Is there a way to estimate an upper bound for the dimension of a vector subspace of the space of $n \times n$ matrices $M_n(\mathbb{C})$ which doesn't intersect any matrices with distinct eigenvalues?

A trivial example of subspace is the scalar matrices $cI$, but I can't seem to find a good way to find larger example. I think one point of confusion was that the set we don't want to intersect may not be a vector space. The other thing I was thinking about was the fact that ``most'' matrices are diagonalizable, but I'm not sure how to turn this into something useful.

If this isn't possible, what happens if we replace the set of matrices with distinct eigenvalues with upper triangular or lower triangular matrices with distinct diagonal entries?

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    $\begingroup$ A bigger example for $n=2$ is the set of upper triangular matrices with repeated eigenvalues. Similar examples exist for all $n\ge2$. $\endgroup$ – Gerry Myerson Oct 31 '17 at 4:13
  • $\begingroup$ @GerryMyerson For $n\geq 2$, your subspace has dimension $\frac{n(n-1)+2}{2}$. This is probably the maximal dimension $\endgroup$ – Ewan Delanoy Oct 31 '17 at 9:47
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    $\begingroup$ About the "distinct eigenvalues", could you please clarify : must the eigenvalues be pairwise distinct (which seems to be the correct interpretation), or does it suffice that not all eigenvalues are equal ? Does $diag(1,1,2)$ qualify ? $\endgroup$ – Ewan Delanoy Oct 31 '17 at 9:49
  • $\begingroup$ @Ewan Delanoy Here, "distinct eigenvalues" means that the eigenvalues are pairwise distinct. Also, why do you think that $\frac{n(n - 1) + 2}{2}$ is the maximal dimension? $\endgroup$ – modnar Oct 31 '17 at 18:23
  • $\begingroup$ That $\frac{n(n-1)+2}{2}$ was a typo actually, I meant $\frac{n(n+1)-2}{2}$ (consider the set of all upper triangular matrices $a_{ij}$ (so $a_{ij}=0$ for $i>j$, with the additional condition $a_{22}=a_{11}$). I believe it to be the maximum dimension because when I tried random examples with larger dimension, I always found a matrix in it with distinct eigenvalues. $\endgroup$ – Ewan Delanoy Oct 31 '17 at 18:59

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