# Diagonalizable matrices with same geometric multiplicity for every eigenvalue similar?

Given $A$ and $B$ diagonalizable matrices similar to a diagonal matrix $C$. That is to say, $A$ and $B$ have the same eigenvalues and for every eigenvalue the geometric multiplicity is the same.

Knowing this, can I determine $A$ and $B$ are similar?

• "Similar"? Sure. They have the same eigenvalues. What else does "similar" mean to you? – David G. Stork Jul 30 '18 at 17:45
• David G. Stork, same eigenvalues do not imply that the matrices are similar. – Mark Jul 30 '18 at 17:49
• @DavidG.Stork "Similar" is a standard term in linear algebra. $A$ and $B$ are similar if there exists an invertible $P$ such that $A=P^{-1}BP$. – David C. Ullrich Jul 30 '18 at 18:04

## 2 Answers

$$A=P^{-1}CP \implies C=PAP^{-1}$$ $$B=Q^{-1}CQ \implies C=QBQ^{-1}$$

$$PAP^{-1}=QBQ^{-1}$$

$$A=P^{-1}QBQ^{-1}P=(Q^{-1}P)^{-1}B(Q^{-1}P)$$

Hence, $A$ and $B$ are similar.

The matrices $A$ and $B$ are similar since both of them are similar to the matrix $C$.

• So determining the basis in which A and B are the matrix C is irrelevant? That was the main thing I wasn't sure about. – InfiGuy Jul 30 '18 at 17:46
• Yes, knowing that both of them are similar to $C$ is enough. – José Carlos Santos Jul 30 '18 at 17:47