Solution. We first need to show that these vectors span $\mathbb{R}^2$, which means that every vector in $\mathbb{R}^2$ must be expressible using a linear combination of $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$. Let $\boldsymbol{x}$ be any vector in $\mathbb{R}^2$ defined as:

This vector can be expressed using $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ like so:

This means that any vector $\boldsymbol{x}$ in $\mathbb{R}^2$ can be expressed as a linear combination of $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$, and so $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ span $\mathbb{R}^2$ by definition.

Next, we must show that $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ are linearly independent. This is obvious because $\boldsymbol{e}_1$ cannot be expressed as some scalar multiple of $\boldsymbol{e}_2$.

Since $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ span $\mathbb{R}^2$ and are linearly independent, they are basis vectors of $\mathbb{R}^2$. Since they are considered to be the simplest basis vectors of $\mathbb{R}^2$, they are referred to as the standard basis vectors of $\mathbb{R}^2$.

Solution. Let's first show that $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ span $\mathbb{R}^2$. Let $\boldsymbol{x}$ be any vector in $\mathbb{R}^2$ defined like so:

Our goal is to show that there exist constants $c_1$ and $c_2$ such that:

This corresponds to solving the following linear system:

The row echelon form of the coefficient matrix is:

By theoremlink, we know that because the row echelon form of the coefficient matrix does not contain a row with all zeros, we conclude that there exist scalars $c_1$ and $c_2$ to give us $\boldsymbol{x}$ in \eqref{eq:jlmhCMRThDPFEAkUl56}.

Next, we must show that $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are linearly independent. This is easy because the vectors clearly cannot be expressed as a scalar multiple of the other.

Because vectors $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ span $\mathbb{R}^2$ and are linearly independent, they are indeed a basis for $\mathbb{R}^2$. This example demonstrates that basis for $\mathbb{R}^2$ is not unique - in fact, there are infinitely many pairs of vectors that span $\mathbb{R}^2$ and are linearly independent. Typically, we use the standard basis vectorslink because they are easy to work with.

Solution. We can easily see that $\boldsymbol{v}_2=3\boldsymbol{v}_1$. This means that:

• $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are linearly dependent.

• since $\boldsymbol{v}_2$ can be expressed as a linear combination of $\boldsymbol{v}_1$, and so the spanning set traces out a line instead of the entire $\mathbb{R}^2$.

Note that both criteria are not satisfied in this case but only one is sufficient to claim that the vectors do not form a basis for $\mathbb{R}^2$.

Let's now look at examples where one criterion is satisfied while the other is not.

Solution. We see that $\boldsymbol{v}_2=2\boldsymbol{v}_1$ so the set of vectors is linearly dependent. This already means that the vectors do not form a basis for $\mathbb{R}^2$. Let's check whether or not the set spans $\mathbb{R}^2$. Let's focus on $\boldsymbol{v}_2$ and $\boldsymbol{v}_3$ and obtain their row echelon form:

By theoremlink, because the row echelon form of the coefficient matrix does not have a row with all zeros, any vector in $\mathbb{R}^2$ can be expressed as a linear combination of $\boldsymbol{v}_2$ and $\boldsymbol{v}_3$. Therefore, the set of vectors spans $\mathbb{R}^2$.

Solution. Since we only have one vector, this vector must be linearly independent. However, $\boldsymbol{v}$ clearly does not span $\mathbb{R}^2$ because some vectors in $\mathbb{R}^2$ cannot be expressed as a scalar multiple of $\boldsymbol{v}$. Therefore, $\boldsymbol{v}$ does not form a basis for $\mathbb{R}^2$.

We have already covered the standard basis of $\mathbb{R}^2$ in the earlier sectionlink. We can also generalize the idea to $\mathbb{R}^n$.