Discover Linear Algebra: A first course in linear algebra

To answer this question, we use Statement 2 of Proposition 16.5.6, which gives us two new questions to answer.

Since both questions have been answered in the affirmative, Statement 2 of Proposition 16.5.6, tells us that $\mathrm{Span}S$ and $\mathrm{Span}{S}^{\prime }$ are the same space.

Consider the set of vectors $S=\{A_1,A_2,A_3,A_4\}$ in ${\mathrm{M}}_2(\mathbb{R})\text{,}$ where

Is this set a spanning set for all of ${\mathrm{M}}_2(\mathbb{R})\text{?}$ That is, is ${\mathrm{M}}_2(\mathbb{R})=\mathrm{Span}S\text{?}$ We already know a spanning set for ${\mathrm{M}}_2(\mathbb{R})$ — the set of standard basis vectors $\mathcal{B}=\{E_{11},E_{12},E_{21},E_{22}\}\text{,}$ where

That is, we already know that ${\mathrm{M}}_2(\mathbb{R})=\mathrm{Span}\mathcal{B}\text{.}$ So we can turn our question into: is $\mathrm{Span}S=\mathrm{Span}\mathcal{B}\text{?}$ With this new version of our problem, we can use the same method as in the previous example. However, we don’t need to explicitly verify that each vector in $S$ can be expressed as a linear combination of the vectors in $\mathcal{B}\text{.}$ Besides being obvious, this fact is already implied by our assertion that ${\mathrm{M}}_2(\mathbb{R})=\mathrm{Span}\mathcal{B}\text{,}$ since clearly each vector in $S$ is a vector in ${\mathrm{M}}_2(\mathbb{R})\text{.}$ So it just remains to verify that each vector in $\mathcal{B}$ can be expressed as a linear combination of the vectors in $S\text{.}$ Let’s begin with vector $E_{11}\text{.}$ We use the same strategy as in the examples in Subsection 16.4.3: express $E_{11}$ as a linear combination of the vectors in $S$ with unknown scalar coefficients, set up equations in those unknown scalars, and determine whether the resulting linear system is consistent.

Comparing entries on left and right sides leads to the system of equations

which can be put in an augmented matrix and reduced.

The reduced augmented matrix above tells us that

and so

though we only cared about the existence of a solution, not the actual solution itself.

In a similar manner, one can calculate that

We have now verified that each vector in $\mathcal{B}$ can be expressed as a linear combination of the vectors in $S\text{.}$ As discussed above, we already knew that each vector in $S$ can be expressed as a linear combination of the vectors in $\mathcal{B}\text{.}$ Therefore, Statement 2 of Proposition 16.5.6 tells us that $\mathrm{Span}S=\mathrm{Span}\mathcal{B}\text{.}$ Since we already knew that $\mathrm{Span}\mathcal{B}$ is equal to the entire space ${\mathrm{M}}_2(\mathbb{R})\text{,}$ we must also have $\mathrm{Span}S$ equal to this entire space.