similar matrices and one-to-oneness of matrix transformation

Question

I need help with this proof

Suppose $A$ and $B$ are $n \times n$ matrices and that there is an invertible matrix $P$ such that $A=PBP^{-1}$. Show that if $x \mapsto Bx$ is one-to-one, then $x \mapsto Ax$ is also one-to-one.

Answer 1

Suppose $v_1$ $\not=$ v$_2$. Then since P$^{-1}$ is invertible, $w_1 = P^{-1}v_1 \not=P^{-1}v_2=w_2.$ Then since B is one-to-one $b_1 = Bw_1 \not= Bw_2 = b_2.$ Again since P is invertible, $a_1 = Pb_1\not=Pb_2=a_2.$ Thus A is one-to-one.

Answer 2

Suppose $$Ax=Ay$$ $$\Longrightarrow PBP^{-1}x=PBP^{-1}y$$ Premultiply both sides by $P^{-1}$, $$\Longrightarrow BP^{-1}x=BP^{-1}y$$ By injectivity of $B$, it follows that, $$\Longrightarrow P^{-1}x=P^{-1}y$$ Premultiply both sides by $P$, $$\Longrightarrow x=y$$

Hence, $A$ is injective.

Hope it helps:)

Answer 3

Suppose that $Av = Aw$ for some vectors $v$ and $w$. That is $$(PBP^{-1})v = (PBP^{-1})w$$ or (multiplying on the left with $P^{-1}$) $$B(P^{-1}v) = B(P^{-1}w)$$ but as we know that the function $x \mapsto Bx$ is one-to-one, then $P^{-1}v = P^{-1}w$ and then $v=w$, as we wanted to prove.