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Is every endomorphism of a $K$-vector space a $K$-linear combination of idempotents?

This question was asked by Jonas Meyer in a comment to this question.

To make sure I earn no points thanks to a question raised by somebody else, I shall put a 500 point bounty on this question (but I think I must wait two days).

George Lowther told me in a comment to this answer that he had a positive answer to Jonas's question. I considered suggesting to George that he posts his answer to Jonas's question as an answer to this question, but I thought Jonas's question deserved to be asked separately.

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Pierre-Yves Gaillard
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  • Sorry, I'm a bit slow responding here, but I'll post my answer later today if I get the chance to log on from home. – George Lowther Aug 04 '14 at 11:28
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    But I'll post a hint as a comment. The ideas in the answer to the question linked above show that 2x2 matrices with zeros on the diagonals are linear combinations of idempotents and, in a similar way, so are diagonal matrices of the form $(a,0;0,-a)$. You can apply this using the Mazur Swindle (http://en.wikipedia.org/w/index.php?title=Mazur_swindle) to show that the matrix (a,0;0;0) is also a linear combination of idempotents. – George Lowther Aug 04 '14 at 11:34
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    Thank you Pierre-Yves and George! – Jonas Meyer Aug 04 '14 at 12:15
  • More generally: are there any nonzero proper subspaces of End(V) invariant under conjugation other than that of the multiples of the identity? – Mariano Suárez-Álvarez Aug 06 '14 at 05:44
  • @MarianoSuárez-Alvarez - Hi Mariano! What about the finite rank endomorphisms? – Pierre-Yves Gaillard Aug 06 '14 at 06:01
  • @MarianoSuárez-Alvarez Is the statement below true? "Any subspaces of $\text{End}_K(V)$ invariant under conjugation is equal to $I$ or to $K+I$, where $I$ is a two-sided ideal." (Note that any two-sided ideal is given by the condition "rank less that $c$", for some infinite cardinal $c$.) – Pierre-Yves Gaillard Aug 06 '14 at 07:04
  • @JonasMeyer - Dear Jonas, I realized with horror that I forgot to give the link to this question of Ewan Delanoy. – Pierre-Yves Gaillard Aug 08 '14 at 14:23
  • @Pierre-YvesGaillard: Thank you. I had seen it. – Jonas Meyer Aug 08 '14 at 15:02

2 Answers2

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For a finite dimensional space, it is straightforward to express any endomorphism as a linear combination of idempotents. For an infinite dimensional space, we can use the fact that $V$ is isomorphic to $V\oplus V$. This follows from the fact that vector spaces are determined up to isomorphism by their dimension, and that $\kappa+\kappa=\kappa$ for any infinite cardinal $\kappa$. So, $$ \mathrm{dim}(V\oplus V)=\mathrm{dim}(V)+\mathrm{dim}(V)=\mathrm{dim}(V). $$ So, the algebra, $\mathrm{End}(V)$, of endomorphisms of $V$ is isomorphic to the endomorphisms of $V\oplus V$, which can be identified with the $2\times2$ matrices over $\mathrm{End}(V)$, $$ \mathrm{End}(V)\cong\mathrm{End}(V\oplus V)\cong\mathrm{M}_2(\mathrm{End}(V)). $$ First, the traceless matrices can be directly expressed as a linear combination of idempotents.

(I) Any matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ in $\mathrm{M}_2(\mathrm{End}(V))$ with $a+d=0$ is a linear combination of idempotents.

Proof: As, $d=-a$, we have $$ \left(\begin{matrix}a&b\\c&d\end{matrix}\right)=\left(\begin{matrix}a&a\\1-a&1-a\end{matrix}\right)-\left(\begin{matrix}1&a-b\\0&0\end{matrix}\right)+\left(\begin{matrix}1&0\\c+a&0\end{matrix}\right)-\left(\begin{matrix}0&0\\1&1\end{matrix}\right), $$ and each of the matrices on the rhs are idempotents. QED

It remains to show that diagonal matrices of the form $(a,0;0,0)$ are linear combinations of idempotents. For this, I will look at the infinite direct sum $V^\omega=V\oplus V\oplus V\oplus\cdots$. This can be identified with the infinite sequences $v=(v_0,v_1,v_2,\ldots)$ for $v_k\in V$, with all but finitely many $v_k$ equal to $0$, under component-wise addition and scalar multiplication. Given any sequence $a_0,a_1,a_2,\ldots$ in $\mathrm{End}(V)$, we use $a_0\oplus a_1\oplus a_2\oplus\cdots$ to represent the linear operator on $V^\omega$ taking $v$ to $(a_0v_0,a_1v_1,\ldots)$.

(II) For any $a\in\mathrm{End}(V)$, the operator $a\oplus 0\oplus 0\oplus\cdots$ is a linear combination of idempotents in $\mathrm{End}(V^\omega)$.

Proof: This is an Eilenberg-Mazur swindle. The operator $a\oplus(-a)$ on $V\oplus V$ is represented by the traceless matrix $(a,0;0,-a)$ so, by (I), is a linear combination of idempotents in $\mathrm{End}(V\oplus V)$. Grouping the factors of $V$ in the definition of $V^\omega$ into pairs, $$ V^\omega\cong(V\oplus V)\oplus(V\oplus V)\oplus(V\oplus V)\oplus\cdots $$ and applying the decomposition of $a\oplus(-a)$ into idempotents to each factor $V\oplus V$ shows that $$ L\equiv a\oplus(-a)\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus\cdots $$ decomposes into a linear combination of idempotents in $\mathrm{End}(V^\omega)$. Similarly, grouping the factors of $V^\omega$ as $$ V^\omega\cong V\oplus(V\oplus V)\oplus(V\oplus V)\oplus\cdots, $$ we can apply the decomposition of $a\oplus(-a)$ into idempotents to each factor of $V\oplus V$ and the zero map to the first factor $V$ to show that $$ M\equiv0\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus\cdots $$ is a linear combination of idempotents. So, $L+M=a\oplus0\oplus0\oplus\cdots$ is a linear combination of idempotents. QED

(III) For any $a\in\mathrm{End}(V)$, the matrix $\left(\begin{matrix}a&0\\0&0\end{matrix}\right)$ is a linear combination of idempotents.

Proof: First, using the fact that $\aleph_0\times\kappa=\kappa$ for any infinite cardinal $\kappa$, we have $$ \mathrm{dim}(V^\omega)=\aleph_0\times\mathrm{dim}(V)=\mathrm{dim}(V), $$ so $V\cong V^\omega$. Identifying the second factor of $V$ with $V^\omega$ in $V\oplus V$ gives the sequence of isomorphisms $$ \begin{align} V\oplus V&\cong V\oplus V^\omega= V\oplus(V\oplus V\oplus V\oplus\cdots)\\ &\cong V\oplus V\oplus V\oplus\cdots=V^\omega. \end{align} $$ Under this map, the operator $a\oplus 0$ represented by the matrix $(a,0;0,0)$ gets taken to $a\oplus0\oplus0\oplus\cdots$, which is a linear combination of idempotents by (II). QED

Putting these together gives,

(IV) Every endomorphism of $V$ is a linear combination of idempotents.

Proof: As above, using the fact that $\mathrm{End}(V)\cong \mathrm{M}_2(\mathrm{End}(V))$, we need to show that any $2\times2$ matrix of endomorphisms of $V$ is a linear combination of idempotents. Writing $$ \left(\begin{matrix}a&b\\c&d\end{matrix}\right)=\left(\begin{matrix}-d&b\\c&d\end{matrix}\right)+\left(\begin{matrix}a+d&0\\0&0\end{matrix}\right), $$ the first term on the rhs is a linear combination of idempotents by (I) and, similarly for the second term using (III). QED

Pierre-Yves Gaillard
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George Lowther
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    Your proof shows that every zero-trace endomorphism of $V$ is a sum of $4$ idempotents, and that every endomorphism of $V$ of the form $(a,0,0,0)$ is a sum of $4+4=8$ idempotents. In the end every endomorphism is the sum of at most $12$ idempotents. One can then ask two questions : 1) can the constant $12$ be improved ? 2) Is it true in finite dimensions also that every endomorphism is the sum of at most $12$ idempotents ? – Ewan Delanoy Aug 05 '14 at 06:15
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    @Ewan: In finite dimensions I can do it with 4 idempotents, and can maybe reduce it to 3. – George Lowther Aug 05 '14 at 15:40
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    Wow! Do you happen to have already written the details somewhere here on MSE ? If not I'll ask it as a new question, as Pierre-Yves did. – Ewan Delanoy Aug 05 '14 at 16:08
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    @EwanDelanoy - Dear Ewan and George: You might be interested by this paper, by Clément de Seguins Pazzis. – Pierre-Yves Gaillard Aug 05 '14 at 18:17
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    @Pierre-Yves: Ha! Interesting. So it can be reduced to 3, and I was along the right lines. Actually, I could express any matrix with 3 idempotents except for certain degenerate cases. By adding another idempotent you can always get out of the degenerate cases, but wasn't sure if the 4th idempotent is actually needed. In fact, by the referenced paper, it isn't. I wonder if the infinite dimensional case can be reduced to a similarly small number. – George Lowther Aug 05 '14 at 21:34
  • Hi George! Before the question became eligible for a bounty I accepted your magnificent answer, but the system told me that the question will never be eligible for a bounty! Fortunately, I could unaccept it! Now it tells me that I must wait 24 hours to award the bounty. I prefer to wait these 24 hours, then to award the bounty, and finally to accept the answer. - You prove that any endomorphism is the difference of two sums of 6 idempotents. The equality ... – Pierre-Yves Gaillard Aug 06 '14 at 08:24
  • $$2\begin{pmatrix}a&0\0&-a\end{pmatrix}=\begin{pmatrix}1+a&1+a\ 1-a&1-a\end{pmatrix}-\begin{pmatrix}1-a&1+a\ 1-a&1+a\end{pmatrix}$$ shows that, if $\text{char} K\neq0$, you can go down from 6 to 4. - Let $S$ be a subring of $K$, and $E_S$ the sub-$S$-module of $E:=\text{End}_KV$ generated by the idempotents. You prove $E_S=E$ if $\dim V=\infty$. Let $n(K,V,S)$ be the least $n$ such any $A$ in $E_S$ is an $S$-linear combination of idempotents if such an $n$ exists, and $\infty$ otherwise. One can try to compute the $n(K,V,S)$... – Pierre-Yves Gaillard Aug 06 '14 at 08:25
  • If $\dim V < \infty$, does the condition $\text{tr} A\in S$ imply $A\in E_S$?
    • – Pierre-Yves Gaillard Aug 06 '14 at 08:26
  • @EwanDelanoy - Dear Ewan: I think it would be great if you asked a new question! – Pierre-Yves Gaillard Aug 06 '14 at 08:29
  • @Pierre-YvesGaillard Dear Pierre-Yves : done ! – Ewan Delanoy Aug 06 '14 at 08:56
  • @Pierre-Yves: Everywhere that I used trace 0 matrices, I could have used trace 1 matrices instead, and these are explicitly linear combinations of 3 idempotents. So, reducing the total number of idempotents required from 12 to 9 is straightforward. – George Lowther Aug 06 '14 at 09:09
  • Thanks! Did you see Ewan's question http://math.stackexchange.com/q/888874/660 ? – Pierre-Yves Gaillard Aug 06 '14 at 09:12
  • Hi George! After your comment, @Ewan changed the 12 to a 9 in his question. I was wondering if, for the sake of clarity, you could edit your answer so that it corresponds better to Ewan's question. - Do you agree that we even get 8 if $char(K)\neq2$? – Pierre-Yves Gaillard Aug 06 '14 at 10:37
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    Actually, we get 7 – George Lowther Aug 06 '14 at 10:59
  • Hi George! I've just seen your we get 7 comment. Wow! (I think I wasn't pinged.) I suggested that you edit this answer to explain the reduction 12 -> 9. Now I think it might be better to leave this answer as it is, but I suggest that you explain how you get 7 in an answer to Ewan's question. – Pierre-Yves Gaillard Aug 07 '14 at 08:01
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    @Pierre-Yves: Actually, you can always do it with just 3 idempotents (using a rather different approach from my answer above). – George Lowther Aug 07 '14 at 12:00
  • Awesome! Looking forward to seeing the details... – Pierre-Yves Gaillard Aug 07 '14 at 12:19
  • @Pierre-Yves: I'll post an answer when I get some time. Unfortunately the proof I have that 3 is enough idempotents is a bit messy at the moment. – George Lowther Aug 08 '14 at 09:38
  • Hi George! Thanks! The following two apparently incompatible statements are both true: (1) I'm impatient to read you post. (2) I'd like you to post your answer at the time that's most convenient for you. – Pierre-Yves Gaillard Aug 08 '14 at 09:53