Prove that diagonalization with real entries is not possible when not all eigenvalues are real

Question

I know that when all the eigenvalues and eigenvectors of a matrix $A$ are real, we have $A = P\Lambda P^{-1}$, where the columns of $P$ are the eigenvectors and the diagonal entries of $A$ are the corresponding eigenvalues.

However, when not all the eigenvalues and eigenvectors are real (i.e. some of them have imaginary part), is it possible to show that there is no such $P$ and $\Lambda$ where the diagonal entries of $\Lambda$ are all real, such that $A = P\Lambda P^{-1}$?

Answer 1

Diagonalization w/ real entries$\implies$ all eigenvalues are real. As pointed out by @Gerry Myerson in the comments, the diagonal entries will be the eigenvalues.

Answer 2

Suppose $A$ has real eigenvalues and $\Lambda$ has complex diagonal entries. We can write $AP=P\Lambda$. Suppose $P=[p_1 \cdots p_n]$ and $\Lambda=diag(\lambda_1,\dots,\lambda_n)$. Then, we have $$ Ap_i = \lambda_i p_i $$ for all $i=1,2,\dots,n$. That is, $\lambda_i$ are the eigenvalues of $A$. But they are the diagonal entries of $\Lambda$ which are complex. Therefore, we have a contradiction.