I'm reviewing a little bit of linear algebra, and I got stuck a little bit on this trivial thing: for finite-dimensional vector spaces we know that injectivity follows from surjectivity and vice-versa.
So basically, if I consider a map between vector spaces of different dimensions, it's hopeless to find a linear bijective map. But if I have
$$T:\mathbb{R} \rightarrow \mathbb{R}^5, \;\text{ described by }\; T = \begin{bmatrix}1\\-1\\1\\0\\0\end{bmatrix},$$
then its kernel would be $\text{Ker }T = \{x \in \mathbb{R} \mid Tx = 0\}.$
Then, it would follow that $\begin{bmatrix}1\\-1\\1\\0\\0\end{bmatrix}*x = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \, \, \, \Longrightarrow \, \, \, \begin{cases}x=0\\ -x=0\\x=0 \\0=0 \\0=0 \end{cases}$.
Basically this would end up implying that the Kernel is the null vector only? What's the point here?
I know this is trivial but it has been a long time since my basic linear algebra course.
Best Answer
Indeed, the kernel is only the zero vector - you are correct. One thing you could learn from this is that the kernel of a linear map must be a subspace of the domain, in this case $\mathbb{R}$. There are only two such subspaces, namely $\{0\}$ and $\mathbb{R}$ itself. So any linear map from $\mathbb{R}$ into anything must either map everything to zero or only $0$ to zero.