UDC 512.874

MATHEMATICS

M. A. GOLDMAN, E. M. LEVICH

ON THE INVARIANT COMPLEMENTABILITY OF SOME SUBSPACES GENERATED BY A LINEAR OPERATOR

(Presented by Academician L. V. Kantorovich on 17 V 1965)

Let \(X\) be a vector space; \(A\) a linear operator mapping \(X\) into \(X\); \(N_A\) the set of all null elements of the operator \(A\) \(\left(N_A=\bigcup_{n=1}^{\infty} Z_{A^n}, \text{ where } Z_{A^n}=\{x \mid A^n x=\theta\}\right)\). We shall call the height \(h(x)\) of an element \(x \in N_A\) the least of the numbers \(n\) for which \(A^n x=\theta\). Put \(H_A=\sup \{h(x)\mid x\in N_A\}\).

Theorem 1. If \(H_A<\infty\), then in \(X\) there exists an \(A\)-invariant complement to \(N_A\).

Proof will be carried out by induction on the magnitude \(H_A\). The case where \(H_A=0\) is trivial. Suppose that for \(H_A=m\) the theorem is valid, and let \(H_A=m+1\). Put \(N_A'=N_A\cap A(X)\), \(N_A''=N_A\ominus N_A'\), and denote by \(X_1\) a complement to \(N_A''\) in \(X\) containing \(A(X)\). Let \(A_1\) be the restriction of the operator \(A\) to \(X_1\); then \(N_{A_1}=N_A\cap X_1=N_A'\). Since \(H_A=m+1\), it follows that \(H_{A_1}=m\). By the induction hypothesis, there exists an \(A_1\)-invariant complement to \(N_{A_1}\) in \(X_1\). This complement will also be an \(A\)-invariant complement to \(N_A\) in \(X\).

Corollary. If \(\dim N_A<\infty\), then in \(X\) there exists an \(A\)-invariant complement to \(N_A\).

Denote by \(\mathfrak{A}(x_0)\) the linear hull of the set \(\{x_0, Ax_0,\ldots,\ldots,A^n x_0,\ldots\}\), \(x_0\in X\).

Theorem 2. Let \(H_A<\infty\). If \(x_0\in N_A\) and \(h(x_0)=H_A\), then in \(X\) there exists an \(A\)-invariant complement to \(\mathfrak{A}(x_0)\).

The proof of this theorem is similar to the proof of Theorem 1.

Theorem 3. If \(X_1\) is a subspace in \(N_A\) and \(A(X_1)=X_1\), then in \(X\) there exists an \(A\)-invariant complement to \(X_1\).

The proof of this theorem is based on the following lemmas.

Lemma 1. If \(X_1\) is a subspace in \(N_A\) and \(A(X_1)=X_1\), then the equation

\[ \sum_{k=1}^{n} \alpha_k A^k x = y, \]

where \(y\in X_1\), has a solution in \(X_1\).

Proof. The case where \(y=\theta\) is trivial. Let \(y\ne\theta\). If \(\alpha_0\ne 0\), then the required solution can be found in the form

\[ x=\sum_{i=0}^{m} \beta_i A^i y, \]

where \(m=h(y)-1\). If \(\alpha_0=\alpha_1=\cdots=\alpha_{j-1}\), \(\alpha_j\ne 0\), then represent our equation in the form \(A^j\left(\sum_{k=j}^{n}\alpha_k A^{k-j}x\right)=y\). Since \(A(X_1)=X_1\), there is a \(y_0\in X_1\) such that \(A^j y_0=y\). We arrive at the equation

\[ \sum_{k=0}^{n-j} \alpha'_k A^k x = y, \]

where \(\alpha'_k=\alpha_{k+j}\). This equation has a solution in \(X_1\), since \(\alpha'_0\ne 0\). The lemma is proved.

Define on \(\mathfrak A(x_0)\) the function \(f_{x_0}\), taking \(f_{x_0}(y)\) to be the least of the numbers \(p\) occurring in all possible representations of the element \(y\) in the form
\[ y=\sum_{k=0}^{p}\alpha_k A^k x_0. \]

Lemma 2. Let \(\mathfrak B\) be some \(A\)-invariant subspace in \(\mathfrak A(x_0)\). If \(y_0\in\mathfrak B\) and
\[ f_{x_0}(y_0)=\min\{f_{x_0}(y)\mid y\in\mathfrak B\}, \]
then \(\mathfrak B=\mathfrak A(y_0)\).

Proof. Obviously, \(\mathfrak A(y_0)\subset\mathfrak B\). We prove the inclusion \(\mathfrak B\subset\mathfrak A(y_0)\). Let
\[ y_0=\sum_{i=0}^{n}\alpha_i A^i x_0,\quad \alpha_n\ne0,\qquad y=\sum_{k=0}^{m}\beta_k A^k x_0,\quad \beta_m\ne0,\quad y\in\mathfrak B. \]
By the condition of the lemma, \(m\ge n\). It is required to show that \(y\in\mathfrak A(y_0)\). We carry out the proof by induction on the quantity \(k(y)=m-n\). If \(k(y)=0\), then \(y=\lambda y_0\in\mathfrak A(y_0)\). Suppose that \(k(y)=l+1\), and suppose the inclusion \(z\in\mathfrak A(y_0)\) is true each time \(z\in\mathfrak B\) and \(0\le k(z)\le l\).

Take
\[ z=y-\frac{\beta_m}{\alpha_n}A^{m-n}y_0. \]
It is clear that \(z\in\mathfrak B\) and \(0\le k(z)\le l\); consequently, \(z\in\mathfrak A(y_0)\), and
\[ y=z+\frac{\beta_m}{\alpha_n}A^{m-n}y_0\in\mathfrak A(y_0). \]
The lemma is proved.

Lemma 3. Let \(X_1, X_2\) be disjoint subspaces in \(X\); \(X_0=X_1\oplus X_2\); \(Q_1\) and \(Q_2\) are the projection operators generated by the decomposition of \(X_0\) into the direct sum of the subspaces \(X_1\) and \(X_2\) (\(Q_1\) and \(Q_2\) are defined on \(X_0\), \(Q_1(X_0)=X_1\), \(Q_2(X_0)=X_2\)).

For any \(E\subset X\) the equality holds
\[ Q_1(X_0\cap E)=(X_2+E)\cap X_1 \quad \bigl(Q_2(X_0\cap E)=(X_1+E)\cap X_2\bigr). \]

Lemma 4. Let \(X_0, X_1, X_2, Q_1, Q_2\) denote the same objects as in Lemma 3. If \(X_1, X_2\) are \(A\)-invariant subspaces, then, whatever the element \(x_0\in X\), the equality
\[ \mathfrak A(Q_1v_0)=(X_2+\mathfrak A(u_0))\cap X_1 \quad \bigl(\mathfrak A(Q_2v_0)=(X_1+\mathfrak A(u_0))\cap X_2\bigr), \]
holds, where \(v_0\) is some element of \(X_0\cap\mathfrak A(u_0)\) for which
\[ f_{u_0}(v_0)=\min\{f_{u_0}(v)\mid v\in X_0\cap\mathfrak A(u_0)\}. \]

Proof. Putting \(E=\mathfrak A(u_0)\) in Lemma 3, we obtain
\[ Q_1(X_0\cap\mathfrak A(u_0))=(X_2+\mathfrak A(u_0))\cap X_1. \]
Since the space \(X_0\cap\mathfrak A(u_0)\) is \(A\)-invariant, by Lemma 2,
\[ X_0\cap\mathfrak A(u_0)=\mathfrak A(v_0). \]
Consequently,
\[ Q_1(\mathfrak A(v_0))=(X_2+\mathfrak A(u_0))\cap X_1. \]
To complete the proof it remains to take into account the equality
\[ Q_1(\mathfrak A(v_0))=\mathfrak A(Q_1v_0), \]
which follows from the commutativity on \(X_0\) of the operators \(A\) and \(Q_1\).

Proof of Theorem 3. Let \(X_2\) be a maximal \(A\)-invariant subspace in \(X\), disjoint from \(X_1\). Suppose that \(X_0=X_1\oplus X_2\ne X\). Let \(x_0\in X_0\). Choose in \(X_0\cap\mathfrak A(x_0)\) an element \(y_0\) satisfying the condition
\[ f_{x_0}(y_0)=\min\{f_{x_0}(y)\mid y\in X_0\cap\mathfrak A(x_0)\}, \]
and denote by \(P_0(A)\) the polynomial in \(A\) for which
\[ P_0(A)x_0=y_0. \]
By Lemma 1, there exists an element \(z_0\in X_1\) such that
\[ P_0(A)z_0=Q_1y_0. \]
Put
\[ u_0=x_0-z_0,\qquad v_0=P_0(A)u_0. \]
It is easy to see that
\[ f_{u_0}(v_0)=\min\{f_{u_0}(v)\mid v\in X_0\cap\mathfrak A(u_0)\}. \]
By Lemma 4, we have the equality
\[ (X_2+\mathfrak A(u_0))\cap X_1=\mathfrak A(Q_1v_0), \]
from which it follows that
\[ (X_2+\mathfrak A(u_0))\cap X_1=\{\theta\} \]
(for \(v_0=y_0-Q_1y_0\)). We have arrived at a contradiction, since \(X_2+\mathfrak A(u_0)\ne X_2\). The theorem is proved.

Corollary. Let
\[ M_A=\bigcap_{n=1}^{\infty}A^n(X). \]
If \(A(M_A)=M_A\), then in \(X\) there exists an \(A\)-invariant complement to \(M_A\cap N_A\).

Theorem 4. If, beginning with some number \(m\), the sets
\[ Z_n=Z_A\cap A^n(X),\quad n=0,1,\ldots, \]
coincide with one another, then in \(X\) there exists an \(A\)-invariant complement to \(N_A\).

Proof. It is known ([1], Theorem 1) that, under the hypotheses of the theorem being proved, \(A(M_A)=M_A\). Hence, by the corollary to Theorem 3, there exists in \(X\) an \(A\)-invariant complement \(X_1\) to \(M_A\cap N_A\). From these same hypotheses it is easy to conclude that
\[ \sup\{h(x)\mid x\in N_A\cap X_1\}=m. \]
Therefore, denoting by \(A_1\) the restriction of the operator \(A\) to \(X_1\), we have \(H_{A_1}=m<\infty\). By Theorem 1, there exists in \(X_1\) an \(A_1\)-invariant complement \(X_2\) to
\[ N_{A_1}=N_A\cap X_1. \]
Obviously, \(X_2\) will also be an \(A\)-invariant complement in \(X\) to \(N_A\).

Corollary. If at least one of the numbers \(\dim Z_A\) or \(\operatorname{codim} A(X)\) is finite, then there exists in \(X\) an \(A\)-invariant complement to \(N_A\) (for from these conditions the hypotheses of Theorem 4 follow (see [1], Theorem 2)).

We give an example of an operator \(A\) for whose null elements there does not exist an \(A\)-invariant complement in \(X\).

Consider the vector space \(X\) with basis
\[ a_{ij},\quad b_k,\quad c_m;\qquad i,j,k=1,2,\ldots;\quad j\le i;\quad m=0,1,\ldots . \]
Define on \(X\) a linear operator \(A\) by setting
\[ Aa_{i1}=\theta\quad (i=1,2,\ldots);\qquad Aa_{ij}=a_{i,j-1}\quad (i=1,2,\ldots;\ 2\le j\le i); \]
\[ Ab_k=b_{k+1}\quad (k=1,2,\ldots);\qquad Ac_0=b_1;\qquad Ac_m=c_{m-1}+a_{mm}\quad (m=1,2,\ldots). \]
The formulas
\[ A^{m+1}c_m=b_1\quad (m=0,1,\ldots),\qquad A^m c_{m+1}=c_1+\sum_{i=2}^{m+1} a_{i2}\quad (m=1,2,\ldots) \]
are valid. It is easy to see that \(a_{ij}\in N_A\). We shall show that the elements \(a_{ij}\) form a basis in \(N_A\). Let \(x\in N_A\) and
\[ x=\sum_{i=1}^{r}\sum_{j=1}^{i}\alpha_{ij}a_{ij} +\sum_{k=1}^{s}\beta_k b_k +\sum_{m=0}^{t}\gamma_m c_m . \]
Choose \(n>\max\{r,t\}\) such that \(A^n x=\theta\). Applying the operator \(A^n\) to the element \(x\), we obtain
\[ \theta=\sum_{k=1}^{s}\beta_k b_{k+n}+\sum_{m=0}^{t}\gamma_m b_{n-m}. \]
It follows that
\[ \beta_k=\gamma_m=0\quad (k=1,2,\ldots,s;\ m=1,2,\ldots,t). \]

Suppose that there exists an \(A\)-invariant complement \(F\) to \(N_A\). We shall show that
\[ c_m\in F\quad (m=1,2,\ldots). \]
For this, in the equality
\[ Ac_{m+1}=c_m+a_{m+1,m+1} \]
replace \(c_{m+1}\) by the sum \(c'_{m+1}+c''_{m+1}\), where
\[ c'_{m+1}\in N_A,\qquad c''_{m+1}\in F; \]
we obtain the equality
\[ Ac''_{m+1}-c_m=-Ac'_{m+1}+a_{m+1,m+1}. \]
Since the equation
\[ Ax=a_{m+1,m+1} \]
is insoluble, we have
\[ \theta\ne -Ac'_{m+1}+a_{m+1,m+1}. \]
Assuming that \(c_m\in F\), we arrive at the relation
\[ \theta\ne (Ac''_{m+1}-c_m)\in N_A\cap F, \]
contradicting the equality
\[ N_A\cap F=\{\theta\}. \]
Thus, \(c_m\in F\). In particular,
\[ c'_1\ne \theta. \]

Let
\[ c'_1=\sum_{i=1}^{r}\sum_{j=1}^{i}\alpha_{ij}a_{ij}. \]
Write the equalities
\[ A^r c''_{r+1}=A^r c_{r+1}-A^r c'_{r+1} =c_1+\sum_{i=2}^{r+1} a_{i2}-A^r c'_{r+1}, \]
from which it follows that
\[ A^r c''_{r+1}-c''_1=c'_1-A^r c'_{r+1}+\sum_{i=2}^{r+1} a_{i2}. \]
The obtained equality is contradictory, for
\[ A^r c''_{r+1}-c''_1\in F,\qquad c'_1-A^r c'_{r+1}+\sum_{i=2}^{r+1} a_{i2}\in N_A, \]
and
\[ c'_1-A^r c'_{r+1}+\sum_{i=2}^{r} a_{i2}\ne 0. \]
The latter follows from the fact that the element \(a_{r+1,2}\), which appears in the sum
\[ \sum_{i=2}^{r+1} a_{i2}, \]
does not occur in the expansions with respect to the basis \(a_{ij}\) of the null elements \(c'_1\) and \(A^r c'_{r+1}\). This proves that there is no \(A\)-invariant complement to \(N_A\).

The example constructed permits us to assert that not every maximal \(A\)-invariant subspace disjoint from \(N_A\) is a complement to \(N_A\) in \(X\). One can show that this assertion remains valid even in the case when
\[ \dim N_A<\infty . \]

However, the following theorem holds.

Theorem 5. If \(A\) is a locally algebraic operator, then every maximal \(A\)-invariant subspace disjoint from \(N_A\) is a complement to \(N_A\).

(The operator \(A\) is called locally algebraic if, for every \(x \in X\), the subspace \(\mathfrak A(x)\) is finite-dimensional.)

The proof of this theorem follows easily from the following lemma.

Lemma 5. Let \(E\) be some invariant subspace disjoint from \(N_A\). If \(x_0 \in E \oplus N_A\) and \(\mathfrak A(x_0)\) is a finite-dimensional subspace, then there exists an \(A\)-invariant subspace \(F \supset E\) such that \(F \cap N_A = \{\theta\}\) and \(x_0 \in F \oplus N_A\).

Proof. Let \(L\) be an \(A\)-invariant complement to \(N_A \cap \mathfrak A(x_0)\) in \(\mathfrak A(x_0)\). Since \(L\) contains no zeros of the operator \(A\) (except \(\theta\)), it follows that \(A(L)=L\).

It is clear that \(E \dot{+} L(=F)\), and \(F\) is an \(A\)-invariant subspace. We show that \(N_A \cap F=\{\theta\}\). To this end, note that the set \(L \setminus E\) is invariant with respect to \(A\) (by the invertibility of \(A\) on \(L\) and the \(A\)-invariance of \(L \cap E\)). Let

\[ y \in N_A \cap F = N_A \cap \{E+(L\setminus E)\}. \]

Then there exists an \(n\) such that \(A^n y=\theta\). Represent the element \(y\) in the form \(y=y' + y''\), where \(y' \in E\), \(y'' \in (L\setminus E)\cup\{\theta\}\). Since \(A^n y' \in E\), \(A^n y'' \in (L\setminus E)\cup\{\theta\}\), and \(A^n y' + A^n y''=\theta\), it follows that \(y=\theta\).

Latvian State University
named after P. Stučka

Received
1 IV 1965

REFERENCES

1. M. A. Gol’dman, S. N. Krachkovskii, DAN, 158, No. 3 (1964).