# Maximum dimension of space of matrices with a real eigenvalue

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 $$g(n)>1$$?

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.
• Thanks! In fact, Adams' result was the motivation for my question, as well as mathoverflow.net/questions/309135. Note that $\dim V+\dim W\leq n^2$, since otherwise $V\cap W\neq \{0\}$. Thus Noam's answer solves my first question for $n=1,2,4,8$, and S. carmeli's answer improves the upper bound in general. – Richard Stanley Mar 31 '19 at 13:46
For the "complementary" problem, $$g(n) for all positive $$n$$, and the upper bound $$g(n) \leq n-1$$ is sharp at least for $$n=1,2,4,8$$.
If $$\dim W \geq n$$ then any nonzero vector $$v \in {\bf R}^n$$ is an eigenvector of some nonzero matrix in $$W$$. Indeed if $$A_1,\ldots,A_n \in W$$ are linearly independent then a linear dependence in the $$n+1$$ vectors $$v$$ and $$A_i v$$, say $$\sum_{i=1}^n c_i A_i v = \lambda v$$, yields nonzero $$A = \sum_{i=1}^n c_i A_i \in W$$ with $$Av = \lambda v$$.
The equality in $$g(n) \leq n-1$$ is trivial for $$n=1$$ and easy for $$n=2$$; for $$n=4$$ and $$n=8$$ we get examples from the traceless quaternions and octonions respectively: $$\left(\begin{array}{cccc} 0 & a & b & c \cr -a & 0 & -c & b \cr -b & c & 0 & -a \cr -c & -b & a & 0 \end{array}\right)$$ has characteristic polynomial $$(x^2+a^2+b^2+c^2)^2$$, and $$\left(\begin{array}{cccccccc} 0 & a & b & c & d & e & f & g \cr -a & 0 & c &-b & e &-d &-g & f \cr -b &-c & 0 & a & f & g &-d &-e \cr -c & b &-a & 0 & g &-f & e &-d \cr -d &-e &-f &-g & 0 & a & b & c \cr -e & d &-g & f &-a & 0 &-c & b \cr -f & g & d &-e &-b & c & 0 &-a \cr -g &-f & e & d &-c &-b & a & 0 \end{array}\right)$$ has characteristic polynomial $$(x^2+a^2+b^2+c^2+d^2+e^2+f^2+g^2)^4$$.