A few questions on matrix diagonalizability I have a few questions on matrix diagonalizability. Suppose that $A \in \mathbb{R}^{n \times n}$.


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*If $A$ is diagonalizable, we can find $P \in \mathbb{R}^{n \times n}$ such that $A = P\Lambda P^{-1}$. Does the definition of diagonalizability in $\mathbb{R}^{n \times n}$ mean that eigenvalues must be real as well? If we are allowed to have complex eigenvalues, can we write more matrices $A = P\Lambda P^{-1}$?

*From the Wikipedia page, I know that the set of diagonalizable matrices on $\mathbb{C}^{n \times n}$ is dense. If $A \in \mathbb{R}^{n \times n}$ is non-diagonalizable, it can be approximated with a diagonalizable matrix $\widetilde{A} \in \mathbb{C}^{n \times n}$ arbitrarily closely. In general $\widetilde{A}$ may have complex eigenvectors. Is it possible to approximate $A$ arbitrarily closely with real eigenvectors and complex eigenvalues?
 A: *

*Continuity of the eigenvalues. 


Let $A\in M_n(\mathbb{C})$ with $spectrum(A)=(\lambda_i)_i$. For every $\epsilon >0$, there is $\alpha >0$ s.t. $||A-X||<\alpha$ implies that there exits a numbering $(\mu_i)_i$ of the eigenvalues of $X$ s.t. $\sum_i|\lambda_i-\mu_i|<\epsilon$.


*One says that $A\in M_n(K)$ is diagonalizable over $K$ IFF there is $P\in GL_n(K)$ s.t. $D=P^{-1}AP$ is diagonal. That implies that the eigenvalues of $A$ are all in $K$. For example, $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is not diagonalizable over $\mathbb{R}$ but it is over $\mathbb{C}$.

*Let $L$ be an extension of the field $K$. Note that if $A\in M_n(K)$ is diagonalizable over $L$, into the diagonal matrix $D\in M_n(K)$, then $A$ is diagonalizable over $K$, into the same matrix $D$.

*Consider a matrix $B\in M_n(\mathbb{R})$ in a neighborhood of $A\in M_n(\mathbb{R})$. According to 1., two conjugate eigenvalues of $A$ give birth to $2$ conjugate eigenvalues of $B$, a simple real eigenvalue of $A$ to a simple real eigenvalue of $B$; the case of multiple real eigenvalues of $A$ is more complicated.

*In any real neighborhood of $A\in M_n(\mathbb{R})$, there is a matrix with distinct complex eigenvalues, then diagonalizable over $\mathbb{C}$ but not necessarily over $\mathbb{R}$.

*Let $A=[a_{i,j}]\in M_n(\mathbb{R})$ be a random matrix (use a normal law for each $a_{i,j}$). Then the eigenvalues are distinct with probability $1$. Yet, when $n$ is a large number, this matrix is "always" not diagonalizable over $\mathbb{R}$ because the mean of the number of real roots of a real polynomial of degree $n$ is in $O(\sqrt{n})$.
