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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.

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

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  • $\begingroup$ 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) $\endgroup$ – mbus2sus Aug 13 at 3:27
  • $\begingroup$ 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? $\endgroup$ – Axion004 Aug 13 at 3:28
  • $\begingroup$ Yes, it is correct. $\endgroup$ – Martund Aug 13 at 3:30
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    $\begingroup$ @Axion004, if $f$ is a 1-1 function, then $f(x)=f(y)\Longrightarrow x=y$ $\endgroup$ – Martund Aug 13 at 3:31
  • $\begingroup$ Thank you for your help! Much appreciated $\endgroup$ – mbus2sus Aug 13 at 3:31
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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.

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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.

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