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Let T be a linear operator on an inner product space $V$. Prove that $\lVert T(x)\rVert = \lVert x\rVert$ for all $x$ in $V$ iff $\langle T(x), T(y)\rangle = \langle x,y\rangle$ for all $x,y$ in $V$.

I tried using $\lVert T(x-y)\rVert = \lVert x-y\rVert$ and I got stuck when I got to $\langle T(x),T(y)\rangle + \langle T(y),T(x)\rangle = \langle x,y\rangle + \langle y,x\rangle $

Please give me some idea how to proceed from there or is there any other way to prove this? Thank you.

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If you are working over $\mathbb{R}$, then you are done, since $\langle v,w\rangle = \langle w,v\rangle$.

If you are working over $\mathbb{C}$, then you now have that the real part of $\langle T(x),T(y)\rangle$ is equal to the real part of $\langle x,y\rangle$.

Now look at $\lVert T(x-iy)\rVert$ versus $\lVert x-iy\rVert$.

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  • $\begingroup$ when working over C, why can we just say the real part is equal? we can just ignore the complex part? I am very confused. $\endgroup$ – Gigi Nov 5 '11 at 5:58
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    $\begingroup$ @Gigi: We are not ignoring anything. For a complex inner product, $$\langle x,y\rangle = \overline{\langle y,x\rangle},$$ where the line signifies the complex conjugate. And if you have a complex number $z$, then $z+\overline{z}$ is twice the real part of $z$; apply that to $\langle x,y\rangle + \langle y,x\rangle$ and to $\langle T(x),T(y)\rangle + \langle T(y),T(x)\rangle$; and for the next part, you'll need to remember that $z-\overline{z}$ is $2i$ times the imaginary part, and the complex inner products are sesquilinear, not bilinear (scalars come out from the second entry conjugated). $\endgroup$ – Arturo Magidin Nov 5 '11 at 6:02
  • $\begingroup$ what about the converse? $\endgroup$ – Hassan Muhammad Nov 5 '11 at 14:10
  • $\begingroup$ @Hassan: The converse is immediate: if $\langle T(x),T(y)\rangle = \langle x,y\rangle$ for all $x$ and $y$, then take $x=y$ to get $\lVert T(x)\rVert^2 = \lVert x\rVert^2$. $\endgroup$ – Arturo Magidin Nov 5 '11 at 20:43

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