Exercise on Vector Space Bases 3
In the vector space V = ℝ3, with $\dim(V) = 3$, find a basis that includes the vectors $v_1 = (3,\,-7,\,4)$ and $v_2 = (2,\,6,\,-5)$.
Solution
Since the space has dimension 3:
$$ \dim(V) = 3 $$
two vectors alone are insufficient to form a basis.
We are given the vectors:
$$ \vec{v}_1 = \begin{pmatrix} 3 \\ -7 \\ 4 \end{pmatrix}, \quad \vec{v}_2 = \begin{pmatrix} 2 \\ 6 \\ -5 \end{pmatrix} $$
We first verify that they are linearly independent. If they were not, it would be impossible to construct a basis containing both vectors.
Two vectors are linearly independent if the only solution to the equation:
$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 = \vec{0} $$
is the trivial one: $k_1 = k_2 = 0$.
Substituting the given vectors:
$$ k_1 \begin{pmatrix} 3 \\ -7 \\ 4 \end{pmatrix} + k_2 \begin{pmatrix} 2 \\ 6 \\ -5 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
Which yields the system:
$$ \begin{cases} 3k_1 + 2k_2 = 0 \\ -7k_1 + 6k_2 = 0 \\ 4k_1 - 5k_2 = 0 \end{cases} $$
The trivial solution always exists. To check for others, we compute the rank of the coefficient matrix:
$$ A = \begin{pmatrix} 3 & 2 \\ -7 & 6 \\ 4 & -5 \end{pmatrix} $$
Verification: A $2 \times 2$ minor is nonzero:
$$ \det \begin{pmatrix} 3 & 2 \\ 4 & -5 \end{pmatrix} = 3(-5) - 2(4) = -15 - 8 = -23 $$
Since $\text{rk}(A) = 2$ and the number of variables is $n = 2$, the system admits only the trivial solution:
$$ |S| = \infty^{n - r} = \infty^{0} = 1 $$
Therefore, $\vec{v}_1$ and $\vec{v}_2$ are linearly independent.
However, since the space has dimension 3, two vectors are not enough to span V. We must add a third vector to complete the basis.
We select a simple candidate from the standard basis, for example:
$$ \vec{v}_3 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $$
We now consider the set:
$$ \{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \} $$
To confirm that this set forms a basis, we check whether the three vectors are linearly independent:
$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + k_3 \vec{v}_3 = \vec{0} $$
Substituting the vectors:
$$ k_1 \begin{pmatrix} 3 \\ -7 \\ 4 \end{pmatrix} + k_2 \begin{pmatrix} 2 \\ 6 \\ -5 \end{pmatrix} + k_3 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
Which leads to the system:
$$ \begin{cases} 3k_1 + 2k_2 + k_3 = 0 \\ -7k_1 + 6k_2 = 0 \\ 4k_1 - 5k_2 = 0 \end{cases} $$
Again, the trivial solution always exists. To verify uniqueness, we compute the rank of the coefficient matrix:
$$ A = \begin{pmatrix} 3 & 2 & 1 \\ -7 & 6 & 0 \\ 4 & -5 & 0 \end{pmatrix} $$
Verification: The determinant is non-zero:
$$ \det(A) = 3 \cdot (6 \cdot 0 - 0 \cdot (-5)) - 2 \cdot (-7 \cdot 0 - 0 \cdot 4) + 1 \cdot (-7 \cdot (-5) - 6 \cdot 4) = 0 + 0 + (35 - 24) = 11 $$
Thus, $\text{rk}(A) = 3$, and since $n = 3$, the system has a unique solution:
$$ |S| = \infty^{n - r} = \infty^0 = 1 $$
Therefore, the vectors $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$ are linearly independent.
Since they are linearly independent and there are three of them, they form a basis of ℝ3.
And so on.
