# similar matrices and one-to-oneness of matrix transformation

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.

• $A^{-1}=PB^{-1}P$ – J. W. Tanner Aug 13 at 2:47

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:)

• This was how I completed the proof -- does this line of reasoning hold? Suppose x --> Bx is one-to-one 1. The equation Bx = 0 has only the trivial solution 2. Bx = 0 has only the trivial solution if and only if the columns of B are linearly independent 3. If follows that B is invertible / det(B) is nonzero Since A and B are similar matrices... A = PBP-1 det(A) = det(P) * det(B) * det(P-1) Since det(P) * det(P-1) = det(PP-1) = det I = 1, it follows that det(A) = det(B) 4. det(A) is nonzero 5. A is invertible 6. x --> Ax is also one-to-one (by the Invertible Matrix Theorem) – mbus2sus Aug 13 at 3:27
• Why does the injectivity of $B$ allow you to form $P^{-1}x=P^{-1}y$? Did you multiply by $B^{-1}$? Can you explain this step? – Axion004 Aug 13 at 3:28
• Yes, it is correct. – Martund Aug 13 at 3:30
• @Axion004, if $f$ is a 1-1 function, then $f(x)=f(y)\Longrightarrow x=y$ – Martund Aug 13 at 3:31
• Thank you for your help! Much appreciated – mbus2sus Aug 13 at 3:31

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.

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.