Question

Prove or give a counterexample:

Answer 1

Answer:

See proof below.

Step-by-step 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.