Matrix Similarity Proof (Show Matrix $B$ Nonsingular)

Question

Show that if A is similar to $B$ and $A$ is nonsingular, then $B$ must also be nonsingular and $A^{-1}$ and $B^{-1}$ are similar:

Work:

$A=S^{-1}BS$

$A^{-1}=(S^{-1}BS)^{-1}$

$A^{-1}=S^{-1}B^{-1}(S^{-1})^{-1}$

$A^{-1}=S^{-1}B^{-1}S$

I can prove that $A^{-1}$ and $B^{-1}$ are similar, but I am unsure of how to show $B$ that is nonsingular.

Answer 1

$$A=S^{-1}BS\iff SAS^{-1}=B$$

Now multiply both sides of the identity with $SA^{-1}S^{-1}$

$$SA^{-1}S^{-1}SAS^{-1}=I=\left(SA^{-1}S^{-1}\right)B$$ $$SAS^{-1}SA^{-1}S^{-1}=I=B\left(SA^{-1}S^{-1}\right)$$

And we have found $C=SA^{-1}S^{-1}$ such that $BC=CB=I$

Answer 2

$$ A=S^{-1}BS \iff B=SAS^{-1} $$ And the product of nonsingular matrices is nonsingular.

Answer 3

Overkill: $$ 0 \neq \det(A) = \det(S^{-1}BS) = \det(S^{-1}) \det(B) \det(S) $$ So $\det(B) \neq 0$.