### A continuation of the previous post

Let $V$ be an n-dimensional vector space, and let $T:V\to W$ be a surjective linear mapping. If $W$ is m-dimensional, then $T$ is a matrix of order $m\times n$.

Is it possible that $m>n$? Let us assume that it is. Then $W$ has a basis $\{b_1,b_2,\dots,b_m\}$, all of which are mapped to by $m$ distinct vectors in $V$. Moreover, these vectors have to be linearly independent, which is impossible as $V$ is n-dimensional. Hence $m\leq n$. To rephrase this, there cannot be a surjective linear mapping from a vector space of lower dimension to that with a higher dimension.

Choose a basis $\{v_1,v_2,\dots,v_n\}$ for $V$, and choose a corresponding representation of the transformation $T$. Note that a difference choice of basis would entail a different representation for the same transformation $T$. How do you visualise multiple matrices denoting the same transformation? Visualize $T$ geometrically. It maps the the same vectors (arrows) to the same vectors (arrows).

The kernel of $K$ will always be a vector subspace of $V$. However, it may or may not be generated by a subset of $\{v_1,v_2,\dots,v_n\}$. But then again,l you can always create a new basis in which a subset generates the kernel, and then formulate the accompanying representation of $T$. Say the new basis is $\{r_1,r_2,\dots,r_n\}$, where $\{r_1,r_2,\dots,r_k\}$ generate the kernel. Then $T$ is of the form $\begin{pmatrix}0&0&\dots &0&a_{1,k+1}&a_{1,k+2}&\dots& a_{1,n}\&0&\dots&0&a_{2,k+1}&a_{2,k+2}&\dots&a_{2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\&0&\dots &0&a_{m,k+1}&a_{m,k+2}&\dots& a_{m,n}\end{pmatrix}$.

That $T$ maps $\{r_{k+1},r_{k+2},\dots,r_{n}\}$ to linearly independent vectors can easily be verified. And as the mapping is onto, $n-k$ is the dimension of $W$.

If the mapping was not surjective, we would have $\dim (\text{Ker }T)+\dim W>n$. But then again, even if the mapping is not surjective, we have $\dim(\text{Ker }T)+\dim(\text{Im T})=n$

Now think of $T$ in terms of the original basis $\{v_1,v_2,\dots,v_n\}$. Note that if a set of vectors are linearly independent in one basis representation, they are linearly independent in EVERY basis representation. Moreover, the kernel of a linear mapping $T$ will consist of the same vectors, regardless of basis. Hence, if $a_{1g}v_1+a_{2g}+\dots+ a_{ng}=r_g$ for $g\leq k$, the vectors $\begin{pmatrix}a_{1g}&a_{2g}&\dots&a_{ng}\end{pmatrix}$ will generate the kernel. What I’m trying to say is that it does not matter whether a subset of the chosen basis generates the kernel or not. The dimensions of the kernel and that of the image will remain the same. It is just that $T$ is easier to visualize if a subset of the basis does indeed generate the kernel.

Just a reminder: the mapping being “onto” is imperative for the aforementioned formula to hold.

So how we have two formulae for the dimension of $V$.  In the first case we chose a subspace of any dimension, and then calculated the dimension of the subspace of $V^*$ which had the former subspace as kernel. In the second case, we first chose the linear mapping, and then calculated the dimensions of the kernel and image (yes dimension of image, which works even if the mapping is not surjective). You get the feeling that you’re pretty much doing the same thing using two different methods. I leave you to further explore whether you’re indeed doing the same thing or not.