Answer
True
Work Step by Step
Consider $B=S^{-1}AS$ and $J$ is Jordan canonical form of $A$.
We have $J=M^{-1}AM$ with $M$ is invertible
Obtain: $B=S^{-1}AS\\
=S^{-1}(MJM^{-1})S\\
=(M^{-1}S)^{-1}J(M^{-1}S)\\
=B(M^{-1}S)(M^{-1}S)^{-1}\\
=J$
We can say that $J$ is Jordan canonical form of $B$
Hence, the statement is true.
