Let $T$ be a linear transformation $T:V\to V$, where $V$ is an infinite dimensional vector space. How can we construct examples such as

$1.$ T is one to one but not onto

$2.$ T is onto but not one to one

$3.$ T is bijective but not invertible

$4.$ T is neither one to one and onto

I feel that for ($4$), the following example is valid.

$T:l^p\to l^p$ given by $T((\xi_1,\xi_2,....))=((0,\xi_1,\xi_2,...))$ but for others I am a bit confused.

  • $\begingroup$ Integration is linear but not one to one on the space of differentiable real values functions. $\endgroup$ – Wintermute Jan 26 '15 at 13:46
  • 1
    $\begingroup$ If I am not mistaken, the map $T$ you defined is an example for (1) and not for (4). It becomes an example for (4) if you map $(\xi_1,\xi_2,\dots)$ to $(0,\xi_2,\xi_3,\dots)$ instead. Note also that (3) is not satisfied by any $T$ (every bijective $T$ is invertible). $\endgroup$ – Matthias Klupsch Jan 26 '15 at 13:53

1) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,x_3,\ldots)$ (which is your idea for (4)). Then $T$ is one to one, because if $T(x)=T(y)$, then $(0,x_1,x_2,x_3,\ldots)=(0,y_1,y_2,y_3\ldots)$ and hence $x_i=y_i$ for each $i\in\mathbb N$. But $T$ is not onto, since there is no sequence $x\in l^p$ such that $T(x)=(1,0,0,0,0,\ldots)$.

2) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(x_2,x_3,\ldots)$. Then $T$ is onto, since if $y=(y_1,y_2,y_3,\ldots)\in l^p$ is given, then choose $x=(0,y_1,y_2,y_3,\ldots)$ to obtain $T(x)=y$. But $T$ is not one to one, since $T((0,0,0,0,\ldots))=T((1,0,0,0,0,0,\ldots))$.

3) There is no such $T$, since if $T$ is bijective, then it is invertible, and since $T$ is linear, its inverse must be linear as well. For the linearity of the inverse, see, for instance, Is the inverse of a linear transformation linear as well?

4) Take your favorite infinite-dimensional vector space $V$ and consider $T(x):=0$ for each $x\in V$. Then $T$ is neither one to one nor onto, which should be obvious.


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