I'm following Linear Algebra Done Right by Axler. I understand the proof, but I wish understand the purpose of a step in the proof.
Why did the author change scalars of $(u_1,\dots ,u_m)$ from $a$ to $d$, where $a$ and $d$ are arbitrary scalars? Is this a stylistic change, for easier reading? Or is there some mathematical meaning behind the change of scalars?
The portion of interest is near the last parts of the proof.
If $V,W$ are subspaces of a finite-dimensional vector space, then $\dim(V+W)=\dim V+\dim W-\dim(V\cap W)$.
Let $V\cap W$ have basis $(u_1,\dots ,u_m)$, with $\dim(V\cap W)=m$.
A basis is linearly independent by definition. Therefore, we can extend the basis of $V\cap W$ to a basis $(u_1,\dots ,u_m,v_1\dots ,v_j)$ of $V$, with $\dim V=m+j$.
We also extend the basis of $V\cap W$ to a basis $(u_1,\dots ,u_m,w_1\dots ,w_k)$ of $W$, with $\dim W=m+k$.
Let $union$ be the list of vectors $(u_1,\dots ,u_m,v_1,\dots ,v_j,w_1,\dots ,w_k)$. We will show that $union$ is a basis of $V+W$. Doing so will complete the proof because:
$$\dim(V+W)=m+j+k=(m+j)+(m+k)-m=\dim V+\dim W-\dim(V\cap W)$$
$span(union)$ contains $V$ and $W$, hence $span(union)$ contains $V+W$. We need to show that $union$ is linearly independent, in order to show that $union$ is a basis of $V+W$.
To show this, suppose all $a,b,c$ are scalars. We need to prove that all the $a=b=c=0$.
$$a_1u_1+\dots +a_mu_m+b_1v_1+\dots +b_jv_j+c_1w_1+\dots +c_kw_k=0$$ $$c_1w_1+\dots +c_kw_k=-a_1u_1-\dots -a_mu_m-b_1v_1-\dots -b_jv_j$$
Hence, $c_1w_1+\dots +c_kw_k\in V$. All $w\in W$, so $c_1w_1+\dots +c_kw_k\in V\cap W$.
Because $(u_1,\dots ,u_m)$ is a basis of $V\cap W$, for some scalars d,
$$c_1w_1+\dots +c_kw_k=d_1u_1+\dots +d_mu_m$$
But $(u_1,\dots ,u_m,w_1,\dots ,w_k)$ is linearly independent, so the last equation implies all $c=d=0$. Our original equation involving $a,b,c$ becomes:
$$a_1u_1+\dots +a_mu_m+b_1v_1+\dots +b_jv_j=0$$
Because $(u_1,\dots ,u_m,v_1\dots ,v_j)$ is linearly independent, all $a=b=0$. Hence, $a=b=c=0$, as desired.