Every endomorphims is a linear combination of how many idempotents in infinite dimensions?

Solution 1:

Every endomorphism of a vector space is a linear combination of three idempotents, even in the infinite dimensional case. I'll give a proof of this, but it is a bit tricky and relies on carefully choosing a basis with respect to which the endomorphism takes on a particular form. I'll also reference some results from the paper mentioned in the question, in order to generalise to infinite dimensions.

I don't know a quick proof of this statement, so this is going to be a very very long answer. [Edit: The proof here is not yet complete. I still have to give a proof of the decomposition in (I), which I have, but takes some time to write out in full, so I'll come back and add this.] First, a bit of notation. I'll consider all vector spaces to be over a field $K$. Next, an ordinal number $\kappa$ is represented as the set of all ordinals less than $\kappa$. So, $\lambda <\kappa$ and $\lambda\in\kappa$ are equivalent statements for ordinals $\kappa,\lambda$. For any ordinal $\kappa$, the smallest ordinal greater than $\kappa$ is written as $\kappa+1$. A limit ordinal, $\kappa$, is an ordinal which cannot be written as $\lambda+1$ for any ordinal $\lambda$. Equivalently, a limit ordinal is an ordinal without a maximum element. The smallest ordinal $0=\emptyset$ is a limit ordinal. The first infinite ordinal, $\omega$, is the set of all finite ordinals and is a limit ordinal. Every ordinal $\kappa$ can be uniquely expressed as $\lambda+n$ for a limit ordinal $\lambda$ and finite ordinal $n$ (i.e., $n$ is a natural number), and we will say that $\kappa$ is even (resp. odd) if $n$ is even (resp. odd).

The idea behind the proof is to find a basis of the vector space under which the decomposition into idempotents can be done explicitly. For example, choosing nonzero $e_0\in V$, then we can set $e_i=A^ie_0$ for each natural number $i$. If these vectors give a basis for $V$, then multiplication by $A$ in $V$ looks like multiplication by $X$ in the polynomial ring $K[X]$, so is the difference of idempotents, as explained in the comment to the original question. If the $e_i$ are linearly independent but don't span $V$, then we continue adding new vectors. Choosing $e_\omega\in V$, but outside of the space generated by $\{e_i\}_{i\in\omega}$, we can set $e_{\omega+i}=A^ie_\omega$, and continue like this. If the sequence is linearly dependent so, for some $n$, $e_{n+1}$ is in the subspace $V_{n+1}$ generated by $\{e_i\}_{i< n+1}$, then we replace $e_{n+1}$ by a vector outside of $V_{n+1}$ and replace $e_{n+1+i}$ by $A^ie_{n+1}$. Continuing in this way (and applying transfinite induction), we end up with a basis $e_i$ with $i$ running over an ordinal $\kappa$, and such that, for each $i$, $Ae_i=e_{i+1}$ or $Ae_i$ is in the subspace generated by $\{e_j\}_{j \le i}$. If $Ae_i=e_{i+1}$ for each $i$, then $A$ can be written as a difference of idempotents. As long as for each pair of consecutive ordinals $i,i+1$, we have $Ae_i=e_{i+1}$ or $Ae_{i+1}=e_{i+2}$, then it is possible to add a multiple of an idempotent to $A$ to put it in the first form, giving $A$ as a linear combination of idempotents. If at each stage, we only choose $e_i$ satisfying this condition on consecutive basis elements, then we could end up at a point where we do not yet have a basis, but cannot continue adding any further basis elements. This can, however, be done in a way such that the action of $A$ on the quotient $V/V_\kappa$ is just multiplication by a scalar, which will be enough to still be able to give a decomposition into three idempotents. Written out, the decomposition is as follows (for any ordinal $\lambda$, I use $V_\lambda$ to denote the subspace generated by $\{e_i\}_{i\in\lambda}=\{e_i\}_{i < \lambda}$).

(I) Let $A$ be an endomorphism of vector space $V$. Then, (at least) one of the following two cases holds

• Case (A): There exists a $c\in K$ such that $A-cI$ has finite rank

• Case (B): There exists a linearly independent sequence $\{e_i\}_{i\in\kappa}$ in $V$, for some limit ordinal $\kappa$, such that the following conditions hold.

  1. For each $i\in\kappa$, either $Ae_i=e_{i+1}$ or $Ae_i\in V_{i+1}$.
  2. For each $i\in\kappa$, either $Ae_i=e_{i+1}$ or $Ae_{i+1}=e_{i+2}$.
  3. There exists a $c\in K$ such that, for all $v\in V$, $(A-c)v\in V_\kappa$.

The final condition here says that on the quotient $V/V_\kappa$, the action of $A$ is just multiplication by a scalar. I'll add the proof of (I) soon, but for now I give a proof of how it shows that $A$ is a linear combination of three idempotents. To approach case (B), we start with a simple situation first.

(II) Suppose that $V$ has a basis $\{e_i\}_{i\in\kappa}$ for some limit ordinal $\kappa$ and, for each $i\in\kappa$, there is a nonzero $c_i\in K$ such that $Ae_i-c_ie_{i+1}\in V_{i+1}$. Then, $A=\pi_1-\pi_2$ for idempotents $\pi_1,\pi_2$. Furthermore, this can be done such that ${\rm ker}(\pi_1)+{\rm ker}(\pi_2)=V$.

Proof: First, define an alternative basis $\{f_i\}_{i\in\kappa}$ as follows. For each limit ordinal $\lambda \in \kappa$ and finite $n$, set $f_{\lambda+n}=A^ne_\lambda$. By induction, it can be seen that for all $i\in\kappa$, there is a nonzero $b_i\in K$ such that $f_i-b_ie_i\in V_i$. For limit ordinals $i$ we have $b_i=1$ and, if this holds for any $i$ then multiplication by $A$ gives $f_{i+1}-b_ic_ie_{i+1}\in V_{i+1}$, so $b_{i+1}=b_ic_i$. Induction shows that this holds for all $i$, from which it follows that $\{f_i\}_{i\in\kappa}$ is a basis. As $Af_i=f_{i+1}$, we have $A=\pi_1-\pi_2$ where $\pi_1,\pi_2$ are the idempotents $$ \begin{align} \pi_1f_i&=\begin{cases} f_{i+1},&\textrm{if }i\textrm{ is even},\\ f_i,&\textrm{if }i\textrm{ is odd}, \end{cases}\\ \pi_2f_i&=\begin{cases} 0,&\textrm{if }i\textrm{ is even},\\ f_i-f_{i+1},&\textrm{if }i\textrm{ is odd}. \end{cases} \end{align} $$ For each even $i\in\kappa$, $f_i\in{\rm ker}(\pi_2)$ and $f_{i+1}-f_i\in{\rm ker}(\pi_1)$. So, both $f_i$ and $f_{i+1}=f_i+(f_{i+1}-f_i)$ are in ${\rm ker}(\pi_1)+{\rm ker}(\pi_2)$. QED

(III) Suppose that $\{e_i\}_{i\in\kappa}$ is a basis for $V$ satisfying properties 1-2 of case (B) above, and $\kappa$ is a limit ordinal. Then, for any nonzero $c\in K$, there exists idempotents $\pi_1,\pi_2,\pi_3$ with $A=c\pi_1+\pi_2-\pi_3$, and ${\rm ker}(\pi_2)+{\rm ker}(\pi_3)=V$.

Proof: Let $S$ be the set of $i\in\kappa$ with $Ae_i=e_{i+1}$ and define the endomorphism $\pi_1$ by $$ \pi_1e_i=\begin{cases} e_i,&\textrm{if }i\in S,\\ e_{i+1},&\textrm{if }i\not\in S. \end{cases} $$ If $i\in S$ then $\pi_1^2e_i=\pi_1e_i=e_i$. If $i\not\in S$ then, by property 2 of case (B), $i+1\in S$. So, $\pi_1^2e_i=\pi_1e_i=e_{i+1}$. This shows that $\pi_1$ is an idempotent. Also, setting $B=A-c\pi_1$, property 1 of case (B) shows that $Be_i-c_ie_{i+1}\in V_{i+1}$ for each $i$, where $c_i=1$ for $i\in S$ and $c_i=-c$ for $i\not\in S$. Then, by (II), $B$ is the difference of two idempotents $\pi_2,\pi_3$ with ${\rm ker}(\pi_2)+{\rm ker}(\pi_3)=V$. QED

I now give the proof that $A$ is a linear combination of three idempotents in case (B).

(IV) In case (B) above, $A=a\pi_1+\pi_2-\pi_3$ for some $a\in K$.

Proof: Let $c\in K$ be as in property 3, so that $(A-c)v\in V_\kappa$ for all $v\in V$. Let $U$ be a complementary subspace of $V_\kappa$, so $V=V_\kappa\oplus U$. With respect to this decomposition, $A$ is given by the matrix $$ A=\left(\begin{matrix}B&C\\0&c1_U\end{matrix}\right), $$ where $B\colon V_\kappa\to V_\kappa$, $C\colon U\to V_\kappa$ are linear maps and $1_U$ is the identity on $U$. If $c$ is nonzero then take $a=c$, otherwise let $a$ be any nonzero element of $K$. By (III), there exists idempotents $\pi^0_1,\pi^0_2,\pi^0_3$ of $V_\kappa$ such that $B=a\pi_1^0+\pi_2^0-\pi_3^0$ and ${\rm ker}(\pi^0_2)+{\rm ker}(\pi^0_3)=V_\kappa$. Then, letting $\{f_j\}_{j\in\lambda}$ be a basis for $U$, we can write $Cf_j=g_j-h_j$ for $g_j\in{\rm ker}(\pi^0_2)$ and $h_j\in{\rm ker}(\pi^0_3)$. Define linear maps $G,H\colon U\to V_\kappa$ by $Gf_j=g_j$ and $Hf_j=h_j$ for $j\in\lambda$. These satisfy $\pi^0_2G=\pi^0_3H=0$ and $C=G-H$. Also, define an endomorphism $J$ on $U$ to be the identity if $c\not=0$ (so $a=c$) and the zero map otherwise. Then, define three endomorphisms of $V$, $$ \begin{align} \pi_1&=\left(\begin{matrix}\pi^0_1&0\\0&J\end{matrix}\right),\\ \pi_2&=\left(\begin{matrix}\pi^0_2&G\\0&1_U\end{matrix}\right),\\ \pi_3&=\left(\begin{matrix}\pi^0_3&H\\0&1_U\end{matrix}\right). \end{align} $$ It can be checked that these are idempotents with $A=a\pi_1+\pi_2-\pi_3$. QED

We now move on to the proof of the decomposition for case (A).

(V) Let $A$ be a $d\times d$ matrix over $K$. Let $c$ be a non-zero element of $K$ such that the $c$-eigenspace, $\{x\in K^d\colon Ax=cx\}$, has dimension at least $d/2+1$. Then, there exists idempotent $d\times d$ matrices $\pi_1,\pi_2,\pi_3$ and $a,b\in K$ such that $A=c\pi_1+a\pi_2+b\pi_3$.

Proof: I'll reference the paper, On decomposing any matrix as a linear combination of three idempotents, by Clément de Seguins Pazzis, which was also linked in the question.

First, a bit of notation from the paper. If $n=m_1+m_2+\cdots+m_r$ and $A_i$ is an $m_i\times m_i$-matrix for $i=1,2,\ldots,r$ then $$ D(A_1,A_2,\ldots,A_r) $$ is the $n\times n$ matrix with $A_1,A_2,\ldots,A_r$ along the diagonal. Two matrices $A,B$ are similar if $A=U^{-1}BU$ for an invertible matrix $U$, and we write this as $A\sim B$. For a monic polynomial $P=X^n+a_{n-1}X^{n-1}+\cdots+a_0$, we use $C(P)$ to represent the $n\times n$ matrix with components $A_{i+1,i}=1$ ($i=1,2,\ldots,n-1$), $A_{i,n}=-a_{i-1}$ ($i=1,2,\ldots,n-1$), and all other entries $0$. This is the matrix representing multiplication by $X$ on $K[X]/(P)$ with respect to the basis $\{1,X,X^2,\ldots,X^{n-1}\}$, is called the companion matrix of $P$, and has characteristic polynomial and minimal polynomial $P$. It has trace $-a_{n-1}$, which I will also write as ${\rm tr}(P)$. For any $n\times n$ matrix $A$, we can consider $K^n$ as a module over $K[X]$ where multiplication by a polynomial $P$ corresponds to multiplying by the matrix $P(A)$. The structure theorem for finitely generated modules over a principal ideal domain implies that $A$ can be decomposed as $$ A\sim D(C(P_1),C(P_2),\ldots,C(P_r)). $$ Furthermore, the number of $P_i$ containing the factor $X-c$ (or, equivalently, vanishing at $c$) is easily seen to be the dimension of the $c$-eigenspace of $A$. Now, on with the proof...

As $A$ has at least one $c$-eigenvalue, its minimal polynomial is a multiple of $X-c$. If the minimal polynomial of $A$ is a power of an irreducible polynomial, then it is a power of $X-c$ so, as stated at the start of section 6.1 of the paper, $A$ is $(c,1,-1)$ composite. This means that there exists idempotents $\pi_1,\pi_2,\pi_3$ with $A=c\pi_1+\pi_2-\pi_3$, as required.

It just remains to consider the case where the minimal polynomial of $A$ is not a power of an irreducible. Lemma 13 of the paper shows that $$ A\sim D(\alpha I_p,\beta I_q,C(Q_1),\ldots,C(Q_r),C(R_1),\ldots,C(R_s)) $$ where $\alpha\not=\beta\in K$, $p,q,r,s$ are non-negative integers with $r,s\ge1$, and $Q_i,R_i$ are monic polynomials satisfying various properties, including the fact that all but at most one of them have degree at least $2$. As the $c$-eigenspace of $A$ has dimension $n\ge d/2+1$, any decomposition of $A$ into cyclic factors of the form $C(P_i)$ must have $n$ of the polynomials having a factor of $X-c$. If $c\not=\alpha$ and $c\not=\beta$ then $n$ of the polynomials $Q_i,R_i$ have a factor of $X-c$, so $r+s\ge n$ giving, $$ \begin{align} d &= p+q+\sum_{i=1}^r{\rm deg}(Q_i)+\sum_{i=1}^s{\rm deg}(R_i)\\ &\ge2(r+s)-1\ge2n-1. \end{align} $$ This contradicts the condition that $n\ge d/2+1$, so we have $c=\alpha$ or $c=\beta$. Without loss of generality, suppose that $c=\alpha$. Letting $t$ be the sum of the degrees of the $P_i$ and $Q_i$, then Lemma 14 of the paper then says that, for any monic polynomial $P$ of degree $t$ with ${\rm tr} P\not=\sum_i{\rm tr}Q_i+\sum_i{\rm tr}R_i$, then there exists $\delta\in K$ and an idempotent $\pi_3$ such that $$ A-\delta\pi_3\sim D(\alpha I_p,\beta I_q,C(P)). $$ Lemma 15 of the paper says that $P$ can be chosen such that, for some $\gamma\in K$, the matrix $D(\alpha I_p,\beta I_q,C(P))$ is $(c,\gamma)$ composite. So, $$ A=c\pi_1+\gamma\pi_2+\delta\pi_3 $$ as required. (More precisely, we can take $\gamma=-c$ if $\beta=0$, and $\gamma=\beta$ if $\beta\not=0$.) QED

Now, we can complete the proof in case (B).

(VI) Let $A$ be an endomorphism of a vector space $V$ such that $A-c1_V$ has finite rank for some $c\in K$. Then, $A$ is a linear combination of three idempotents.

Proof: If $V$ is finite dimensional, then $A$ is a linear combination of three idempotents by the paper On decomposing any matrix as a linear combination of three idempotents. So, consider the case where $V$ is infinite dimensional.

As $F\equiv A-c1_V$ is of finite rank, we can decompose $V=V_1\oplus V_2$ where $m={\rm dim}(V_1)$ is finite, $F$ is the zero operator on $V_2$ and $F$ maps $V_1$ into itself. Decompose $V_2=V_3\oplus V_4$ where $V_3$ has dimension $m+2$. Then, acting on the space $V_1\oplus V_3$ of dimension $d=2(m+1)$, $A$ has $c$-eigenspace containing $V_1$, so has dimension at least $m+2=d/2+1$. Hence, by (IV) we can put the restriction of $A$ to $V_1\oplus V_3$ in the form $c\pi_1+a\pi_2+b\pi_3$ where $\pi_i$ are idempotents on $V_1\oplus V_3$. They can be extended to idempotents on $V=V_1\oplus V_3\oplus V_4$ by setting $\pi_1v=v$ and $\pi_2v=\pi_3v=0$ on $V_4$, which satisfies the required conditions. QED

So, assuming that either of the cases in (I) holds, we have shown that the endomorphism $A$ is a linear combination of three idempotents. All that remains is to give a proof of (I), which I will do in two stages. First, I'll prove the following statement. Note, here we do not claim that $\kappa$ is a limit ordinal, so, there could exist a maximal element $i\in\kappa$, so that $i+1=\kappa$. By statement 1, if $i+1=\kappa$ then we have $Ae_i\in V_\kappa$ and, if $i+1\in\kappa$, then $Ae_i\in V_{i+2}\subseteq V_\kappa$. So, $A$ maps $V_\kappa$ into itself.

(VII) Let $A$ be an endomorphism of vector space $V$. Then, there exists a linearly independent sequence $\{e_i\}_{i\in\kappa}$ in $V$, for some ordinal $\kappa$, such that the following statements hold.

1. For each $i\in\kappa$, either $Ae_i=e_{i+1}$ (and $i+1\in\kappa$), or $Ae_i\in V_{i+1}$.
2. For each $i\in\kappa$ with $i+1\in\kappa$, either $Ae_i=e_{i+1}$ or $Ae_{i+1}=e_{i+2}$ (and $i+2\in\kappa$).
3. There exists a $c\in K$ such that, for all $v\in V$, $(A-c)v\in V_\kappa$.

Proof: We can choose the sequence by applying Zorn's lemma. Let $S$ be the set of linearly independent sequences $\{e_i\}_{i\in\kappa}$ for ordinals $\kappa$ and satisfying 1-2 above. We also order $S$ by writing $\{e_i\}_{i\in\lambda}\le\{f_i\}_{i\in\kappa}$ if $\lambda\le\kappa$ and $e_i=f_i$ for all $i\in\lambda$. It can be seen that this satisfies the requirements of Zorn's lemma (every chain in $S$ has an upper bound by taking the least upper bound of the ordinals and forming the linearly independent sequence which is the union of the sequences in the chain). So, there exists a maximal element $\{e_i\}_{i\in\kappa}$ of $S$. It just needs to be shown that this satisfies property 3.

Choose any element $v\in V$. If $\{A^rv\colon r=0,1,\ldots\}$ was a linearly independent sequence in $V/V_\kappa$, then we could find a larger element in $S$ as follows. Set $\kappa^\prime=\kappa+\omega$, then $e^\prime_i=e_i$ for $i < \kappa$ and $e^\prime_{\kappa+i}=A^iv$ for $i\in\omega$. The sequence $\{e^\prime_i\}_{i\in\kappa^\prime}$ is a strictly larger sequence in $S$, contradicting maximality.

So, we have shown that $\{A^rv\colon r=0,1,\ldots\}$ is a linearly dependent sequence in $V/V_\kappa$. Let $s\ge0$ be the smallest nonnegative integer such that $v,Av,\ldots,A^sv$ is linearly dependent over $V/V_\kappa$. Suppose that $s\ge2$. Then, we can find a larger element of $S$ as follows. Set $\kappa^\prime=\kappa+s$, then $e^\prime_i=e_i$ for $i < \kappa$ and $e_{\kappa+i}=A^iv$ for $0\le i < s$. The sequence $\{e^\prime_i\}_{i\in\kappa^\prime}$ is a strictly larger sequence in $S$, contradicting maximality (note, we need $s\ge2$ otherwise condition 2 could fail and the new sequence wouldn't be in $S$). So, $s\le1$.

We have now shown that, for all $v\in V$, $v,Av$ are linearly dependent over $V/V_\kappa$. Let us use $B$ to denote the action of $A$ on $W\equiv V/V_\kappa$. Then, $w,Bw$ are linearly dependent for all $w\in W$. This means that, if $w\not=0$, then $Bw=cw$ for some $c\in K$. Now choose any two nonzero $w_1,w_2\in W$, so that $Bw_1=c_1w_1$ and $Bw_2=c_2w_2$ for some $c_1,c_2\in K$. If $w_1,w_2$ are linearly dependent then we immediately have $c_1=c_2$. If they are independent, consider $v=w_1+w_2$. Then, $Bv=cv$ for some $c\in K$, so $cw_1+cw_2=c_1w_1+c_2w_2$. By linear independence, $c_1=c_2=c$. Hence $Bw=cw$ for all $w\in W$, and condition 3 is satisfied. QED

A simple corollary of this is,

(VIII) Let $A$ be an endomorphism of $V$. Then, there exists a linearly independent sequence $\{e_i\}_{i\in\kappa}$, for some limit ordinal $\kappa$, such that the following hold.

1. For each $i\in\kappa$, either $Ae_i=e_{i+1}$ or $Ae_i\in V_{i+1}$.
2. For each $i\in\kappa$, either $Ae_i=e_{i+1}$ or $Ae_{i+1}=e_{i+2}$.
3. There exists a $c\in K$ such that the action of $A-c1$ on $V/V_\kappa$ has finite rank.

Proof: By (VII), there exists a linearly independent sequence $\{e_i\}_{i\in\lambda}$, for some ordinal $\lambda$, satisfying properties 1-3 of (VII), with $\lambda$ in place of $\kappa$. We can then uniquely write $\lambda=\kappa+n$ for a limit ordinal $\kappa$ and $n\in\omega$. Properties 1-2 of (VII) still hold for this subsequence and, as $\kappa$ is a limit ordinal, this is equivalent to 1-2 above. Also, by 3 of (VII), there is a $c\in K$ such that $(A-c)v\in V_\lambda$ for all $v\in V$. Hence the action of $A-c1$ on $V/V\kappa$ has image contained in $V_\lambda/V_\kappa$, which has dimension $n$, so $A-c1$ is of finite rank acting on $V/V_\kappa$. QED

(The proof is to be continued...)