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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.

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  • $\begingroup$ It's a mathematical change. It's not true a priori, that the set of scalars a and d are the same. $\endgroup$ – астон вілла олоф мэллбэрг Aug 15 '17 at 14:20
  • $\begingroup$ This question means that you don't understand the proof! $\endgroup$ – Cauchy Aug 15 '17 at 14:35
  • $\begingroup$ @астонвіллаолофмэллбэрг Ah of course... after some thinking I finally got it. Thanks! Should I remove this question, since the answer is so trivial? $\endgroup$ – Zhengqun Koo Aug 15 '17 at 15:37
  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Aug 15 '17 at 15:39
  • $\begingroup$ 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! $\endgroup$ – астон вілла олоф мэллбэрг Aug 16 '17 at 23:26
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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 $V\cap W$.)

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