Linear transformation: $T(T(x)) = T(x) + x$ for any $x$ in $\mathbb{R}^3$. Prove that $T$ is injective and surjective.

Question

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$T(T(x)) = T(x) + x$ for any $x$ in $\mathbb{R}^3$. Prove that $T$ is injective and surjective.

Answer 1

If

$T(T(x)) = T(x) + x = x + T(x), \tag 1$

and

$Tx = 0, \tag 2$

then

$x = x + 0 = x + T(x) = T(T(x)) = T(0) = 0; \tag 3$

that is

$\ker T = \{0\}, \tag 4$

and so $T$ is injective. We further note that for any $y \in \Bbb R^3$, (1) yields

$y = T(T(y)) - T(y) = T(T(y)) - y), \tag 5$

which shows that every $y$ is the image under $T$ of $T(y) - y$, so $T$ is surjective as well, which may also be concluded directly from (4), since $T$ operates on the finite-dimensional vector space $\Bbb R^3$; in finite dimensions we have "injective $\Longleftrightarrow$ surjective", as is well-known.

N.B: It is perhaps worth observing that there is nothing intrinsically $3$-dimensional about this result; it apparently holds for $T:V \to V$, where $V$ is any vector space over any field $\Bbb F$, whether or not $\dim V < \infty$; of course, in that case we can't us our little "injective $\Longleftrightarrow$ surjective" trick; we may then instead rely on (5). End of Note.

Answer 2

Given any $v,w\in\mathbb R^3$, let $x=T(v), y=T(w)$ then

\begin{array}{} T(x)=x+v\\ T(y)=y+w \end{array}

Now assume that $T(v)=T(w),$ then $x=y,$ so

\begin{align} T(x)=x+v=y+w=T(y)\implies v=w. \end{align}

So $T$ is injective, and since $T$ linear and $\dim\mathbb R^3\lt\infty$, one-one implies onto.