Matrix Similarity Question

Question

Show that similarity is an equivalence relation. More specifically, recall that we say $A, B \in M_{n \times n}(F)$ (set of $n\times n$ matrices) are similar if there exists an invertible $Q$ such that $B = Q^{−1}AQ$. More compactly, $A \sim B \iff ∃Q,\ B = Q^{−1}AQ$.

Show that $\sim$ is an equivalence relation.

I get why they are similar but I don't understand at all how that is an equivalence statement.

That $Q^{-1}$ is supposed to be $Q$ inverse. I'm new to this and haven't learned how to format. Formatting advice would also be appreciated. Thank you.

Answer 1

You need to show that similarity satisfies all $3$ parts of the definition of an equivalence:

• reflexivity: $A$ is always similar to itself
• symmetry: If $A$ is similar to $B$, then $B$ is similar to $A$
• transitivity: If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

Answer 2

The matrices $A,B$ are similar if $B=Q^{-1}AQ$.

You have to prove that:

$$ 1) \qquad \forall A \rightarrow A \sim A $$ and this is done with $Q=I$.

$$ 2) \qquad B\sim A \Rightarrow A\sim B $$ and this is done by: $$ B=Q^{-1}AQ \Rightarrow QB=AQ \Rightarrow A=QBQ^{-1}=(Q^{-1})^{-1}AQ^{-1} $$

$$ 3) \qquad (B\sim A \;\land \; B\sim C) \Rightarrow A\sim C $$ done by: $$ (B=Q^{-1}AQ \;\land\; B=P^{-1}CP) \Rightarrow C= PBP^{-1}=PQ^{-1}AQP^{-1}=(QP^{-1})^{-1}A(QP^{-1}) $$