Confusion about span in Linear Algebra

Sep 2019
Hi there, this is my first post here. I am currently studying some control theory related stuff and it's basically linear algebra. However, I have some problems. I have given this exercise on which I already got an answer in Mathematics Stackoverflow:

The controllability matrix $P=\left[\begin{array}{lll}{A^{2} B} & {A B} & {B}\end{array}\right]$ of a state-space model $(A,B,C,D)$ has the
following SVD. Determine the kernel and image space of $P$. $$
P=\left[\begin{array}{lll}{u_{1}} & {u_{2}} & {u_{3}}\end{array}\right]\left[\begin{array}{ccc}{\sigma_{1}} & {0} & {0} \\ {0} & {\sigma_{2}} & {0} \\ {0} & {0} & {\sigma_{3}}\end{array}\right]\left[\begin{array}{c}{v_{1}^{T}} \\ {v_{2}^{T}} \\ {v_{3}^{T}}\end{array}\right]
the vectors $u_1,\dots,u_r$ will form an orthonormal basis for the image of $P$ and the vectors $v_{r+1},\dots,v_{n}$ will form a basis for the kernel of $P$.
However, when checking other sources I found something which does in my eyes not fit the answer above:
M=\left[\begin{array}{cc}{U_{1}} & {U_{2}}\end{array}\right]\left[\begin{array}{cc}{\Sigma_{1}} & {0} \\ {0} & {0}\end{array}\right]\left[\begin{array}{c}{V_{1}^{\prime}} \\ {V_{2}^{\prime}}\end{array}\right]
\text { Nullspace: } \mathcal{N}=\operatorname{span}\left[V_{2}\right]

If I got this right, the span is the spanning set of a matrix. I see how I get it despite a lack of intuition about it. What I don't get is why they are writing those indices. Isn't $V_2$ a vector? What's the span of a vector and why the second one? The same for $U$. The answer I received and this definition are somehow similar but not identical. The answer mentions a set of vectors, whilst this definition is about the span of a (single) vector. I am a little bit confused.

Thank you guys. I am looking forward to be part of this community. :)

Regards, Bob
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