Let $T$ be a linear operator on an inner product space $V$, and suppose that $||T(x)|| = ||x||$ for all $x$. Prove $T$ is one-to-one.

Question

Let $T$ be a linear operator on an inner product space $V$, and suppose that $||T(x)|| = ||x||$ for all $x$. Prove that $T$ is one-to-one.

$\textbf{Solution:}$ Assume $T(x_1) = T(x_2).$ Using the linear property of $T$, $$||T(x_1 - x_2)|| = ||T(x_1)-T(x_2)||$$ $$=||0|| \hspace{2pt} \text{since } T(x_1) = T(x_2)$$ $$=0.$$ Where do I go from here? Can I please have help proving this?

Answer 1

The first part of your solution is correct. So taking that,

Assume $T(x_1) = T(x_2).$ Using the linear property of $T$, $$||T(x_1 - x_2)|| = ||T(x_1)-T(x_2)||$$ $$=||0|| \hspace{5pt} \text{since } T(x_1) = T(x_2)$$ $$=0.$$ This implies, $$||T(x_1 - x_2)|| = 0. \hspace{35pt} (1)$$ Also, since $||T(x)|| = ||x||$ for all $x\in V,$ $$||T(x_1 - x_2)|| = ||x_1 -x_2||. \hspace{35pt} (2)$$

From (1) and (2), $$||x_1 -x_2|| = 0. \hspace{35pt} (3)$$ By the property of the norm of a vector in the vector space, $||x|| = 0$ if and only if $x=0.$ Hence, (3) gives us, $x_1 - x_2 = 0$, and $x_1 = x_2.$ Therefore, the given linear operator $T$ is one-to-one, and we are done.

Answer 2

We have to prove that $T$ is nonsingular, that is to say, $T(x) = 0$ implies that $x = 0$.

Indeed, according to the given assumption, one has that \begin{align*} T(x) = 0 \Longrightarrow \|T(x)\| = \|x\| = 0 \Longrightarrow x = 0 \end{align*}

and we are done.

Hopefully this helps.

Answer 3

Since $\lVert T(x_1-x_2)\rVert=0$, $\lVert x_1-x_2\rVert=0$. But this means that $x_1=x_2$. So, this proves that $T(x_1)=T(x_2)\implies x_1=x_2$.