Describe the kernel, and determine whether the given vector space linear transformation is invertible: Let $F$ be the vector space of all functions mapping $\Bbb R$ into $\Bbb R$
$T:F\rightarrow\Bbb R$ defined by $T(f)=f(-4)$

$ker(T)=$ {$f\in F|f(-4)=0$}, by definition of kernel.
To prove that $T$ is invertible, we have to show that $T$ is one-to-one and onto. However, I never really understand how to prove onto (although I did understand the concept). So, I didn't know how to continue from here.
 A: $T$ is not invertible. If it were one could reconstruct a function from its value at a single point, $-4$.
A: To prove that a given function is onto, you need to prove that every element of the codomain is the output of the function for some input, so in this case that means you're proving that $\forall y \in \mathbb{R}\ \exists f \in F : f(-4) = y$. So if you were given a random real number, could you construct (or at least prove the existence of) a function in F that takes that real number as its value at -4?
On the other side of the fence, you also need to prove that $T$ is one-to-one, which means that each valid output in $\mathbb{R}$ comes from only one possible input in $F$. If there are two functions that give the same output, then you haven't got a one-to-one transformation. (Does the kernel contain more than one element?)
A: To prove that $T$ is onto, let $a\in\Bbb R$ and let $f:\Bbb R\to\Bbb R$ be defined by $f(x)=a$. Then $T(f)=f(-4)=a$. Hence $T$ is onto.
To see that $T$ is not one-to-one, let $f,g:\Bbb R\to\Bbb R$ be defined by
\begin{align*}
f(x) &= 0 &
g(x) &= 4+x
\end{align*}
Then $f\neq g$ since $f(0)=0\neq 4=g(0)$. However,
$$
T(f)=f(-4)=0=4-4=g(-4)=T(g)
$$
Hence $T$ is not one-to-one. Consequently, $T$ is not invertible.
