Exercise on Vector Space Bases 1

In the vector space V = ℝ³, find a basis that includes the vectors v₁ = (2, -1, 0) and v₂ = (1, 1, 3).

Solution

The two given vectors are:

$$ \vec{v}_1 = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} \quad \vec{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} $$

These vectors are clearly linearly independent, as neither is a scalar multiple of the other - this can be seen without formal computation.

The dimension of the vector space V = ℝ³ is:

$$ \dim V = 3 $$

Thus, any basis of V must consist of three linearly independent vectors.

While v₁ and v₂ are linearly independent, they do not span ℝ³. Therefore, they do not yet form a basis.

To complete the basis, we need to add a third vector that is linearly independent of both.

For convenience, we can select v₃ = (1, 0, 0), a standard basis vector:

$$ \vec{v}_3 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $$

We now consider the set:

$$ \{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \} $$

To confirm that this set forms a basis, we must verify that the three vectors are linearly independent. That is, we must check whether the equation

$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + k_3 \vec{v}_3 = \vec{0} $$

has only the trivial solution $k_1 = k_2 = k_3 = 0$.

Substituting the vectors into the equation, we get:

$$ k_1 \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} + k_3 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

Computing the linear combination yields:

$$ \begin{pmatrix} 2k_1 + k_2 + k_3 \\ -k_1 + k_2 \\ 3k_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

Which corresponds to the following system of equations:

$$ \begin{cases} 2k_1 + k_2 + k_3 = 0 \\ -k_1 + k_2 = 0 \\ 3k_2 = 0 \end{cases} $$

From the third equation, we immediately get:

$$ k_2 = 0 $$

Substituting into the second equation:

$$ -k_1 = 0 \quad \Rightarrow \quad k_1 = 0 $$

And into the first equation:

$$ 2k_1 + k_2 + k_3 = 0 \quad \Rightarrow \quad k_3 = 0 $$

Therefore, the only solution is:

$$ k_1 = k_2 = k_3 = 0 $$

Since the trivial combination is the only solution, the vectors are indeed linearly independent.

Given that the dimension of V is 3, any set of three linearly independent vectors automatically forms a basis.

We can thus conclude that the set $\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \}$ is a valid basis for the vector space V:

$$ B_V = \{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \} $$

Note: There’s no need to check whether the set spans ℝ³ separately. Since the space is 3-dimensional, any set of three linearly independent vectors automatically spans it and forms a basis.

And so on.