# Are the two matrices similar?

Consider the matrices $$A= \begin{bmatrix} 2 & 2 & 1 \\0 & 2 & -1\\ 0 & 0 & 3\end{bmatrix}$$ and $$B= \begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 0\\ 0 & 0 & 3\end{bmatrix}.$$

Which are true:

1. $A$ and $B$ are similar over $\Bbb Q$.
2. $A$ is diagonalizable over $\Bbb Q$.
3. $B$ is Jordan Canonical Form of $A$.
4. The minimal and characteristic polynomial of $A$ are same.

The characteristic polynomial of $A$ is coming as $(x-2)^2(x-3)$. Since minimal polynomial and characteristic polynomial have the same roots so minimal polynomial must be $(x-2)(x-3)$ or $(x-2)^2(x-3)$.$A$ does not satisfy $(x-2)(x-3)$ so it's minimal polynomial must be $(x-2)^2(x-3)$.Obviously $B$ is JCF of $A$.

And since roots of minimal polynomial are not distinct so $A$ is not diagonalisable.Hence $3,4$ are true; $1$ is false.

How to prove/disprove $1$

The characteristic polynomials of $A$ and $B$ are obviously the same and factors over $\mathbb Q$.

Since it it also easy to see that the eigenspace corresponding to the eigenvalue $2$ for $A$ is one-dimensional, we can conclude that $B$ is in fact the Jordan canonical form of $A$.

This implies the answers to (1) and (2), which again gives information about the minimal polynomials.

• How can we say that $A$ and $B$ are similar? – Learnmore Dec 23 '15 at 14:42
• @Amartya: By definition a matrix and its Jordan canonical form are always similar. Otherwise it wouldn't be its Jordan form. – Henning Makholm Dec 23 '15 at 14:43

You have the following result:

If $\mathbb{F} \subseteq \mathbb{K}$ are two fields and $A,B \in M_n(\mathbb{F})$ are similar as matrices in $M_n(\mathbb{K})$ (so there exists an invertible $P \in M_n(\mathbb{K})$ such that $P^{-1}AP = B$) then they are also similar as matrices in $M_n(\mathbb{F})$.

In your case, $A$ and $B$ are clearly similar over $\mathbb{C}$ and have rational entries and so they are also similar over $\mathbb{Q}$. The matrix $A$ is not diagonalizable over $\mathbb{Q}$ so (1) is true and (2) is false.