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

diagonalizationeigenvalues-eigenvectorslinear algebramatrices

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.

Best Answer

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.

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