Proof: 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

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:

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

Work:

$A=S^{-1}BS$

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

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

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

Also, my testbook says that if $B$ is similar to $A$ then exists a $S$ such that $B=S^{-1}AS$. The "placement" of $B$ and $A$ cannot be swapped right? For example if $B$ is similar to $A$, I can't say that $A=S^{-1}BS$?

Answer 1

Prove $B$ Nonsingular

A matrix $B$ is nonsingular/invertible if exists a matrix $A$ such that $AB=BA=I$

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

• $$SAS^{-1}=B$$

  $$SAS^{-1}S=BS$$

  $$SAA^{-1}=BSA^{-1}$$ $$SS^{-1}=BSA^{-1}S^{-1}$$ $$I =B(SA^{-1}S^{-1})$$

• $$SAS^{-1}=B$$ $$S^{-1}SAS^{-1}=S^{-1}B$$

$$A^{-1}AS^{-1}=A^{-1}S^{-1}B$$ $$SS^{-1}=SA^{-1}S^{-1}B$$ $$I = (SA^{-1}S^{-1})B$$

Answer 2

A is similar to B if and only if A=R^-1BR for a regular matrix R. From there RAR^-1=B. The product of multiplication of regular matrices is also a regular matrix.

Answer 3

To show that $ B $ has an inverse, you need to find a matrix, call it $ X $, such that $ B X = I $, the identity.

Now, if $ B $ has an inverse, then $ B^{-1} = ( S A S^{-1} )^{-1} = S A^{-1} S^{-1} $.

Why? If $ A = S B S^{-1} $ then you can multiply both sides from the left by $ S^{-1} $ and both sides on the right by .... and get that $ B

So you take $ X = S A^{-1} S^{-1} $ and you check that $ B X = I $.