# If $A$ is a $2\times 2$ non-diagonal diagonalizable matrix then have two distinct eigenvalues

Let $$A$$ be a $$2\times 2$$ non-diagonal real matrix. I know that if it has two distinct real eigenvalues then it is diagonalizable. I would like to know if the reverse is also true, that is, matrix $$A$$ is digitizable if it has two distinct (real) eigenvalues.

Obviously if I remove the hypothesis to be non-diagonal, then it is not true, as can be seen in these other questions. Q1 and Q2

I tried to create an example of a non-diagonal matrix with equal eigenvalues that is diagonalizable, but I couldn't. But, in this comment, it is said that my statement is true,but I couldn't prove it.

I thought this theorem could help me, but I was unable to apply it in the direction I want.

Let $$T:V\to V$$ be a linear operator, in a space of finite dimension and let $$\lambda_1,\cdots \lambda_t$$ be its distinct eigenvalues. The following statements are equivalent:

1. A is diagonalizable
2. $$p_T(x)=(x-\lambda_1)^{n_1}\cdots (x-\lambda_t)^{n_t},n_i\ge 1$$ and $$\gamma_{T}(\lambda_i)=\mu_{T}(\lambda_i)$$
3. $$\dim V=\sum_{i=1}^{t}\dim Aut_A(\lambda_i)$$

where $$\gamma_{T}(\lambda_i)$$ and $$\mu_{T}(\lambda_i)$$ are geometric multiplicity and algebraic multiplicity, respectively.

Is it possible to prove that if $$A$$, (not diagonal) is diagonalizable, then it has two distinct real eigenvalues with the theorem above? Or does anyone have any suggestions for another useful result.

• Any $2\times 2$ diagonalizable matrix with repeated eigenvalues is a multiple of the identity Oct 27, 2020 at 22:11

Suppose $$A$$ is non-diagonal and diagonalizable. By contradiction, suppose also that it has only one eigenvalue $$\lambda$$ repeated two times.
Since $$A$$ is diagonalizable, you have a base-change matrix $$M$$ such that $$A=MDM^{-1}$$ with $$D$$ a diagonal matrix containing the eigenvalue. Since $$A$$ has only one eigenvalue with algebraic multiplicity 2, then $$D=\lambda I$$. As a consequence, $$A = M \lambda IM^{-1} = \lambda MM^{-1} = \lambda I$$ so $$A$$ is diagonal. Contradiction.
• I'd reformulate the proof without contradiction. If $A$ is a diagonalizable $2\times2$ matrix with an eigenvalue of multiplicity $2$, then it is diagonal. Oct 27, 2020 at 22:19