Linear Algebra

Does an Extra Vector Change the Span?

linear combination problems and solutions in linear algebra

Problem 706

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.

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Proof.

We prove that $S_2$ is also a spanning set for $V$, that is, we prove that
\[\Span(S_2)=V.\]

Prove $\Span(S_2) \subset V$

We first show that $\Span(S_2)$ is contained in $V$. Let $\mathbf{x}$ be an element in $\Span(S_2)$. Then there exist scalars $c_1, c_2, c_3, c_4$ such that
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3 + c_4 \mathbf{v}_4.\] Since $\Span(S_1)=V$, we know that $c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3$ is a vector in $V$. As $\mathbf{v}_4\in V$, we have $c_4\mathbf{v}_4 \in V$.
Since $V$ is a vector space, the sum of two elements in $V$ is in $V$.
So, \[\mathbf{x}=(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3 \mathbf{v}_3) + (c_4 \mathbf{v}_4) \in V.\] This proves that $\Span(S_2) \subset V$.

Prove $\Span(S_2) \supset V$

Note that since $S_1$ is a spanning set for $V$, every element of $S_1$ can be written as a linear combination of the vectors $\mathbf{v}_1, \mathbf{v}_2$, and $\mathbf{v}_3$.
That is, for any $\mathbf{v}\in V$, there exist scalars $c_1, c_2, c_3$ such that
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3.\] Observe that this can be written as follows.
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3+0\mathbf{v}_4.\] This tells us that $\mathbf{v}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, and $\mathbf{v}_4$.
Hence, any vector in $V$ can be written as a linear combination of the vectors in $S_2$.
Thus, $V\subset \Span(S_2)$.

Putting these inclusion together yields that $V=\Span(S_2)$, and hence $S_2$ is a spanning set for $V$.

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