TITLE: Maximum dimension of space of matrices with a real eigenvalue
QUESTION [15 upvotes]: Let $M_n(\mathbb{R})$ denote the space of all $n\times n$ real
matrices. What is the maximum dimension $f(n)$ of a subspace $V$ of
$M_n(\mathbb{R})$ such that every matrix in $V$ has at least one real
eigenvalue? It is easy to see that if $n$ is odd then $f(n)=n^2$,
while if $n$ is even then $n^2-n+1\leq f(n)\leq n^2-1$.
There is also the ``complementary'' problem: what is the maximum
dimension $g(n)$ of a subspace $W$ of $M_n(\mathbb{R})$ such that
every nonzero matrix in $W$ has no real eigenvalues? Clearly if $n$ is
odd then $g(n)=0$. When $n$ is even, can one have $g(n)>1$?

REPLY [8 votes]: The "complementary" problem is strongly related to the problem of independent vector fields on $S^{n-1}$. Indeed, if $W$ is a space of matrices none of them having a real eigen-value and $(A_1,...,A_k)$ is its basis, it means that for every $v$, the vectors $(v,A_1(v),...,A_k(v))$ are linearly independent. Hence, the projections of $(A_1v,...,A_kv)$ to the plane orthogonal to $v$ give rise to $k$ independent vector fields on the sphere $S^{n-1}$. This problem was solved by Adams using non-trivial algebraic topology, and the exact possible number is $\rho(n)-1$ where $\rho((2k+1)2^{4a+b})=2^b+8a$. Note that this is consistent with the previous answer by Noam, since $S^1,S^3,S^7$ are really the only parallelizable spheres.  
Now, to see that this is the answer in our case, it suffices to note that the construction of the examples in Adams work actually arrises from a solution to the question you ask: they all come from construction of linear examples. Namely, consider a Clifford algebra $Cl_k$ having a representation of dimension $n$. Then the imaginary clifford elements $(e_1,...,e_k)$ give example to a space of matrices as you wish, and it is known that this is a maximal example for independent vector fields on on the $n-1$-sphere as well.