Let $A$ be a matrix $n\times n$ and let be $\bar\lambda$ an eingenvalue of $A$ such that $m_a(\bar\lambda)=m_g(\bar\lambda)=1$.
With $m_a$ and $m_g$ I mean the algebric and the geometric molteplicity.
If I want to know the Jordan canonical form of the matrix $A$ I know that I will have for sure just one block respect the eingenvalue $\bar\lambda$.
Can I conclude for some reason that the dimension of this block is $1\times 1$?
The geometric multiplicity of an eigenvalue is equal to the number of Jordan blocks, and as mentioned by@bharb, the algebraic multiplicity is equal to the sum of the sizes of the Jordan blocks relative to $\lambda $. If both multiplicities are equal to $1$, there can be only one block of size $1$.