Subspaces of End(V) that can fix any vector

Suppose V is a finite-dimensional vector space and I have a linear subspace of its endomorphisms $$W \subseteq \mbox{End}(V).$$ How can I easily check if every vector of $V$ is fixed by some element of $W$? I would also be interested in any nice conditions on $W$ that imply $$\forall v \in V: \;\;\; v \in Wv,$$ even if they aren't biconditional.

• If the identity matrix happens to be in $W$, then the condition is satisfied trivially. However, it is easy to find examples where W omits the identity but can still fix any vector. For example, the two-by-two traceless matrices have this property.

• Unfortunately, $W$ can be large without satisfying the property. For example, take all matrices which have a zero in the upper left. These can never fix the first basis vector.

Side note:

In my particular situation, $W$ happens to be closed under multiplication. If this turns out to be a helpful condition, please don't hesitate to use it.

• Somewhat tangential: for algebraically closed fields, any proper subalgebra of End$\,(V)$ stabilizes some proper subspace (Burnside). Will study that proof and see if I can steal some ideas for your question. – P Vanchinathan Dec 5 '12 at 0:42
• I think one requires these proper subalgebras to be unital, if I'm not mistaken? – George Melvin Dec 5 '12 at 0:51
• @George Melvin: True. As I had always worked with rings with unity (commutative or not) I did not think of that as a requirement. – P Vanchinathan Dec 5 '12 at 5:45
• If $W$ is closed under multiplication and contains any invertible element $\varphi$, then $1\in W$: 1 can be written as a linear combination of positive powers of $\varphi$, which is follows from the fact that $\varphi$ satisfies a polynomial with non-zero constant term (namely its characteristic polynomial) – Julian Rosen Dec 5 '12 at 23:12
• So I'm going to just consider for now the case where $\dim W=n^2-1$ (here $n=\dim V$), so $W$ is described by a single linear constraint (the coefficients of which we put in a matrix). For $n=2$, I found that the property holds iff either the determinant of this matrix is nonzero, or if the trace is 0. I don't know if this generalizes. The trace condition is easy to understand -- the trace is 0 iff $1\in W$. Not so clear on what's going on with the determinant. – Harry Altman Dec 5 '12 at 23:16

The question is closely related to the notion of algebraic reflexivity. The (algebraic) reflexive closure of $W$ is the space of all operators $g \in End(V)$ such that $g(x) \in Wx$ for all $x \in V$. The author asks when the identity operator belongs to the reflexive closure of the operator space $W$.

I will consider $W$ as a matrix space to simplify things. By a lemma of Azoff (On Finite Rank Operators and Preannihilators), the reflexive closure of $W$ is the orthogonal of the span of the rank $1$ matrices in $W^\bot$ (the orthogonal of $W$ for the trace bilinear form $(A,B) \mapsto tr(AB)$). Thus, the identity is in the reflexive closure of $W$ if and only if it is orthogonal to all the rank $1$ matrices in $W^\bot$. Put differently, every vector of $V$ is fixed by at least one operator in $W$ if and only if all the rank $1$ operators in $W^\bot$ have trace zero.

Using this insight, one recovers the case of hyperplanes that was explained earlier by Harry Altman.

Now, if we add the condition that $W$ be stable under multiplication and that the underlying field $F$ be algebraically closed, then the identity belongs to the reflexive closure of $W$ if and only if it belongs to $W$! This combines ideas from other answers: if $W$ does not contain the identity, it contains only singular operators. One can use the Burnside theorem as it actually applies to every linear subspace that is stable under multiplication (provided that the space $V$ has dimension larger than $1$), not only to those which contain the identity. Using this repeatedly, one finds a basis of $V$ in which $W$ is represented by a space of block-upper-triangular matrices, where each diagonal block space is either a full matrix space $M_p(F)$ or $\{0\}$ (with the latter consisting of the zero $1 \times 1$ matrix). As the underlying field $F$ is infinite, one sees that at least one of those diagonal block spaces must be $\{0\}$ (otherwise, one shows that $W$ contains an invertible operator). With such a block space, the corresponding vector $x$ of $V$ is fixed by no operator in $W$.

With a similar method, but relying on Wedderburn's structure theorem for irreductible subalgebras instead of Burnside's theorem (which is just a special case of it), one can extend the above result to the more general situation where $W$ is stable under multiplication and the underlying field has cardinality greater than or equal to $\dim V$.

• This completely answers the question and equips me to handle future questions too. Magnifique! – John Wiltshire-Gordon Jun 14 '13 at 2:00
• Amazing answer! – Mariano Suárez-Álvarez Jun 14 '13 at 5:43

For subspaces that are stable under multiplication, there's an even more profound result that holds for all fields :

Given a finite-dimensional vector space $V$ and a linear subspace $W$ of $End(V)$ that is stable under multiplication but does not contain the identity, there are linear subspaces $V_1$ and $V_2$ of $V$ such that $V_1$ is a proper subspace of $V_2$ and every operator in $W$ maps $V_2$ into $V_1$. In particular, every operator in the reflexive closure of $W$ must map $V_2$ into $V_1$, whence the reflexive closure of $W$ consists only of singular operators!

The proof is essentially similar to the one I have outlined in my first answer. The key is a theorem of Wedderburn which states that a subalgebra of $End(V)$ is either reducible or simple. If we have a linear subspace $W$ which is stable under multiplication, then we note that $W +Span(id)$ is a subalgebra of $End(V)$ and $W$ is a two-sided ideal of it: thus, either $W$ is irreducible, or $W$ is a simple algebra, or $W$ is limited to the zero operator and $V$ has dimension $1$. Using this repeatedly, one finds a basis in which all the operators in $W$ are represented by block-upper-triangular matrices, and, on the diagonal, each block space is an algebra or consists of the $1 \times 1$ zero matrix. If there is at least one block of the latter time, then we have found the subspaces $V_1$ an $V_2$ we were looking for.

If we now assume that all the diagonal block spaces are algebras, then we prove that $W$ contains a non-singular matrix, which yields, as explained in some answers, that $W$ contains the identity. To do this, it suffices to prove the following lemma :

Let $A_1,\dots,A_n$ be algebras, and $H$ be a linear subspace of $A_1\times \cdots \times A_n$ that is stable under multiplication and projects surjectively onto each $A_i$. Then, $H$ contains the unit element of $A_1\times \cdots \times A_n$.

This is proved by induction over n. The case $n=1$ is trivial. Assume that this holds for $n-1$, with $n \geq 2$ fixed. Then, we write $A_1\times \cdots \times A_{n-1}=B$; by induction, $H$ contains $(1_B,x)$ for some $x \in A_n$. Then, as some $b \in B$ satisfies $(b,1_{A_n}) \in B \times A$, we see that $(1_B,1_{A_n})=(1_B,x)+(b,1_{A_n})-(1_B,x)*(b,1_{A_n}) \in H$.

This is a partial answer to the case where we aren't requiring that $W$ be closed under multiplication. Specifically, this is the case when $W$ has codimension $1$. I'm going to assume $V=F^n$, where $F$ is our base field, so $\mbox{End}(V)=M_n(F)$.

In this case, we can describe $W$ by a nonzero matrix $A$, i.e., $M\in W \iff A\cdot M=0$, where by $\cdot$ I mean we are actually "dotting" the matrices as if they were vectors, $A\cdot B=\sum_{i,j=1}^n a_{ij}b_{ij}$.

Claim: $W$ can fix any vector if and only if $A$ either has trace $0$ or has rank greater than $1$.

First, some observations: Obviously, scaling $A$ doesn't change $W$, so we can freely scale $A$. Furthermore, if we conjugate $A$, this doesn't change whether $W$ can fix any vector.

This is because for any matrix $S$, we have $SA\cdot M=A \cdot S^T M$, and $AS\cdot M=A \cdot MS^T$. The first is because because $A\cdot M$ is just the sum of the dot products of the columns of $A$ with the corresponding columns of $M$, and the second is because of the same fact with the rows. So if $W'$ is the space corresponding to $SAS^{-1}$, then $W'=S^T W (S^T)^{-1}$. So if $v\in V$ and $W$ can fix any vector, then $W$ can fix $(S^T)^{-1} v$, which means that $W'$ can fix $v$.

So we can freely scale and conjugate $A$. Assume for now that $F$ is algebraically closed, to make things easier; then we can assume that $A$ is in Jordan normal form.

Suppose $W$ fails to fix some vector. Then $A$ must have nonzero trace, as otherwise we would have $I\in W$. Since $A$ has nonzero trace, $A$ has some nonzero eigenvalue, and so $A$ has some row consisting of a nonzero element on the diagonal and zeroes elsewhere. (Really, this was all we needed algebraic closure for.) Conjugating and scaling, we'll assume this element is a $1$ and is in the lower right corner.

So now we have the question, for any $v=(x_1,\ldots,x_n)\in F^n$, does there exist $B=(b_{ij})_{i,j=1}^n\in M_n(F)$ such that $B$ fixes $v$ and $A\cdot B=0$? Well, for any given $v$, this is just a system of linear equations in the $b_{ij}$. Since we assumed $a_{jj}=1$, we'll exclude $b_{jj}$ as a parameter, writing it in terms of the others instead.

The $n \times n^2-1$ matrix for this system of linear equations (where the columns $b_{ij}$ are put in order first by row and then by column) is as follows: In row $i$, for $i\lt n$, the nonzero entries run from $n(i-1)+1$ to $ni$, being $x_1, \ldots, x_n$ in turn. In row $n$, we have $x_1,\ldots,x_{n-1}$ in positions $n(n-1)+1$ to $n^2-1$ as before, but of course there is no column $n^2$, and also there are scattered $-a_{ij}$ in the other columns. (These don't affect the final $x_1,\ldots,x_{n-1}$ because of the assumption that $a_{nj}=0$ for $j\lt n$.)

If any of $x_1,\ldots,x_{n-1}$ are nonzero, this matrix clearly has full rank, and so $W$ can fix $v$. Whereas if $x_1=\ldots=x_{n-1}=0$ and $x_n\ne 0$, then $W$ can clearly again fix $v$... except possibly if each one of the scattered $-a_{ij}$ is in the same column as some $x_n$ above, i.e., if the only nonzero entries in $A$ are in the last column. Since we assumed $W$ does not fix some vector, we conclude that $A$ has rank $1$.

Thus, if $W$ fails to fix some vector, $A$ has nonzero trace and rank $1$. For the converse, if $A$ has nonzero trace and rank $1$, then once again we can put $A$ in Jordan normal form. There are only two possible Jordan normal forms for a matrix of rank $1$ -- a lone nonzero element on the diagonal, and a lone $1$ just above the diagonal. Since $A$ has nonzero trace, we're in the former case. Scaling and conjugating again, we can assume $A$ is the matrix with a lone $1$ in the upper left corner, so $W$ is the space of all matrices with a $0$ in the upper left corner. But then $W$ cannot fix $(1,0,\ldots,0)$, so $W$ fails to fix some vector.

This concludes the proof when $F$ is algebraically closed. If $F$ is not algebraically closed, well, all the equations we're trying to solve are linear, so this makes no difference. More concretely, if $A$ has nonzero trace and rank $1$, it's still similar to the matrix with a lone $1$ in the upper left corner (similarity is invariant under field extension), and so $W$ still fails to fix some vector. Otherwise, for a given vector $v$, finding $B$ that fixes $v$ and satisfies $A\cdot B=0$ is still just solving a system of linear equations in the $b_{ij}$, so if there is a solution over $\overline{F}$, there is a solution over $F$ (e.g. take a solution over $\overline{F}$ and average it with its conjugates). This proves the claim.

I'm going to add another partial answer, this time regarding the case where we do require that $W$ be closed under multiplication.

Namely: If $W$ is closed under multiplication, and $\dim W>n^2-2n+2$, then the only way $W$ can fix every vector is if $W$ contains the identity.

As Julian Rosen pointed out in the comments, if $W$ is closed under multiplication and contains any invertible element, it then contains the identity. So in order for $W$ to fail to fix some vector, it must have no invertible elements.

Fortunately, large subspaces of matrices containing no invertible elements (more generally, having elements of bounded rank) have been much studied! Here I'll use the results of "The classification of large spaces of matrices with bounded rank", by Clément de Seguins Pazzis. This will get us a converse to the above statement. For according to this, if $\dim W>n^2-2n+2$ and $W$ has no invertible elements, then either there exists a nonzero $v\in V$ which is in the kernel of all elements of $W$ (and so $W$ cannot fix every vector), or there is some $n-1$-dimensional subspace of $V$ which contains the images of all elements of $W$ (and so $W$ cannot fix every vector).

Thus, for $\dim W> n^2-2n+2$, being able to fix every vector is equivalent to containing some invertible element, which is equivalent to containing the identity.

(Note, by the way, that if $\dim W > n^2-n$, then automatically $W$ must contain some invertible element.)

Now the paper also proves results about what happens when $\dim W=n^2-2n+2$, but I don't see how to apply these, because while the resulting subspaces either A. don't fix every vector or B. aren't closed under multiplication, I don't think these properties are preserved under the notion of equivalence used in the paper, and so it's not so clear what happens then. Still, I think one ought to be able to do better than what I've written here -- we've only used closure under multiplication to go from "$W$ contains some invertible element" to $I \in W$, and I think one ought to be able to do much more with it than that.