Linear map between the same dimensional spaces

Question

Let $V,W$ be a vector spaces of the same dimension $m$ and $f\colon V\to W$ be a linear map.

I know that for finite $m$, $f$ is injective $\Leftrightarrow$ $f$ is surjective $\Leftrightarrow$ $f$ is isomorphism.

Is it true for infinite $m$?

Answer 1

Well, no. Take $V = W = k^\mathbb{N}$, which we read as the $k$ sequences indexed by $n$ with $n$ ranging over all of $\mathbb{N}$. Then define the shift operator

$$ S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$$

The map is clearly injective, and equally clearly not surjective.

Answer 2

No. Let $V=l_2$ (this is the space of sequences $(x_1,x_2,\ldots)$ of complex numbers such that $\Sigma_i |x_i|^2$ is finite), $S:V\to V$ $S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)$. $S$ is injective, but not surjective. See this answer.

Answer 3

No.

$\varphi:\begin{array} \Bbb R \left[X\right] \to \Bbb R\left[X\right]\\P \mapsto P'\end{array}$

$\Bbb R \left[X\right]$ represents the set of polynomials with coefficients in $\Bbb R$. $P'$ is the formal derivative of $P$.

It's clearly surjective but not injective since all constant polynomials have $\Bbb 0$ as image.

Answer 4

As a counterpart to exitingcorpse's injective-not-surjective map, I'd also like to add $T(x_1,x_2,x_3\dots):=(x_2,x_3,\dots)$ as a surjective-not-injective map.

In fact, you can easily see that $S(T(x))=x$ for all $x$, showing that $ST$ is the identity map, but $TS$ is not the identity map! It's a useful example to have in your pocket...