Confirming surjectivity at a glance

Question

Learning how to tell if a function is surjective at a glance. There are a few linear transformations in quotes below that I tried to show (informally) are/n't surjective. Please, see if any of that makes sense.

$T: \mathbb R^2 \to \mathbb R^3 \text { given as } T \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 2x + 3y \\ x + y \\ 0 \end{pmatrix}$

$(a, b, 0)^T$ is an arbitrary vector in the range of $T$ and $(a, b, c)$ is an arbitrary vector in $\mathbb R^3$ where $c \in \mathbb R$ and so the range of $T$ has fewer vectors than its codomain. Thus the range and the codomain are not equal. $T$ is not surjective.

$T: P_3 \to P_2$ given by $T(p) = p'$ where $p'$ is the derivative of $p$

$p' = ax^2 + bx + c$ is in arbitrary vector in both the range and codomain of $T$ so $T$ is surjective

$T: \mathbb R^3 \to \mathbb R^2 \text { given as } T \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} y \\ z \\ \end{pmatrix}$

Arbitrary vector in the range is an ordered pair. Ordered pairs make up the codomain as well. So this transformation is surjective

$T: P_2 → P_3$ given by $T(ax^2 + bx + c) = ax + (b + c)$

The range of $T$ contains linear polynomials only and so it contains fewer elements than $P_3.$ Thus $T$ is not surjective.

Answer 1

For finite-dimensional vector spaces $V$ and $W$, a map $T: V \to W$ cannot be surjective if $\dim V < \dim W$. This is a consequence of the rank-nullity theorem: this theorem states that if $T: V \to W$ is a linear transformation of finite-dimensional vector spaces, then $\dim V = \dim \text{im } T + \dim \ker T$. $T$ is surjective if and only if $\dim \text{im } T = \dim W$, and $\dim \ker T$ is nonnegative.