Defined that $\langle x,y \rangle=\langle Ax,Ay\rangle$

Prove that above given define a inner product iff A is invertible.

I know that $\langle Ax,Ay\rangle=xA^{t}Ay$ and for inner product we have to show that $\langle Ax,Ay \rangle > 0$ but further i don't know how to proceed

Please help!


You should use different symbol for the inner product on the left. I will use a prime. $\langle Ax, Ax \rangle=0$ iff $\|Ax\|=0$ iff $Ax=0$. So the condition that $\langle x, x \rangle '=0$ implies $x=0$ is satisfied only when $A$ satisfies the property that $Ax=0$ implies $x=0$ which means $A$ is invertible. The other properties of inner product are straightforward.

  • $\begingroup$ sir,if $Ax=0$ implies $x=0$ then how can we say A is invertible? $\endgroup$ – gaurav saini May 10 at 9:04
  • $\begingroup$ If $Ax=0$ implies $x=0$ then $A$ is one-to-one, hence invertible. I am assuming that you are working in a finite dimensional vector space. $\endgroup$ – Kavi Rama Murthy May 10 at 9:12
  • $\begingroup$ sir, $Ax=0$ implies $x=0$ so the kernal of $A={0}$ which implies A is one -one and in finite dimension space if A is one -one then it becomes invertivle.. i got it Thnakyou sir. $\endgroup$ – gaurav saini May 10 at 9:22

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