"If for every eigenvalue of matrix A - the algebraic multiplicity equals 1 so A is diagonalizable" True/False?

Question

I can't find the answer. I know that $A$ is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors , but I can't determine if it's true according to the given information.

Answer 1

Hint. If $F$ is a field and $A$ is a square matrix over $F$, then the followings are equivalent:

1. $A$ is diagonalizable over $F$
2. The minimal polynomial of $A$ has distinct zeros in $F$.

Now if $m(x)$ is the minimal polynomial of $A$, then $m(x)$ divides the characteristic polynomial of $A$. In your case, what can you say about $m(x)$?

Answer 2

Hint: the geometric multiplicity of an eigenvalue (i.e. the dimension of the associated eigenspace) is always $\ge 1$ and $\le$ the algebraic multiplicity.