Which one is more? diagonalizable matrix or non-diagonalizable matrix?

Question

Are there more diagonalizable matrixes or non-diagonalizable matrixes?

Answer 1

As Gae. S. stated, the answer depends on which field your matrices are defined over. My answer is concerned with matrices over the field $\mathbb{C}$, the answer for $\mathbb{R}$ is analogous. Moreover, it depends on how you define "size". If by size you mean cardinality, the answer is that there are equally many if $n>1$ (if $n=1$, all matrices are diagonal). This is what I show below.

Let $\mathcal{M}$ be the set of matrices in $\mathbb{C}^{n\times n}$ and $\mathcal{D}\subset \mathcal{M}$ be the set of diagonalizable matrices. Let $\Lambda\subset\mathcal{D}$ be the set of diagonal matrices and $\Lambda'\subset\mathcal{M}$ be the set of matrices whose entries are zero, besides on the main diagonal and the superdiagonal (the entries above the diagonal).

By definition, matrices in $\mathcal{D}$ are similar to matrices in $\Lambda$. By the Jordan normal form, matrices in $\mathcal{M}$ are similar to matrices in $\Lambda'$.

Now for the cardinality ("size") argument. If $n>2$, $\Lambda'\setminus\Lambda$ is nonempty. We have: $$ Card(\mathbb{C})=Card(\mathbb{C}^{n-1})=Card(\Lambda'\setminus\Lambda)\leq Card(\mathcal{M}\setminus\mathcal{D})\leq Card(\mathcal{M})=Card(\mathbb{C}^{n^2})=Card(\mathbb{C}) $$ Similarly, $$ Card(\mathbb{C})=Card(\mathbb{C}^{n})=Card(\Lambda)\leq Card(\mathcal{D})\leq Card(\mathcal{M})=Card(\mathbb{C}^{n^2})=Card(\mathbb{C}) $$ So, we conclude that $$ Card(\mathbb{C})=Card(\mathcal{M}\setminus\mathcal{D})=Card(\mathcal{D}) $$