- $A-2I$ and $B-2I$ are similar.
- $A$ and $B$ have the same trace.
- $A^{-1}$ and $B^{-1}$ are similar matrices.

- 1 only.
- 2 only.
- 3 only.
- 1 and 2 only.
- 1, 2, and 3 only.

Solution :

Suppose $A$ and $B$ are invertible $n\times n$ matrices. If $A$ and $B$ are similar, which of the following statements are true?

**1, 2, and 3 are all true.**

Moreover, we have \[B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P.\] But by the definition of being similar, this means that $A^{-1}$ and $B^{-1}$ are similar. So 3 is true. Since only the last choice offers 2 and 3 as true, we know that the answer is the last choice:**1, 2, and 3 are all true**.

Can you prove why 1 must be true?

- $A-2I$ and $B-2I$ are similar.
- $A$ and $B$ have the same trace.
- $A^{-1}$ and $B^{-1}$ are similar matrices.

- 1 only.
- 2 only.
- 3 only.
- 1 and 2 only.
- 1, 2, and 3 only.

Solution :

Recall that $A$ and $B$ are similar if $B=P^{-1}AP$ for some invertible matrix $P$.

Since $\text{tr}(AB)=\text{tr}(BA)$, we know that 2 is true due to: \[\text{tr}(B)=\text{tr}(P^{-1}AP)=\text{tr}(APP^{-1})=\text{tr}(A).\]

Moreover, we have \[B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P.\] But by the definition of being similar, this means that $A^{-1}$ and $B^{-1}$ are similar. So 3 is true. Since only the last choice offers 2 and 3 as true, we know that the answer is the last choice:

Can you prove why 1 must be true?

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After thinking about it, we have

\[A-2I=PBP^{-1}-2PIP^{-1}=P(B-2I)P^{-1}\]

so $A-2I$ and $B-2I$ are similar by the same matrix $P$.

Yup! That was exactly the way I thought about it too!

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