=\min_{ \stackrel{\|b-Ax\| \leqslant \rho}{\|x\| \leqslant \|z_0\|}} \|x\|.
\end{equation*}
\notag
$$
We set $c=1+\|z_0\|$. If $\|x\| \leqslant \|z_0\|$, then
$$
\begin{equation*}
\|b-Ax\| \leqslant\|b_0-A_0x\| + c \varepsilon\quad\text{and} \quad \|b_0-A_0x\| \leqslant \|b-Ax\| + c \varepsilon.
\end{equation*}
\notag
$$
Thus, if $\|b_0-A_0x\| \leqslant \rho-c\varepsilon$, then $\|b-Ax\| \leqslant \rho$. Consequently,
$$
\begin{equation*}
\begin{aligned} \, &f(\rho+c\varepsilon,A_0,b_0) \leqslant f(\rho,A,b), \qquad f(\rho,A_0,b_0) \leqslant f(\rho-c\varepsilon,A_0,b_0) \\ & \qquad\Longrightarrow |f(\rho,A,b)-f(\rho,A_0,b_0)| \leqslant f(\rho-c\varepsilon,A_0,b_0) - f(\rho+c\varepsilon,A_0,b_0). \end{aligned}
\end{equation*}
\notag
$$
The lemma is proved. Theorem 3.1. Under the assumptions of Theorem 2.1 the Tikhonov solution $x(A_0,b_0,\mu,\delta)$ depends continuously on $A_0$, $b_0$, $\mu$ and $\delta$. Proof. As we know already, $x(A_0,b_0,\mu,\delta)=z(\rho,A_0,b_0)$ and the quantity $\rho$ satisfies the equation where for $\rho > R_0(A_0,b_0)$ the function depends continuously on $\rho$, $A_0$ and $b_0$ and decreases strictly monotonically in $\rho$ for any fixed $A_0$ and $b_0$ until it attains the value zero. Thus, the function $F(\rho)=\mu f(\rho,A_0,b_0) + \delta - \rho$ is continuous and strictly monotonically decreasing. Therefore, there exists unique $\rho_*$ such that $F(\rho_*)=0$.
We prove that the function $\rho_*=\rho_*(\mu,\delta,A_0,b_0)$ depends continuously on its arguments. We take an arbitrary sufficiently small $\varepsilon > 0$. Then
$$
\begin{equation*}
F(\rho_*-\varepsilon) > 0, \qquad F(\rho_*+\varepsilon) < 0.
\end{equation*}
\notag
$$
These inequalities are preserved under all sufficiently small perturbations of $A_0$ and $b_0$; therefore, the root of the equation remains on the interval $[\rho_*-\varepsilon, \rho_*+\varepsilon]$. By Lemma 3.2 the vector $z(\rho)$ depends continuously on $\rho$ and therefore on $A_0$ and $b_0$. The theorem is proved. § 4. Final observationsProving that the Tikhonov solution is well defined is related directly to its calculation algorithm, which consists of two steps: $\bullet$ find the root $\rho_*$ of the equation $\rho=\delta + \mu f(\rho)$; $\bullet$ find a vector of minimum norm in the set $\|b_0-A_0 x\| \leqslant \rho$. The problem at the second step is to find a point $x$ such that $f(\rho)=\|x\|$. The corresponding value of $\rho$ is specified at the first step. In the case of the Frobenius norms these are quadratic programming problems. We emphasize that the individual system $A_0x=b_0$ is not assumed to have a solution. However, if this system is consistent, then the Tikhonov solution can be regarded as a two-parameter regularization algorithm. The following theorem generalizes the result in [5], stated there for the Frobenius norms. Theorem 4.1. Let the assumptions of Theorem 2.1 hold. If the system $A_0x=b_0$ is consistent, then the Tikhonov solutions $x(A_0,b_0,\mu,\delta)$ converge to its minimum norm solution as $\delta, \mu \to 0$. Proof. Let $\widehat{x}$ be a minimum norm solution of the system $A_0x\,{=}\,b_0$. Given $A_0$, $b_0$, $\mu$ and $\delta$, the Tikhonov solution $x$ is equal to $z(\rho,A_0,b_0)$ for $\rho \leqslant \delta + \mu \|\widehat{x}\|$. Therefore, $\rho \to R_0=0$ as $\mu,\delta \to 0$. By Lemma 3.2, $z(\rho,A_0,b_0) \to z(R_0,A_0,b_0)=\widehat{x}$. The theorem is proved. How restrictive is the requirement that the class $\Sigma$ must contain at least one consistent system? For $m \leqslant n$ this is certainly fulfilled: it suffices to note that there is a nonsingular matrix in an arbitrarily small neighbourhood of a singular matrix. For $m > n$ it can occur that for any small perturbations of the matrix and the right-hand side the resulting systems are inconsistent. However, we can switch from $A_0x=b_0$ to the extended system with square matrix
$$
\begin{equation*}
\widetilde{A}_0 \widetilde{x}=b_0, \qquad \widetilde{A}_0=\begin{bmatrix} A_0, &0_{m \times (m-n)} \end{bmatrix}.
\end{equation*}
\notag
$$
The normal pseudosolution of the original system obviously coincides with the normal pseudosolution of the extended system. Now we can seek a Tikhonov solution $\widetilde{z}$ for the class $\Sigma=\{ \widetilde{A}_0, b_0, \mu, \delta \}$ and take the vector formed by the first $n$ components of $\widetilde{z}$ as a generalized solution of the original system. The fact that only compatible systems are considered in the definition of a Tikhonov solution is apparently essential. Even for a class containing consistent systems we can consider a vector of minimum norm among all normal pseudosolutions of systems equivalent in accuracy; however, such a vector can be distinct from the Tikhonov solution. Introducing additional restrictions on matrices (for example, symmetry, sparseness, Toeplitz property and so on) when we define a Tikhonov solution requires a careful analysis, since perturbations due to the supercompatibility of the matrix norm with the vector norms can violate these restrictions. 
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\Bibitem{Tyr22} |
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