This is an elementary question, but a little subtle so I hope it is suitable for MO.


Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$. 


The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda\_iI$, and we have the Jordan canonical form:


$$ J = \begin{bmatrix} J\_1 \\\ & J\_2 \\\ & & \ddots \\\ & & & J\_n \end{bmatrix}$$


where each block $J\_i$ corresponds to the eigenvalue $\lambda\_i$ and is of the form


$$ J\_i = \begin{bmatrix} \lambda\_i & 1 \\\ & \lambda\_i & \ddots \\\ & & \ddots & 1 \\\ & & & \lambda\_i \end{bmatrix}$$


and each $J\_i$ has the property that $J\_i - \lambda\_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J\_i - \lambda\_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. Thus, Jordan canonical form gives the closest possible to a diagonal matrix. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization.


So far, so good. What I want to prove is the assertion that "Almost all square matrices over $\mathbb{C}$ is diagonalizable". The measure on the space of matrices is obvious, since it can be identified with $\mathbb{C}^{n^2}$. How to prove, perhaps using the above Jordan canonical form explanation, that almost all matrices are like this?


I am able to reason out the algebra part as above, but is finding difficulty in the analytic part. All I am able to manage is the following. The characteristic equation is of the form


$$(x - \lambda\_1)(x - \lambda\_2) \cdots (x - \lambda\_n)$$


and in the space generated by the $\lambda\_i$'s, the measure of the set in which it can happen that $\lambda\_i = \lambda\_j$ when $i \neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$.


But here I have cheated, I used only the characteristic equation instead of using the full matrix. How do I prove it rigorously?



Proving "almost all matrices over C are diagonalizable".
The discriminant of the characteristic polynomial of a matrix depends polynomially on the coefficients of the matrix, and its vanishing detects precisely the existence of multiple eigenvalues. Therefore the set where the discriminant does not vanish is contained in the set of diagonalizable matrices.


Now the set where a non-zero polynomial vanishes is very, very thin (in many senses: it does not contain open sets, it has zero Lebesgue measure, etc) so in consequence the set of diagonalizable matrices is correspondingly thick.


