# Is every endomorphism of a vector space a linear combination of idempotents?

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.

• 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. Commented Aug 4, 2014 at 11:28
• 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 (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. Commented Aug 4, 2014 at 11:34
• Thank you Pierre-Yves and George! Commented Aug 4, 2014 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? Commented Aug 6, 2014 at 5:44
• @MarianoSuárez-Alvarez - Hi Mariano! What about the finite rank endomorphisms? Commented Aug 6, 2014 at 6:01

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

• 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 ? Commented Aug 5, 2014 at 6:15
• @Ewan: In finite dimensions I can do it with 4 idempotents, and can maybe reduce it to 3. Commented Aug 5, 2014 at 15:40
• 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. Commented Aug 5, 2014 at 16:08
• @EwanDelanoy - Dear Ewan and George: You might be interested by this paper, by Clément de Seguins Pazzis. Commented Aug 5, 2014 at 18:17
• @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. Commented Aug 5, 2014 at 21:34

Here is a corollary to George's answer:

(a) If $V$ is an infinite dimensional $K$-vector space, then the set of sub-vector spaces of $V$ has cardinal $\operatorname{card}(K)^{\dim(V)}$.

Proof. Let $\mathcal S$ be the set in question, and let $\mathcal E$ and $\mathcal I$ be respectively the set of endomorphisms and of idempotent endomorphisms of $V$. We have $$\operatorname{card}(\mathcal E)=\operatorname{card}(K)^{\dim(V)}.$$ George proved that any endomorphism is a $\mathbb Z$-linear combination of idempotents. This implies $$\operatorname{card}(\mathcal E)=\aleph_0\operatorname{card}(\mathcal I).$$ The two above displays imply $$\operatorname{card}(\mathcal I)=\operatorname{card}(K)^{\dim(V)}.$$ The map $$\mathcal I\to\mathcal S,\quad e\mapsto\operatorname{Im}(e)$$ being surjective, and the map $$\mathcal I\to\mathcal S\times\mathcal S,\quad e\mapsto(\operatorname{Im}(e),\operatorname{Ker}(e))$$ being injective, $\mathcal I$ and $\mathcal S$ are equipotent, and the proof is complete.

Note that George's previous answer was already sufficient for the above argument.

Questions:

(b) Is there a simpler proof of (a)?

(c) Was Statement (a) previously recorded? Where?

Thank you very much in advance, dear reader, if you can tell me what the answers to these questions are.