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

See proof below.

Step-by-step explanation:

True

For this case we need to use the following theorem "If are eigenvectors of an nxn matrix A and the associated eigenvalues are distinct, then 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

From definition of eigenvector for each one needs to have associated an eigenvector for

And using the theorem from before , the n eigenvectors are linearly independent since the are distinct so then we ensure that A is diagonalizable.

30

step-by-step explanation:

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