Prove or give a counterexample:
If A is an n x n matrix with n distinct (real)
eigenvalues,then A is diagonalizable.

For this case we need to use the following theorem "If [tex] v_1, v_2,....v_k[/tex] are eigenvectors of an nxn matrix A and the associated eigenvalues [tex]\lambda_1, \lambda_2,...,\lambda_k [/tex] are distinct, then [tex] v_i's[/tex] are linearly independent". Now we can proof the statement like this:

Proof

Let A a nxn matrix and we can assume that A has n distinct real eingenvalues let's say [tex] \lambda_1, \lambda_2, ....,\lambda_n[/tex]

From definition of eigenvector for each one [tex] \lambda_i[/tex] needs to have associated an eigenvector [tex] v_i[/tex] for [tex] 1 \leq i \leq n[/tex]

And using the theorem from before , the n eigenvectors [tex] v_1,....,v_n[/tex] are linearly independent since the [tex]\lambda_i [/tex] [tex] 1\leq i \leq n[/tex] are distinct so then we ensure that A is diagonalizable.

Prove or give a counterexample: If A is an n x n matrix with n distinct (real) eigenvalues,then A is diagonalizable.

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Prove or give a counterexample:
If A is an n x n matrix with n distinct (real)
eigenvalues,then A is diagonalizable.