# Determining whether or not the subset S is a subspace

So I'm reading my linear algebra textbook where it says:

Theorem 4.2:
Let $S = \text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}$ be a subset of $\mathbb R^{n}$. Then $S$ is a subspace of $\mathbb R^{n}$.

Then later on the page it says:

If you can show that $S$ is generated by a set of vectors, then by Theorem 4.2 $S$ is a subspace.

I'm really having trouble understanding this theorem and how I should apply this to a problem. I was hoping if anyone could explain and/or give examples that demonstrate this theorem since I can't seem to find any further explanations on my textbook.

• "$S = \text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}$" means exactly "$S$ is generated by a set of vectors $\{u_1,u_2,\ldots,u_n\}$." May 13, 2015 at 9:49

To approach such questions it is typically crucial to recall the definitions. You do not give the definition of $\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}$ you use.

There are basically two; I assume the one which you ought to use (else you question is a bit void); see below for the other.

Definition: $\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}= \{\sum_{i=1}^n \lambda_iu_i \colon \lambda_i \in \mathbb{R} \}$.

Assuming this definition one can show that $S=\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}$ is a subspace by checking that:

• for all $v,v' \in S$ one has $v+v' \in S$.
• for all $v \in S$ and $\mu \in \mathbb{R}$ one has $\mu v \in S$.

(If you want to be very precise you'd also have to assert $S\neq \emptyset$ but that's a detail.)

For the first write $v= \sum_{i=1}^n \lambda_iu_i$ and $v'= \sum_{i=1}^n \lambda_i'u_i$ (this is possible by the definition of $S$). Then $v+ v' = \sum_{i=1}^n (\lambda_i+ \lambda_i')u_i$ is again in $S$ as $\lambda_i+\lambda_i'$ is a real number again. The second is quite similar to check, I skip the details.

So the theorem asserts that whenever you have a set of the form $\{\sum_{i=1}^n \lambda_iu_i \colon \lambda_i \in \mathbb{R} \}$ for some vectors $u_1, \dots, u_n$ then this set is a subspace.

In fact it is the smalles subspace that contains $u_1, \dots, u_n$; one thus calls it the subspace generated by $u_1, \dots, u_n$. One could also use this as the defintion (compare a comment on the question).

Alternative definition: $\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}$ is the smallest subspace that contains $u_1, \dots, u_n$.

If you were to use that definition then the theorem would be obvious. However, you then would want to prove that $\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}= \{\sum_{i=1}^n \lambda_iu_i \colon \lambda_i \in \mathbb{R} \}$.

Either way the main thing to retain is: $\text{span}\{u_{1}, u_{2}, \cdots, u_{n}\}= \{\sum_{i=1}^n \lambda_iu_i \colon \lambda_i \in \mathbb{R} \}$ is the smallest subspace that contains $u_1, \dots, u_n$, one calls it the subspace generated by $u_1, \dots, u_n$.