[Math] Showing that the spans of two sets of vectors are equal

Question

Vector $v_k$ is a linear combination of vectors $v_1,v_2,\dots,v_{k-1}$. How do I prove that the span of set of vectors $v_1,v_2,\dots,v_{k-1}$ is equal to the span of set of vectors $v_1,v_2,\dots,v_k$ by containment?

Answer 1

This will mainly be an exercise in notation, but here goes...

Let $V=span \{ v_1, ~v_2 ,\dots, v_{k-1} \}$ and let $W=span \{v_1, v_2, \dots, v_k\}$. Seek to show that $V=W$, so we need to show that $V \subseteq W$ and $W \subseteq V$.

Claim $V \subseteq W$:

• Let $v \in V$. Then $v= a_1 v_1 + a_2 v_2 + \cdots + a_{k-1} v_{k-1} $. Then $v= a_1 v_1 + a_2 v_2 + \cdots + a_{k-1} v_{k-1} + 0 v_k$, so $v$ is in $W$.

Claim $W \subseteq V$:

• Since $v_k \in V = span \{v_1, v_2, \dots, v_{k-1} \}$, we have that $v_k = b_1 v_1 + b_2 v_2 + \cdots + b_{k-1} v_{k-1}$. Let $w \in W$ be given. Then \begin{align*} w&= c_1 v_1 + c_2 v_2 + \cdots + c_{k-1} v_{k-1} + c_k v_k \\ &= c_1 v_1 + c_2 v_2 + \cdots + c_{k-1} v_{k-1} + c_k (b_1 v_1 + b_2 v_2 + \cdots + b_{k-1} v_{k-1}) \\ &= (c_1 + c_k b_1) v_1 + (c_2 + c_k b_2) v_2 + \cdots + (c_{k-1} + c_k b_{k-1}) v_{k-1} \end{align*} So $w \in span \{v_1, v_2, \dots, v_{k-1} \} = V$.

Edit

Note that this proof is relatively atrocious notation--waay too many characters and subscripts floating around, leading to a very bloated proof. A much more elegant proof can be given using summation notation ($\sum$) or with matrices. However, I am inferring that you are in an introductory linear algebra course, so I went with the "brute force" approach to highlight what is really going on, without getting to lost using fancy notation.