Question

Prove or give a counterexample:

If A is an n x n matrix with n distinct (real)
eigenvalues,then A is diagonalizable.

Answer 1

See proof below.

Explanation:

True

For this case we need to use the following theorem "If
`v_1, v_2,....v_k` are eigenvectors of an nxn matrix A and the associated eigenvalues
`\lambda_1, \lambda_2,...,\lambda_k` are distinct, then
`v_i's` 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
`\lambda_1, \lambda_2, ....,\lambda_n`

From definition of eigenvector for each one
`\lambda_i` needs to have associated an eigenvector
`v_i` for
`1 \leq i \leq n`

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