Method of proving counting formula of subspace addition

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.

Theorem:

If $$V,W$$ are subspaces of a finite-dimensional vector space, then $$\dim(V+W)=\dim V+\dim W-\dim(V\cap W)$$.

Proof:

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.

• It's a mathematical change. It's not true a priori, that the set of scalars a and d are the same. – Teresa Lisbon Aug 15 '17 at 14:20
• This question means that you don't understand the proof! – Cauchy Aug 15 '17 at 14:35
• @астонвіллаолофмэллбэрг Ah of course... after some thinking I finally got it. Thanks! Should I remove this question, since the answer is so trivial? – Zhengqun Koo Aug 15 '17 at 15:37
• It's good that you thought yourself and got it. If you have got it, write an answer below, and call me, so that I can critique it. Then, you can accept the answer yourself, and close this question, rather than have it float around since it is resolved. – Teresa Lisbon Aug 15 '17 at 15:39
• You are welcome. When you can do so, accept your answer and close this question, since it is resolved. Also, if you like this site, do use it more often! – Teresa Lisbon Aug 16 '17 at 23:26

1 Answer

Because both $(w_1,\dots ,w_k)$ and $(u_1,\dots ,u_m)$ are bases of same subspace $V\cap W$, each basis must be linearly independent.

Hence there is a unique set of scalars $c,d$ for each basis to represent a linear combination of any vector in the subspace. Let this vector be $x$.

$$c_1w_1+\dots +c_kw_k=x=d_1u_1+\dots +d_mu_m$$

The set of scalars $d$ could be, but is not constrained to, the set of scalars $a$. Again using the unique representation of vectors,

$$d_1u_1+\dots +d_mu_m=-a_1u_1-\dots -a_mu_m-b_1v_1-\dots -b_jv_j$$

(The linear combination of vectors $u\in V\cap W$ and $v\in V$, is in subspace $V\cap W$, by analogy to the similar proof for vector $w\in W$ being in subspace $V\cap W$.)

• This answer is fine. – Teresa Lisbon Aug 16 '17 at 22:54