Why Equality Requires Two Directions
Two different lists of vectors can generate the same subspace. One list may contain vectors that are absent from the other, or may have extra vectors that add no new combinations. To prove that the generated subspaces are equal, it is not enough to show that one list can be built from the other: that proves only that one span is contained in the other. The complete argument checks both directions, using explicit linear combinations.
We work in a vector space \(V\) over a field \(\mathbb{F}\). The examples use \(\mathbb{R}^n\) with real scalars. A list matters here because it records which vectors are being used as generators; a list may contain repeated or redundant vectors.
The span is a subspace: the all-zero coefficients produce \(\mathbf{0}\); adding two combinations adds their corresponding coefficients; and multiplying a combination by a scalar multiplies each coefficient by that scalar. In particular, saying that \(\mathbf{x}\) belongs to the span of a list means precisely that there are coefficients expressing \(\mathbf{x}\) as a linear combination of that list.
One List of Generators Gives One Inclusion
Suppose every vector in a list \(\mathbf{u}_1,\dots,\mathbf{u}_p\) belongs to the span of \(\mathbf{v}_1,\dots,\mathbf{v}_q\). Any vector in the first span is a linear combination of the \(\mathbf{u}_i\). Replacing each \(\mathbf{u}_i\) by its expression using the \(\mathbf{v}_j\) shows that the original vector is also a linear combination of the \(\mathbf{v}_j\). This proves the inclusion of the first span in the second.
Proof. Let \(\mathbf{x}\in\operatorname{span}\{\mathbf{u}_1,\dots,\mathbf{u}_p\}\). By the definition of span, there are scalars \(a_1,\dots,a_p\in\mathbb{F}\) such that \(\mathbf{x}=\sum_{i=1}^{p}a_i\mathbf{u}_i\). For each \(i\), the hypothesis gives scalars \(b_{i1},\dots,b_{iq}\in\mathbb{F}\) such that \(\mathbf{u}_i=\sum_{j=1}^{q}b_{ij}\mathbf{v}_j\). Substituting these expressions yields \(\mathbf{x}=\sum_{i=1}^{p}a_i\sum_{j=1}^{q}b_{ij}\mathbf{v}_j =\sum_{j=1}^{q}\left(\sum_{i=1}^{p}a_i b_{ij}\right)\mathbf{v}_j\). The final expression is a linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_q\), so \(\mathbf{x}\) belongs to their span. Thus the stated inclusion holds. \(\square\)
The lemma explains the generator-by-generator method: to establish an inclusion, it is enough to check the generators of the span on the left. To establish equality, that check must be made in both directions.
Proof. First suppose the spans are equal. Each \(\mathbf{u}_i\) belongs to its own span, so equality of the spans places \(\mathbf{u}_i\) in the span of the \(\mathbf{v}_j\). Likewise, each \(\mathbf{v}_j\) belongs to the span of the \(\mathbf{u}_i\). This proves the required generator memberships. Conversely, suppose both sets of memberships hold. By the lemma, the first set of memberships gives \(\operatorname{span}\{\mathbf{u}_1,\dots,\mathbf{u}_p\}\subseteq\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_q\}\). Applying the lemma with the lists exchanged gives \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_q\}\subseteq\operatorname{span}\{\mathbf{u}_1,\dots,\mathbf{u}_p\}\). Two sets contained in each other are equal, proving the reverse implication. The argument also covers an empty list, using the convention that its span is \(\{\mathbf{0}\}\). \(\square\)
A Practical Proof Plan
For a span equality, keep the two directions separate. Write out the expression for each generator in terms of the other list, rather than asserting that the lists “generate the same vectors.” Once all the expressions are established, cite the lemma or theorem to draw the set-inclusion or equality conclusion.
To prove \(\operatorname{span}\{\mathbf{u}_i\}\subseteq\operatorname{span}\{\mathbf{v}_j\}\), express every \(\mathbf{u}_i\) as a combination of the \(\mathbf{v}_j\).
Express every \(\mathbf{v}_j\) as a combination of the \(\mathbf{u}_i\). The coefficients may be different from those in the first direction.
State that the two generator checks give opposite inclusions, hence the spans are equal.
Worked Examples of the Two Checks
Worked Example: Two Coordinate Directions
Let \(\mathbf{u}_1=(1,1)\), \(\mathbf{u}_2=(1,-1)\), \(\mathbf{v}_1=(1,0)\), and \(\mathbf{v}_2=(0,1)\) in \(\mathbb{R}^2\). Prove that the two lists have equal spans.
First express each \(\mathbf{u}_i\) using the \(\mathbf{v}_j\):
For the reverse direction, use halves of the \(\mathbf{u}_i\):
Each generator of either list is in the other span. The theorem therefore gives \(\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2\}=\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}\). The displayed coordinate calculations verify every expression used in both inclusions.
Worked Example: A Redundant Generator
In \(\mathbb{R}^3\), let \(\mathbf{a}_1=(1,0,1)\), \(\mathbf{a}_2=(0,1,1)\), \(\mathbf{a}_3=(1,1,2)\), and let \(\mathbf{b}_1=(1,1,2)\), \(\mathbf{b}_2=(1,-1,0)\), \(\mathbf{b}_3=(2,0,2)\). Show that the spans are equal.
For the first inclusion, express each \(\mathbf{a}_i\) using the \(\mathbf{b}_j\). In fact, only the first two \(\mathbf{b}\)'s are needed:
For the reverse inclusion, express every \(\mathbf{b}_j\) using the \(\mathbf{a}_i\):
Both generator checks are complete, so the spans are equal. The extra vector \(\mathbf{b}_3\) causes no difficulty: the proof does not require either list to be free of redundant entries.
Worked Example: A Third Description of a Subspace
Let \(\mathbf{c}_1=(1,1,0)\), \(\mathbf{c}_2=(0,1,1)\), \(\mathbf{c}_3=(1,2,1)\), and let \(\mathbf{d}_1=(1,0,-1)\), \(\mathbf{d}_2=(0,1,1)\) in \(\mathbb{R}^3\). Prove equality of the spans.
Write the \(\mathbf{c}\)-vectors using the \(\mathbf{d}\)-vectors:
Now express both \(\mathbf{d}\)-vectors using the \(\mathbf{c}\)-vectors:
Every vector in either generating list has an explicit expression in the other list. Thus \(\operatorname{span}\{\mathbf{c}_1,\mathbf{c}_2,\mathbf{c}_3\} =\operatorname{span}\{\mathbf{d}_1,\mathbf{d}_2\}\). In particular, the expression for \(\mathbf{c}_3\) shows directly why that generator does not require a third vector in the second list.
Common Mistake: Stopping After One Inclusion
A frequent error is to prove that the generators of one list belong to the other span, then conclude equality. That establishes only one inclusion. For a concrete example, let \(U=\operatorname{span}\{(1,0)\}\) and \(W=\operatorname{span}\{(1,0),(0,1)\}\) in \(\mathbb{R}^2\). The generator \((1,0)\) of \(U\) belongs to \(W\), so \(U\subseteq W\). It is wrong to conclude \(U=W\) from that calculation alone.
To check the missing direction, ask whether the generator \((0,1)\) of \(W\) belongs to \(U\). A purported expression \((0,1)=c(1,0)\) would require \((c,0)=(0,1)\), hence \(c=0\) from the first coordinate and \(0=1\) from the second. No such scalar exists. Therefore \((0,1)\notin U\), so \(W\not\subseteq U\) and the spans are not equal. The correction is not to find a more persuasive argument for the first inclusion; it is to verify the reverse inclusion as well.
When writing a proof, make the logical structure visible: first list the combinations proving one inclusion, then list the combinations proving the other. If a generator is zero, its expression is the combination with all coefficients zero; if a list has redundant vectors, each still needs an expression, though those expressions need not use every vector in the other list.
Check Your Understanding
For each question, identify exactly which generator memberships would establish the required conclusion.
- In \(\mathbb{R}^2\), express \((2,1)\) and \((1,2)\) as combinations of \((1,0)\) and \((0,1)\). What inclusion does this prove?
- Suppose every generator in a list \(A\) belongs to \(\operatorname{span}(B)\). State the inclusion that follows, and explain why this alone may not give equality.
- Find coefficients expressing \((1,0)\) and \((0,1)\) using \((1,1)\) and \((1,-1)\).
- Two lists each contain the zero vector. Does that fact by itself prove their spans are equal? Explain what must still be checked.
- Write the two generator-membership conditions that are equivalent to \(\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2\}=\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}\).
