Reports of the Academy of Sciences of the USSR
1969. Volume 189, No. 5

UDC 512.874

MATHEMATICS

M. A. GOLDMAN

ON A DECOMPOSITION OF A LINEAR OPERATOR

(Presented by Academician A. N. Tikhonov on 8 V 1969)

Let \(X\) be a vector space over a field \(K\); \(H(X)\) the set of all linear operators acting in \(X\). Denote by \(D(A)\) and \(R(A)\), respectively, the domain of definition and the range of the operator \(A \in H(X)\). Put
\[ Z(A)=\{x:\ x\in D(A),\ Ax=\theta\}, \]
where \(\theta\) is the zero element of the space \(X\);
\[ N(A)=\bigcup_{n=1}^{\infty} Z(A^n); \]
\(A_\lambda=\lambda E-A\), where \(E\) is the identity operator on \(X\), \(\lambda\in K\);
\[ \sigma(A)=\{\lambda:\ \lambda\in K,\ Z(A_\lambda)\ne\{\theta\}\}; \]
\[ \mathscr E(K_0;A)=\bigoplus_{\lambda\in K_0} N(A_\lambda), \]
where \(K_0\) is an arbitrary nonempty part of \(K\). An element \(x\) of \(\bigcap_{n=1}^{\infty}D(A^n)\) is called \(A\)-algebraic if the linear subspace \(\mathscr L(x;A)\), generated by the elements \(x, Ax,\ldots,A^n x,\ldots\), has finite dimension. The operator \(A\) is called locally algebraic if the set \(M(A)\) of all \(A\)-algebraic elements is equal to \(D(A)\).

Let \(A\in H(X)\), \(K_0\ne K\).

Theorem 1 (on the decomposition of an operator). For every \(\mu\) from \(K\setminus K_0\) there exists a decomposition \(A=B+C\), where \(B,C\in H(X)\), and they have the following properties: a) \(\sigma(B)\cap K_0=\varnothing\); b) \(D(C)=X\); c) \(R(C)=\mathscr E(K_0;A)\); d) \(C\) is a locally algebraic operator; e) \(BC=\mu C\).

Theorem 2 (on an invariant complement—necessary conditions). If in \(X\) there exists an \(A\)-invariant algebraic complement to \(\mathscr E(K_0;A)\), then for any \(\mu\) from \(K\setminus K_0\) a decomposition \(A=B+C\) is possible, where \(B,C\in H(X)\), and they have the properties b), c), d), e) (see Theorem 1), and also a′) \(\sigma(A)\setminus K_0\subseteq\sigma(B)\subseteq\{\mu\}\cup\sigma(A)\setminus K_0\) and f) \(CB=\mu C\).

Theorem 3 (on an invariant complement—sufficient conditions). If, for some \(\mu\) from \(K\setminus K_0\), the operator \(A\) admits a decomposition \(A=B+C\), where \(B,C\in H(X)\) and have the properties a), b), d), e) and f) (see Theorems 1 and 2), then in \(X\) there exists an \(A\)-invariant algebraic complement to \(\mathscr E(K_0;A)\).

Theorem 4 (on normal separability). Suppose that for some \(\mu\) from \(K\setminus K_0\) there is a decomposition \(A=B+C\), where \(B,C\in H(X)\) and they have properties a), b), d), e) and f) (the condition of Theorem 3). If \(F\) is some \(A\)-invariant algebraic complement in \(X\) to \(\mathscr E(K_0;A)\) and \(\lambda\in K_0\), then the equality
\[ A_\lambda(D(A)\cap F)=F \]
holds if and only if
\[ R(B_\lambda)=X. \]

For the proof of these theorems we first give several auxiliary propositions. Let \(A,B,C\in H(X)\), \(\lambda,\mu,\nu\in K\).

I. If \(BC=\mu C\), then \(R(C)\subseteq D(B)\) and \(Bx=\mu x\) for \(x\in R(C)\).

II. Let \(A=B+C\), \(T=\mu E+C\). If \(D(A)=D(B)\) and \(BC=\mu C\), then \(Ax=Tx\) for \(x\in R(C)\).

III. If \(A=B+C\), \(T=\mu E+C\), \(D(C)=X\) and \(BC=\nu C=CB\), then \(D(AT)=D(A)\), \(D(TA)=D(A)\) and \(ATx=TAx\) for \(x\in D(A)\).

IV. If \(A=B+C\), \(Z(B)=\{\theta\}\) and \(BC=\nu C\) \((\nu\ne0)\), then \(N(A)\subseteq R(C)\).

Proof. It is required to show that \(Z(A^n)\subseteq R(C)\) \((n=1,2,\ldots)\). We shall establish this by induction. Let \(x\in Z(A)\), i.e. \(Bx+Cx=\theta\). Since \(Cx=\nu^{-1}BCx\), we have \(B(x+\nu^{-1}Cx)=\theta\); hence, taking into account the condition \(Z(B)=\{\theta\}\), we obtain \(x+\nu^{-1}Cx=\theta\), \(x=-\nu^{-1}Cx\in R(C)\). This proves the inclusion \(Z(A)\subseteq R(C)\). Suppose now that \(Z(A^n)\subseteq R(C)\) and \(x\in Z(A^{n+1})\); then \(Ax\in Z(A^n)\subseteq R(C)\), i.e. \(Bx+Cx\in R(C)\), whence \(Bx\in R(C)\), i.e. \(Bx=Cx_1\). Further we have: \(Bx=\nu^{-1}BCx_1\), \(B(x-\nu^{-1}Cx_1)=\theta\), \(x=\nu^{-1}Cx_1\in R(C)\), which proves the inclusion \(Z(A^{n+1})\subseteq R(C)\).

V. If \(A=B+C\), \(T=\nu E+C\) \((\nu\ne0)\), \(D(A)=D(B)\), \(Z(B)=\{\theta\}\), and \(BC=\nu C\), then \(N(A)=N(T)\).

Proof. Consider the restrictions \(A_0\) and \(T_0\) of the operators \(A\) and \(T\) to \(R(C)\) (according to I, \(R(C)\subseteq D(A)\)). Since \(N(A)\subseteq R(C)\) and \(N(T)\subseteq R(C)\) (see IV), we have \(N(A_0)=N(A)\) and \(N(T_0)=N(T)\). It remains to use the equality \(A_0=T_0\) (see II).

VI. If \(C\) is a locally algebraic operator and \(Z(C)=\{\theta\}\), then \(R(C)=D(C)\).

VII. If \(C\) is a locally algebraic operator and \(D(C)=X\), then for any \(K_0\subseteq K\) in \(X\) there exists a \(C\)-invariant algebraic complement to \(\mathscr E(K_0;C)\).

The proof can be obtained by somewhat modifying the proof of Theorem 5 in \((^1)\).

VIII. If \(A_\lambda x\in\mathscr E(K_0;A)\), where \(\lambda\in K_0\), then \(x\in\mathscr E(K_0;A)\).

Proof. The condition \(A_\lambda x\in\mathscr E(K_0;A)\) means that

\[ A_\lambda x=\sum_{k=1}^n x_k,\quad \text{where } x_k\in N(A_{\lambda_k}),\ \lambda_k\in K_0\ (k=1,\ldots,n). \]

Let \(A_{\lambda_k}^{r_k}x_k=\theta\) \((k=1,\ldots,n)\); then

\[ \left(\prod_{k=1}^n A_{\lambda_k}^{r_k}\right) A_\lambda x=\theta, \]

which means \(x\in M(A)\). Denote by \(\mathscr E_1\) an \(A\)-invariant algebraic complement in \(M(A)\) to \(\mathscr E(K_0;A)\) (see VII), and let \(x=u+v\), where \(u\in\mathscr E(K_0;A)\), \(v\in\mathscr E_1\); then \(A_\lambda v=A_\lambda x-A_\lambda u\in\mathscr E(K_0;A)\). But \(A_\lambda v\in\mathscr E_1\); consequently, \(A_\lambda v=\theta\), and, since \(\lambda\in K_0\), \(v\in\mathscr E(K_0;A)\). At the same time \(v\in\mathscr E_1\), hence \(v=\theta\), \(x=u\in\mathscr E(K_0;A)\).

IX. Let \(A=B+C\), \(R(C)\subseteq D(B)\), \(B(R(C))\subseteq R(C)\). If the restriction \(B_0\) of the operator \(B\) to \(R(C)\) is a locally algebraic operator and \(Z(B_0)=\{\theta\}\), then \(R(A)\subseteq R(B)\).

Proof. From the equality \(A=B+C\) follows the inclusion \(R(A)\subseteq R(B)+R(C)\). But \(R(B)+R(C)=R(B)\), since \(R(B)\supseteq R(C)\) \((R(B)\supseteq R(B_0)\), and according to VI \(R(B_0)=D(B_0)(=R(C)))\).

X. Let \(B_1, B_2, C_1, C_2\in H(X)\), \(Z(B_1)=Z(B_2)=\{\theta\}\), \(D(C_1)=D(C_2)=X\), \(B_1+C_1=B_2+C_2\), \(B_1C_1=\nu C_1\), \(B_2C_2=\nu C_2\). If \(C_1\) and \(C_2\) are locally algebraic operators, then \(R(B_1)=R(B_2)\).

Proof. We shall show that \(R(B_2)\subseteq R(B_1)\). For this, consider the equality \(B_2=B_1+C\), where \(C=C_1-C_2\), and apply proposition IX, putting \(A=B_2\), \(B=B_1\).

It is necessary to verify the three conditions appearing in IX: 1) \(R(C)\subseteq D(B_1)\); 2) \(B_1(R(C))\subseteq R(C)\); 3) the restriction of the operator \(B_1\) to \(R(C)\) is a locally algebraic operator. Condition 1) follows from the inclusions \(R(C_1)\subseteq D(B_1)\), \(R(C_2)\subseteq D(B_2)\) (see I) and the equality \(D(B_1)=D(B_2)\), which holds by virtue of the hypotheses \(B_1+C_1=B_2+C_2\), \(D(C_1)=D(C_2)=X\). Condition 2) follows from the easily proved equality \(B_1C=C(\nu E+C_2)\). To verify condition 3), write an arbitrary element of \(R(C)\) in the form \(y=Cx\) and consider \(\mathscr L(y;B_1)\). It is not hard to see, using the equality \(B_1C=C(\nu E+C_2)\), that \(B_1^n y=C(\nu E+C_2)^n x\) \((n=1,2,\ldots)\). Hence we conclude that \(\mathscr L(y;B_1)=C(\mathscr L(x;\nu E+C_2))\). Since the operator \(\nu E+C_2\) is locally algebraic (as the sum of commuting locally algebraic-

... operators \(\nu E\) and \(C_2\), then \(\dim \mathscr L(x;\nu E+C_2)<\infty\), and hence \(\dim \mathscr L(y;B_1)<\infty\). Thus condition 3) is verified. Consequently, \(R(B_2)\subseteq R(B_1)\). The converse inclusion, of course, also holds.

Proof of Theorem 1. To construct the desired decomposition of the operator \(A\), take in \(X\) some algebraic complement \(F\) to the subspace \(\mathscr E(K_0;A)\). Let \(P\) and \(Q\) be the projectors generated by the decomposition
\[ X=\mathscr E(K_0;A)\oplus F, \]
where \(R(P)=\mathscr E(K_0;A)\), \(R(Q)=F\). Put \(B=AQ+\mu P\), \(C=AP-\mu P\), where \(\mu\) is a prescribed value from \(K\setminus K_0\), and show that the operators \(B\) and \(C\) have the required properties. Properties b) and e) are verified easily. We prove a), c), and d).

a) It is necessary to show that \(Z(B_\lambda)=\{\theta\}\) if \(\lambda\in K_0\). Let \(B_\lambda x=\theta\), i.e.,
\[ AQx+\mu Px=\lambda x; \]
then \(A_\lambda Qx=(\mu-\lambda)Px\in \mathscr E(K_0;A)\). By VIII, \(Qx\in \mathscr E(K_0;A)\). But \(Qx\in F\); hence \(Qx=\theta\). Therefore, in view of the equality \(A_\lambda Qx=(\mu-\lambda)Px\) and the condition \(\mu-\lambda\ne 0\), we obtain \(Px=\theta\). Hence \(Px+Qx=x=\theta\).

c) Let \(\lambda\in K_0\); then \(\lambda-\mu=\nu\ne 0\) and, by what has just been proved, \(Z(B_\lambda)=\{\theta\}\). Since \(A_\lambda=B_\lambda-C\) and \(B_\lambda(-C)=\nu(-C)\), by IV, \(N(A_\lambda)\subseteq R(C)\). Consequently, \(\mathscr E(K_0;A)\subseteq R(C)\). The reverse inclusion is obvious.

d) Let \(y=Cx\); then \(y\in \mathscr E(K_0;A)\), \(Py=y\), \(Cy=(A-\mu E)y\),
\[ C^n y=\sum_{k=0}^{n}(-\mu)^k {n\choose k} A^{\,n-k}y\quad (n=1,2,\ldots). \]
Hence we conclude that
\[ \mathscr L(Cy;C)\subseteq \mathscr L(y;A). \]
But \(\dim \mathscr L(y;A)<\infty\); therefore, \(\dim \mathscr L(x;C)<\infty\) for every \(x\in X\).

Proof of Theorem 2. The required decomposition of the operator \(A\) is effected by the operators \(B\) and \(C\), which are constructed in the same way as in Theorem 1, but in the present case the algebraic complement \(F\) to \(\mathscr E(K_0;A)\) is chosen to be \(A\)-invariant, owing to which the additional properties a′) and f) appear (note that a′) \(\Rightarrow\) a)). Property f) is verified easily. We prove a′). Let \(\lambda\in\sigma(A)\setminus K_0\), \(A_\lambda x=\theta\), \(x\ne\theta\); then \(x\notin\mathscr E(K_0;A)\), \(Qx=\nu\ne\theta\), \(B_\lambda\nu=A_\lambda\nu(=A_\lambda Qx)\). Since \(AQ=QA\) (by the \(A\)-invariance of \(\mathscr E(K_0;A)\) and \(F\)), we have \(A_\lambda Qx=QA_\lambda x=\theta\), hence \(B_\lambda\nu=\theta\), i.e. \(\lambda\in\sigma(B)\). Let now \(\lambda\in\sigma(B)\), \(B_\lambda x=\theta\), \(x\ne\theta\), \(\nu=Qx\); then
\[ A_\lambda\nu=B_\lambda x+(\mu-\lambda)Px=(\mu-\lambda)Px. \]
If \(Px=\theta\), then \(A_\lambda\nu=\theta\) and \(\nu=x\ne\theta\); consequently, \(\lambda\in\sigma(A)\). If \(Px\ne\theta\), then, applying the operator \(P\) to the equality
\[ A_\lambda Qx=(\mu-\lambda)Px, \]
we obtain \((\mu-\lambda)Px=\theta\), whence \(\lambda=\mu\). Thus,
\[ \sigma(B)\subseteq \{\mu\}\cup\sigma(A). \]
But \(\sigma(B)\cap K_0=\varnothing\); hence
\[ \sigma(B)\subseteq \{\mu\}\cup\sigma(A)\setminus K_0. \]

Proof of Theorem 3. Let \(T=\mu E+C\); then \(T_\lambda=(\lambda-\mu)E-C\), and since \(A_\lambda=B_\lambda-C\), \(B_\lambda(-C)=(\lambda-\mu)(-C)\), \(Z(B_\lambda)=\{\theta\}\) for \(\lambda\in K_0\), it follows, by V, that \(N(A_\lambda)=N(T_\lambda)\) for every \(\lambda\in K_0\). Consequently, \(\mathscr E(K_0;A)=\mathscr E(K_0;T)\). Let \(F\) be some \(T\)-invariant algebraic complement in \(X\) to \(\mathscr E(K_0;T)\) (see VII). We shall establish the inclusion \(A(D(A)\cap F)\subseteq F\), whereby the theorem will be proved. Fix \(x_0\in D(A)\cap F\). Since
\[ T(D(A))\subseteq \mu E(D(A))+C(D(A))\subseteq D(A)+R(C), \]
and \(R(C)\subseteq D(B)(=D(A))\) (see I), we have \(T(D(A))\subseteq D(A)\); hence, taking into account the \(T\)-invariance of \(F\), we obtain the inclusion
\[ \mathscr L(x_0;T)\subseteq D(A)\cap F. \]
Choose a basis \(x_1,\ldots,x_n\) in \(\mathscr L(x_0;T)\) and put
\[ Ax_k=y_k=y_k'+y_k'', \]
where \(y_k'\in\mathscr E(K_0;T)\), \(y_k''\in F\) \((k=1,\ldots,n)\). Since \(y_k'\in\mathscr E(K_0;A)\), there exists an operator \(S\) of the form
\[ \prod_{i=1}^{m} T_{\lambda_i}^{\,m_i}, \]
where \(\lambda_i\in K_0\) \((i=1,\ldots,m)\), such that \(Sy_k'=\theta\). Therefore, \(Sy_k=Sy_k''\in F\) \((k=1,\ldots,n)\). It is easy to see, using Proposition VIII, that \(Z(S)\subseteq \mathscr E(K_0;T)\), and hence
\[ Z(S)\cap \mathscr L(x_0;T)=\{\theta\} \]
(since \(\mathscr L(x_0;T)\subseteq D(A)\cap F\)). From this we conclude that
\[ S(\mathscr L(x_0;T))=\mathscr L(x_0;T), \]
from which it follows that the equation

\(Sx=x_0\) has a solution \(x\in \mathscr{L}(x_0;T)\). Writing \(x\) in the form \(\sum_{k=1}^{n}\alpha_k x_k\), we find

\[ SAx=\sum_{k=1}^{n}\alpha_k SAx_k=\sum_{k=1}^{n}\alpha_k Sy_k\in F. \]

But \(SAx=ASx\) for \(x\in D(A)\), since \(TAx=ATx\) for \(x\in D(A)\) (see III); consequently, \(Ax_0=ASx=SAx\in F\). In view of the arbitrariness of \(x_0\in D(A)\cap F\), this proves that \(A(D(A)\cap F)\subseteq F\).

Proof of Theorem 4. Suppose first that the given decomposition \(A=B+C\) has been obtained in the way it was done in the proof of Theorem 2: \(B=AQ+\mu P,\ C=AP-\mu P\).

Let the equality \(A_\lambda(D(A)\cap F)=F\) be given. It is required to prove that the equation \(B_\lambda x=y\) has a solution for every \(y\in X\). To this end take \(\bar{x}\in D(A)\cap F\) so that \(A_\lambda\bar{x}=Qy\). Then, taking into account that \(QP=0,\ P^2=P,\ P\bar{x}=\theta\), and \(Q\bar{x}=\bar{x}\), we obtain:
\[ B_\lambda\bigl(Py/(\mu-\lambda)+\bar{x}\bigr) =\bigl[A_\lambda Q+(\mu-\lambda)P\bigr]\bigl(Py/(\mu-\lambda)+\bar{x}\bigr) =A_\lambda\bar{x}+Py=Qy+Py=y. \]
Consequently, the required solution is \(Py/(\mu-\lambda)+\bar{x}\).

Let the equality \(R(B_\lambda)=X\) be given. It is required to prove that the equation \(A_\lambda x=y\) has a solution for every \(y\in F\). Take \(\bar{x}\in D(B_\lambda)(=D(A))\) so that \(B_\lambda\bar{x}=y\). Then we obtain \(Q\bar{x}\in D(A)\cap F\),
\[ A_\lambda Q\bar{x}=B_\lambda\bar{x}-(\mu-\lambda)P\bar{x}=y-(\mu-\lambda)P\bar{x}. \]
Since \(A_\lambda Q\bar{x}\in F\) and \(y\in F\), it follows that \((\mu-\lambda)P\bar{x}\in F\). Consequently, \(P\bar{x}=\theta,\ A_\lambda Q\bar{x}=y\). Thus the required solution is \(Q\bar{x}\).

It remains to remove the assumption made at the beginning of the proof. For this it is necessary to take into account VIII and the fact that the operator \(A_\lambda\) does not depend on the choice of the decomposition \(A=B+C\).

Latvian State University
named after P. Stuchka

Received
27 V 1968

REFERENCES

1. M. A. Gol’dman, E. M. Levich, DAN, 166, No. 2 (1966).