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# 17 When are two spans the same ?
Before we start, let us have some terminologies. Let us consider a set of vectors (or objects) called $S$. Suppose we have a list of vectors (or objects) $v_{1},\ldots,v_{n}$ such that $$
\operatorname{span} (v_{1},\ldots,v_{n}) = S,
$$then we say $\{v_{1},\ldots,v_{n}\}$ is a **spanning set** or **generating set** for $S$. We sometimes also write and say the list of vectors (or objects) $v_{1},\ldots,v_{n}$ **spans** or **generates** $S$, if $\operatorname{span}(v_{1},\ldots,v_{n}) = S$, forgoing the curly set braces.
Of course, by this definition, a spanning set (or generating set) for $\operatorname{span}(v_{1},\ldots,v_{n})$ is $\{v_{1},\ldots,v_{n}\}$.
---
Now as we have seen quite often enough, we have span of a set equals to the span of another set. That is to say, **a set of vectors may have different generating sets**. For example, consider the set of vectors in the plane that are along the line $L: y = 2x$. This line $L$ can be written as $$
L=\operatorname{span} (\begin{bmatrix}
1\\2
\end{bmatrix}) = \operatorname{span} (\begin{bmatrix}
1 \\
2
\end{bmatrix},\begin{bmatrix}
5\\10
\end{bmatrix}) = \operatorname{span} (\begin{bmatrix}
-2\\-4
\end{bmatrix})
$$So we see that the same $L$ may have many different spanning sets! That is, many different collection of vectors span to the same line $L$. How do we know that all these spans are really the same set?
Well often enough in simple scenarios we can appeal to **geometry** and rely on our geometric intuition. However **sometimes our geometric intuition fails or even betrays us**. In that case, we implore **algebra** to help us.
So let us tackle this question, given two lists of vectors $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$, how to check whether $\operatorname{span}(v_{1},\ldots,v_{n)}$ is equal to $\operatorname{span}(w_{1},\ldots w_{m})$ or not?
To answer this we have the following proposition, a characterization of this span equality.
> **Proposition. A characterization for span equality.**
> Given two lists of vectors (or objects) $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$. We have $$
\operatorname{span} (v_{1},\ldots,v_{n}) = \operatorname{span} (w_{1},\ldots,w_{m})
$$ if and only if for each $v_{i}$ in the list, we have $v_{i}\in \operatorname{span}(w_{1},\ldots,w_{m})$; and for each $w_{j}$ in the list, we have $w_{j} \in \operatorname{span}(v_{1},\ldots,v_{n})$.
We shall prove this proposition, and note that this is an **if and only if statement**, hence a **characterization**. So this is a characterization of when two spans are equal.
To prove such if and only if statement we need to show **both directions of the implication**. Let us do it.
$\blacktriangleright$ Proof of proposition.
In this proof, we have two lists of vectors (or objects) $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$.
$(\implies)$ Assume $\operatorname{span}(v_{1},\ldots v_{n}) = \operatorname{span}(w_{1},\ldots,w_{m})$. We wish to show each vector $v_{i}$ in the list $v_{1},\ldots,v_{n}$ can be found in $\operatorname{span}(w_{1},\ldots,w_{m})$; and each vector $w_{j}$ in the list $w_{1},\ldots,w_{m}$ can be found in $\operatorname{span}(v_{1},\ldots,v_{n})$.
Indeed, take any $v_{i}$ in the list $v_{1},\ldots,v_{n}$, observe that $$
v_{i} = 0v_{1}+\cdots +1v_{i}+\cdots +0v_{n},
$$so $v_{i}$ is a linear combination of $v_{1},\ldots,v_{n}$. In other words, $v_{i} \in \operatorname{span}(v_{1},\ldots,v_{n})$. But $\operatorname{span}(v_{1},\ldots v_{n}) = \operatorname{span}(w_{1},\ldots,w_{m})$, so $$
v_{i}\in \operatorname{span} (w_{1},\ldots,w_{m}).
$$
Symmetrically, if we take any $w_{j}$ in the list $w_{1},\ldots,w_{m}$, then as $w_{j}$ is a linear combination of $w_{1},\ldots,w_{m}$, we have $$
w_{j} \in \operatorname{span} (w_{1},\ldots,w_{m})= \operatorname{span} (v_{1},\ldots,v_{n}).
$$
Hence we have established the $(\implies)$ direction of this proof.
$(\impliedby)$ Now assume each vector $v_{i}$ in the list $v_{1},\ldots,v_{n}$ is in $\operatorname{span}(w_{1},\ldots,w_{m})$, and each vector $w_{j}$ in the list $w_{1},\ldots,w_{m}$ is in $\operatorname{span}(v_{1},\ldots,v_{n})$. We wish to show $\operatorname{span}(v_{1},\ldots,v_{n}) = \operatorname{span}(w_{1},\ldots w_{m})$.
Since we want to establish two sets are the same, we should show **mutual containment of those sets**.
Let us take $x \in \operatorname{span}(v_{1},\ldots,v_{n})$. This means $x$ is a linear combination of $v_{1},\ldots,v_{n}$. In other words, $$
x = c_{1}v_{1} + \cdots +c_{n}v_{n}\quad\quad(\ddagger)
$$Now in our assumption, each $v_{i}$ is in $\operatorname{span}(w_{1},\ldots,w_{m})$. This means each $v_{i}$ is a linear combination in $w_{1},\ldots,w_{m}$. Let us write all of these out: $$
\begin{align*}
v_{1} & = \lambda_{11} w_{1} +\cdots + \lambda_{1m} w_{m} \\
v_{1} & = \lambda_{21} w_{1} +\cdots + \lambda_{2m} w_{m} \\
& \vdots \\
v_{n} & = \lambda_{n1} w_{1} +\cdots + \lambda_{nm} w_{m}
\end{align*}
$$for some coefficients $\lambda_{ij}$. Now if we plug into the expression $(\ddagger)$ of $x$ above, we get $$
\begin{align*}
x & = c_{1}(\lambda_{11} w_{1} +\cdots + \lambda_{1m} w_{m}) + \cdots + c_{n}(\lambda_{n1} w_{1} +\cdots + \lambda_{nm} w_{m}) \\
& = (c_{1}\lambda_{11}+c_{2}\lambda_{21}+\cdots +c_{n}\lambda_{n1})w_{1} + \cdots + (c_{1}\lambda_{1m}+\cdots+c_{n}\lambda_{nm}) w_{m} \\
& =\tilde c_{1}w_{1}+\cdots +\tilde c_{m}w_{m},
\end{align*}
$$for some scalars $\tilde c_{1},\ldots,\tilde c_{m}$. So we see that $x$ is a linear combination of $w_{1},\ldots,w_{m}$. Hence $$
x \in \operatorname{span} (w_{1},\ldots,w_{m}),
$$and we have $\operatorname{span}(v_{1},\ldots,v_{n}) \subset \operatorname{span}(w_{1},\ldots w_{m})$.
By a symmetric argument, we can also show the other containment $\operatorname{span}(v_{1},\ldots,v_{n}) \supset \operatorname{span}(w_{1},\ldots w_{m})$. As an exercise for yourself, try writing it out.
Hence indeed $\operatorname{span}(v_{1},\ldots,v_{n}) = \operatorname{span}(w_{1},\ldots w_{m})$, we have as desired. $\blacksquare$
---
So how do we use it? **Exactly as the proposition prescribed!** If you want to check if the span of two sets are the same, you check each element in the spanning set of one is in the other span, and vice versa.
**Example.**
Consider the two sets $\operatorname{span}(\begin{bmatrix}1\\2 \end{bmatrix} , \begin{bmatrix}2\\3 \end{bmatrix})$ and $\operatorname{span}(\begin{bmatrix}7\\1 \end{bmatrix} , \begin{bmatrix}7\\2 \end{bmatrix})$. Are they the same?
By the proposition above, these two spans are the same provided that we can show the following: $$
\begin{align*}
\begin{bmatrix}1\\2 \end{bmatrix} & \in \operatorname{span}(\begin{bmatrix}7\\1 \end{bmatrix} , \begin{bmatrix}7\\2 \end{bmatrix}) \\
\begin{bmatrix}2\\3 \end{bmatrix} & \in \operatorname{span}(\begin{bmatrix}7\\1 \end{bmatrix} , \begin{bmatrix}7\\2 \end{bmatrix}) \\
\begin{bmatrix}7\\1 \end{bmatrix} & \in\operatorname{span}(\begin{bmatrix}1\\2 \end{bmatrix} , \begin{bmatrix}2\\3 \end{bmatrix}) \\
\begin{bmatrix}7\\2 \end{bmatrix} & \in\operatorname{span}(\begin{bmatrix}1\\2 \end{bmatrix} , \begin{bmatrix}2\\3 \end{bmatrix})
\end{align*}
$$Now recall computationally, checking whether a vector is in the span of the others amounts to checking whether some linear system is consistent, which amounts to setting up an augmented matrix and row reducing to an echelon form that give a consistent result. In this case, we would get $$
\begin{align*}
\begin{bmatrix}
7 & 7 & \vdots & 1 \\
1 & 2 & \vdots & 2
\end{bmatrix} \stackrel{\text{row}}\sim \text{a conssistent echelon form} \\
\begin{bmatrix}
7 & 7 & \vdots & 2 \\
1 & 2 & \vdots & 3
\end{bmatrix} \stackrel{\text{row}}\sim \text{a conssistent echelon form} \\
\begin{bmatrix}
1 & 2 & \vdots & 7 \\
2 & 3 & \vdots & 1
\end{bmatrix} \stackrel{\text{row}}\sim \text{a conssistent echelon form} \\
\begin{bmatrix}
1 & 2 & \vdots & 7 \\
2 & 3 & \vdots & 2
\end{bmatrix} \stackrel{\text{row}}\sim \text{a conssistent echelon form}
\end{align*}
$$So yes, indeed, $\operatorname{span}(\begin{bmatrix}1\\2 \end{bmatrix} , \begin{bmatrix}2\\3 \end{bmatrix}) = \operatorname{span}(\begin{bmatrix}7\\1 \end{bmatrix} , \begin{bmatrix}7\\2 \end{bmatrix})$. $\blacklozenge$